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{{task|Arbitrary precision}} [[Category:Arbitrary precision]] [[Category:Arithmetic operations]]
;Task: Explicitly implement [[wp:long multiplication|long multiplication]].
This is one possible approach to arbitrary-precision integer algebra.
For output, display the result of 264 * 264.
The decimal representation of 264 is: 18,446,744,073,709,551,616
The output of 264 * 264 is 2128, and is: 340,282,366,920,938,463,463,374,607,431,768,211,456
360 Assembly
For maximum compatibility, we use only the basic 370 instruction set (use of MVCL). Pseudo-macro instruction XPRNT can be replaced by a WTO.
LONGINT CSECT
USING LONGINT,R13
SAVEAREA B PROLOG-SAVEAREA(R15)
DC 17F'0'
DC CL8'LONGINT'
PROLOG STM R14,R12,12(R13)
ST R13,4(R15)
ST R15,8(R13)
LR R13,R15
MVC XX(1),=C'1'
MVC LENXX,=H'1' xx=1
LA R2,64
LOOPII ST R2,RLOOPII do for 64
MVC X-2(LL+2),XX-2 x=xx
MVC Y(1),=C'2'
MVC LENY,=H'1' y=2
BAL R14,LONGMULT
MVC XX-2(LL+2),Z-2 xx=longmult(xx,2) xx=xx*2
L R2,RLOOPII
ELOOPII BCT R2,LOOPII loop
MVC X-2(LL+2),XX-2
MVC Y-2(LL+2),XX-2
BAL R14,LONGMULT
MVC YY-2(LL+2),Z-2 yy=longmult(xx,xx) yy=xx*xx
XPRNT XX,LL output xx
XPRNT YY,LL output yy
RETURN L R13,4(0,R13) epilog
LM R14,R12,12(R13)
XR R15,R15 set return code
BR R14 return to caller
RLOOPII DS F
*
LONGMULT EQU * function longmult z=(x,y)
MVC LENSHIFT,=H'0' shift=''
MVC LENZ,=H'0' z=''
LH R6,LENX
LA R6,1(R6) from lenx
XR R8,R8
BCTR R8,0 by -1
LA R9,0 to 1
LOOPI BXLE R6,R8,ELOOPI do i=lenx to 1 by -1
LA R2,X
AR R2,R6 +i
BCTR R2,0
MVC CI,0(R2) ci=substr(x,i,1)
IC R0,CI ni=integer(ci)
N R0,=X'0000000F'
STH R0,NI
MVC LENT,=H'0' t=''
SR R0,R0
STH R0,CARRY carry=0
LH R7,LENY
LA R7,1(R7) from lenx
XR R10,R10
BCTR R10,0 by -1
LA R11,0 to 1
LOOPJ1 BXLE R7,R10,ELOOPJ1 do j=leny to 1 by -1
LA R2,Y
AR R2,R7 +j
BCTR R2,0
MVC CJ,0(R2) cj=substr(y,j,1)
IC R0,CJ
N R0,=X'0000000F'
STH R0,NJ nj=integer(cj)
LH R2,NI
MH R2,NJ
AH R2,CARRY
STH R2,NKR nkr=ni*nj+carry
LH R2,NKR
LA R1,10
SRDA R2,32
DR R2,R1
STH R2,NK nk=nkr//10
STH R3,CARRY carry=nkr/10
LH R2,NK
O R2,=X'000000F0'
STC R2,CK ck=string(nk)
MVC TEMP,T
MVC T(1),CK
MVC T+1(LL-1),TEMP
LH R2,LENT
LA R2,1(R2)
STH R2,LENT t=ck!!t
B LOOPJ1 next j
ELOOPJ1 EQU *
LH R2,CARRY
O R2,=X'000000F0'
STC R2,CK ck=string(carry)
MVC TEMP,T
MVC T(1),CK
MVC T+1(LL-1),TEMP
LH R2,LENT
LA R2,1(R2)
STH R2,LENT t=ck!!t
LA R2,T
AH R2,LENT
LH R3,LENSHIFT
LA R4,SHIFT
LH R5,LENSHIFT
MVCL R2,R4
LH R2,LENT
AH R2,LENSHIFT
STH R2,LENT t=t!!shift
IF1 LH R4,LENZ
CH R4,LENT if lenz>lent
BNH ELSE1
LH R2,LENZ then
LA R2,1(R2)
STH R2,L l=lenz+1
B EIF1
ELSE1 LH R2,LENT else
LA R2,1(R2)
STH R2,L l=lent+1
EIF1 EQU *
MVI TEMP,C'0' to
MVC TEMP+1(LL-1),TEMP
LA R2,TEMP
AH R2,L
SH R2,LENZ
LH R3,LENZ
LA R4,Z
LH R5,LENZ
MVCL R2,R4
MVC LENZ,L
MVC Z,TEMP z=right(z,l,'0')
MVI TEMP,C'0' to
MVC TEMP+1(LL-1),TEMP
LA R2,TEMP
AH R2,L
SH R2,LENT
LH R3,LENT
LA R4,T
LH R5,LENT
MVCL R2,R4
MVC LENT,L
MVC T,TEMP t=right(t,l,'0')
MVC LENW,=H'0' w=''
SR R0,R0
STH R0,CARRY carry=0
LH R7,L
LA R7,1(R7) from l
XR R10,R10
BCTR R10,0 by -1
LA R11,0 to 1
LOOPJ2 BXLE R7,R10,ELOOPJ2 do j=l to 1 by -1
LA R2,Z
AR R2,R7 +j
BCTR R2,0
MVC CZ,0(R2) cz=substr(z,j,1)
IC R0,CZ
N R0,=X'0000000F'
STH R0,NZ nz=integer(cz)
LA R2,T
AR R2,R7 -j
BCTR R2,0
MVC CT,0(R2) ct=substr(t,j,1)
IC R0,CT
N R0,=X'0000000F'
STH R0,NT nt=integer(ct)
LH R2,NZ
AH R2,NT
AH R2,CARRY
STH R2,NKR nkr=nz+nt+carry
LH R2,NKR
LA R1,10
SRDA R2,32
DR R2,R1
STH R2,NK
STH R3,CARRY nk=nkr//10; carry=nkr/10
LH R2,NK
O R2,=X'000000F0'
STC R2,CK ck=string(nk)
MVC TEMP,W
MVC W(1),CK
MVC W+1(LL-1),TEMP
LH R2,LENW
LA R2,1(R2)
STH R2,LENW w=ck!!w
B LOOPJ2 next j
ELOOPJ2 EQU *
LH R2,CARRY
O R2,=X'000000F0'
STC R2,CK ck=string(carry)
MVC Z(1),CK
MVC Z+1(LL-1),W
LH R2,LENW
LA R2,1(R2)
STH R2,LENZ z=ck!!w
LA R7,0 from 1
LA R10,1 by 1
LH R11,LENZ to lenz
LOOPJ3 BXH R7,R10,ELOOPJ3 do j=1 to lenz
LA R2,Z
AR R2,R7 j
BCTR R2,0
MVC ZJ(1),0(R2) zj=substr(z,j,1)
CLI ZJ,C'0' if zj^='0'
BNE ELOOPJ3 then leave j
B LOOPJ3 next j
ELOOPJ3 EQU *
IF2 CH R7,LENZ if j>lenz
BNH EIF2
LH R7,LENZ then j=lenz
EIF2 EQU *
LA R2,TEMP to
LH R3,LENZ
SR R3,R7 -j
LA R3,1(R3)
STH R3,LENTEMP
LA R4,Z from
AR R4,R7 +j
BCTR R4,0
LR R5,R3
MVCL R2,R4
MVC Z-2(LL+2),TEMP-2 z=substr(z,j)
LA R2,SHIFT
AH R2,LENSHIFT
MVI 0(R2),C'0'
LH R3,LENSHIFT
LA R3,1(R3)
STH R3,LENSHIFT shift=shift!!'0'
MVC TEMP,Z
LA R2,TEMP
AH R2,LENZ
MVC 0(2,R2),=C' '
B LOOPI next i
ELOOPI EQU *
MVI TEMP,C' '
LA R2,Z
AH R2,LENZ
LH R3,=AL2(LL)
SH R3,LENZ
LA R4,TEMP
LH R5,=H'1'
ICM R5,8,=C' '
MVCL R2,R4 z=clean(z)
BR R14 end function longmult
*
L DS H
NI DS H
NJ DS H
NK DS H
NZ DS H
NT DS H
CARRY DS H
NKR DS H
CI DS CL1
CJ DS CL1
CZ DS CL1
CT DS CL1
CK DS CL1
ZJ DS CL1
LENXX DS H
XX DS CL94
LENYY DS H
YY DS CL94
LENX DS H
X DS CL94
LENY DS H
Y DS CL94
LENZ DS H
Z DS CL94
LENT DS H
T DS CL94
LENW DS H
W DS CL94
LENSHIFT DS H
SHIFT DS CL94
LENTEMP DS H
TEMP DS CL94
LL EQU 94
YREGS
END LONGINT
{{out}}
18446744073709551616
340282366920938463463374607431768211456
Ada
===Using properly range-checked integers===
(The source text for these examples can also be found on [https://bitbucket.org/ada_on_rosetta_code/solutions Bitbucket].)
First we specify the required operations and declare our number type as an array of digits (in base 2^16):
package Long_Multiplication is
type Number (<>) is private;
Zero : constant Number;
One : constant Number;
function Value (Item : in String) return Number;
function Image (Item : in Number) return String;
overriding
function "=" (Left, Right : in Number) return Boolean;
function "+" (Left, Right : in Number) return Number;
function "*" (Left, Right : in Number) return Number;
function Trim (Item : in Number) return Number;
private
Bits : constant := 16;
Base : constant := 2 ** Bits;
type Accumulated_Value is range 0 .. (Base - 1) * Base;
subtype Digit is Accumulated_Value range 0 .. Base - 1;
type Number is array (Natural range <>) of Digit;
for Number'Component_Size use Bits; -- or pragma Pack (Number);
Zero : constant Number := (1 .. 0 => 0);
One : constant Number := (0 => 1);
procedure Divide (Dividend : in Number;
Divisor : in Digit;
Result : out Number;
Remainder : out Digit);
end Long_Multiplication;
Some of the operations declared above are useful helper operations for the conversion of numbers to and from base 10 digit strings.
Then we implement the operations:
package body Long_Multiplication is
function Value (Item : in String) return Number is
subtype Base_Ten_Digit is Digit range 0 .. 9;
Ten : constant Number := (0 => 10);
begin
case Item'Length is
when 0 =>
raise Constraint_Error;
when 1 =>
return (0 => Base_Ten_Digit'Value (Item));
when others =>
return (0 => Base_Ten_Digit'Value (Item (Item'Last .. Item'Last)))
+ Ten * Value (Item (Item'First .. Item'Last - 1));
end case;
end Value;
function Image (Item : in Number) return String is
Base_Ten : constant array (Digit range 0 .. 9) of String (1 .. 1) :=
("0", "1", "2", "3", "4", "5", "6", "7", "8", "9");
Result : Number (0 .. Item'Last);
Remainder : Digit;
begin
if Item = Zero then
return "0";
else
Divide (Dividend => Item,
Divisor => 10,
Result => Result,
Remainder => Remainder);
if Result = Zero then
return Base_Ten (Remainder);
else
return Image (Trim (Result)) & Base_Ten (Remainder);
end if;
end if;
end Image;
overriding
function "=" (Left, Right : in Number) return Boolean is
begin
for Position in Integer'Min (Left'First, Right'First) ..
Integer'Max (Left'Last, Right'Last) loop
if Position in Left'Range and Position in Right'Range then
if Left (Position) /= Right (Position) then
return False;
end if;
elsif Position in Left'Range then
if Left (Position) /= 0 then
return False;
end if;
elsif Position in Right'Range then
if Right (Position) /= 0 then
return False;
end if;
else
raise Program_Error;
end if;
end loop;
return True;
end "=";
function "+" (Left, Right : in Number) return Number is
Result : Number (Integer'Min (Left'First, Right'First) ..
Integer'Max (Left'Last , Right'Last) + 1);
Accumulator : Accumulated_Value := 0;
Used : Integer := Integer'First;
begin
for Position in Result'Range loop
if Position in Left'Range then
Accumulator := Accumulator + Left (Position);
end if;
if Position in Right'Range then
Accumulator := Accumulator + Right (Position);
end if;
Result (Position) := Accumulator mod Base;
Accumulator := Accumulator / Base;
if Result (Position) /= 0 then
Used := Position;
end if;
end loop;
if Accumulator = 0 then
return Result (Result'First .. Used);
else
raise Constraint_Error;
end if;
end "+";
function "*" (Left, Right : in Number) return Number is
Accumulator : Accumulated_Value;
Result : Number (Left'First + Right'First ..
Left'Last + Right'Last + 1) := (others => 0);
Used : Integer := Integer'First;
begin
for L in Left'Range loop
for R in Right'Range loop
Accumulator := Left (L) * Right (R);
for Position in L + R .. Result'Last loop
exit when Accumulator = 0;
Accumulator := Accumulator + Result (Position);
Result (Position) := Accumulator mod Base;
Accumulator := Accumulator / Base;
Used := Position;
end loop;
end loop;
end loop;
return Result (Result'First .. Used);
end "*";
procedure Divide (Dividend : in Number;
Divisor : in Digit;
Result : out Number;
Remainder : out Digit) is
Accumulator : Accumulated_Value := 0;
begin
Result := (others => 0);
for Position in reverse Dividend'Range loop
Accumulator := Accumulator * Base + Dividend (Position);
Result (Position) := Accumulator / Divisor;
Accumulator := Accumulator mod Divisor;
end loop;
Remainder := Accumulator;
end Divide;
function Trim (Item : in Number) return Number is
begin
for Position in reverse Item'Range loop
if Item (Position) /= 0 then
return Item (Item'First .. Position);
end if;
end loop;
return Zero;
end Trim;
end Long_Multiplication;
And finally we have the requested test application:
with Ada.Text_IO;
with Long_Multiplication;
procedure Test_Long_Multiplication is
use Ada.Text_IO, Long_Multiplication;
N : Number := Value ("18446744073709551616");
M : Number := N * N;
begin
Put_Line (Image (N) & " * " & Image (N) & " = " & Image (M));
end Test_Long_Multiplication;
{{out}}
18446744073709551616 * 18446744073709551616 = 340282366920938463463374607431768211456
Using modular types
The following implementation uses representation of a long number by an array of 32-bit elements:
type Long_Number is array (Natural range <>) of Unsigned_32;
function "*" (Left, Right : Long_Number) return Long_Number is
Result : Long_Number (0..Left'Length + Right'Length - 1) := (others => 0);
Accum : Unsigned_64;
begin
for I in Left'Range loop
for J in Right'Range loop
Accum := Unsigned_64 (Left (I)) * Unsigned_64 (Right (J));
for K in I + J..Result'Last loop
exit when Accum = 0;
Accum := Accum + Unsigned_64 (Result (K));
Result (K) := Unsigned_32 (Accum and 16#FFFF_FFFF#);
Accum := Accum / 2**32;
end loop;
end loop;
end loop;
for Index in reverse Result'Range loop -- Normalization
if Result (Index) /= 0 then
return Result (0..Index);
end if;
end loop;
return (0 => 0);
end "*";
The task requires conversion into decimal base. For this we also need division to short number with a remainder. Here it is:
procedure Div
( Dividend : in out Long_Number;
Last : in out Natural;
Remainder : out Unsigned_32;
Divisor : Unsigned_32
) is
Div : constant Unsigned_64 := Unsigned_64 (Divisor);
Accum : Unsigned_64 := 0;
Size : Natural := 0;
begin
for Index in reverse Dividend'First..Last loop
Accum := Accum * 2**32 + Unsigned_64 (Dividend (Index));
Dividend (Index) := Unsigned_32 (Accum / Div);
if Size = 0 and then Dividend (Index) /= 0 then
Size := Index;
end if;
Accum := Accum mod Div;
end loop;
Remainder := Unsigned_32 (Accum);
Last := Size;
end Div;
With the above the test program:
with Ada.Strings.Unbounded; use Ada.Strings.Unbounded;
with Ada.Text_IO; use Ada.Text_IO;
with Interfaces; use Interfaces;
procedure Long_Multiplication is
-- Insert definitions above here
procedure Put (Value : Long_Number) is
X : Long_Number := Value;
Last : Natural := X'Last;
Digit : Unsigned_32;
Result : Unbounded_String;
begin
loop
Div (X, Last, Digit, 10);
Append (Result, Character'Val (Digit + Character'Pos ('0')));
exit when Last = 0 and then X (0) = 0;
end loop;
for Index in reverse 1..Length (Result) loop
Put (Element (Result, Index));
end loop;
end Put;
X : Long_Number := (0 => 0, 1 => 0, 2 => 1) * (0 => 0, 1 => 0, 2 => 1);
begin
Put (X);
end Long_Multiplication;
Sample output:
340282366920938463463374607431768211456
Aime
data b, c, v;
integer d, e, i, j, s;
b = 1.argv;
b.dump(',');
v = 2.argv;
v.dump(',');
c.run(~b + ~v + 1, 0);
for (i, d in b) {
b[i] = d - '0';
}
for (j, d of v) {
d = v[j] - '0';
s = 0;
for (i, e of b) {
s += e * d + c[i + j];
c[i + j] = s % 10;
s /= 10;
}
while (s) {
s += c[i + j];
c[i + j] = s % 10;
s /= 10;
i -= 1;
}
}
c.delete(-1);
c.bf_drop0("");
for (i, d in c) {
c[i] = d + '0';
}
o_form("~\n", c);
ALGOL 68
The long multiplication for the golden ratio has been included as half the digits cancel and end up as being zero. This is useful for testing.
Built in or standard distribution routines
{{works with|ALGOL 68G|Any - tested with release mk15-0.8b.fc9.i386}} [[ALGOL 68G]] allows any precision for '''long long int''' to be defined when the program is run, e.g. 200 digits.
PRAGMAT precision=200 PRAGMAT
MODE INTEGER = LONG LONG INT;
LONG INT default integer width := 69;
INT width = 69+2;
INT fix w = 1, fix h = 1; # round up #
LONG LONG INT golden ratio w := ENTIER ((long long sqrt(5)-1) / 2 * LENG LENG 10 ** default integer width + fix w),
golden ratio h := ENTIER ((long long sqrt(5)+1) / 2 * LENG LENG 10 ** default integer width + fix h);
test: (
print((
"The approximate golden ratios, width: ", whole(golden ratio w,width), new line,
" length: ", whole(golden ratio h,width), new line,
" product is exactly: ", whole(golden ratio w*golden ratio h,width*2), new line));
INTEGER two to the power of 64 = LONG 2 ** 64;
INTEGER neg two to the power of 64 = -(LONG 2 ** 64);
print(("2 ** 64 * -(2 ** 64) = ", whole(two to the power of 64*neg two to the power of 64,width), new line))
)
Output:
The approximate golden ratios, width: +618033988749894848204586834365638117720309179805762862135448622705261
length: +1618033988749894848204586834365638117720309179805762862135448622705261
product is exactly: +1000000000000000000000000000000000000000000000000000000000000000000001201173450350400438606015942314498798603569682901026716145698077078121
2 ** 64 * -(2 ** 64) = -340282366920938463463374607431768211456
Implementation example
{{works with|ALGOL 68|Standard - no extensions to language used}} {{works with|ALGOL 68G|Any - tested with release mk15-0.8b.fc9.i386}}
MODE DIGIT = INT;
MODE INTEGER = FLEX[0]DIGIT; # an arbitary number of digits #
# "digits" are stored in digit base ten, but 10000 & 2**n (inc hex) can be used #
INT digit base = 1000;
# if possible, then print the digit with one character #
STRING hex digit repr = "0123456789abcdefghijklmnopqrstuvwxyz"[AT 0];
INT digit base digit width = ( digit base <= UPB hex digit repr + 1 | 1 | 1 + ENTIER log(digit base-1) );
INT next digit = -1; # reverse order so digits appear in "normal" order when printed #
PROC raise value error = ([]STRING args)VOID:
( print(("Value Error: ", args, new line)); stop );
PROC raise not implemented error = ([]STRING args)VOID:
( print(("Not implemented Error: ", args, new line)); stop );
PROC raise integer not implemented error = (STRING message)INTEGER:
( raise not implemented error(("INTEGER ", message)); SKIP );
INT half max int = max int OVER 2;
IF digit base > half max int THEN raise value error("INTEGER addition may fail") FI;
INT sqrt max int = ENTIER sqrt(max int);
IF digit base > sqrt max int THEN raise value error("INTEGER multiplication may fail") FI;
# initialise/cast a INTEGER from a LONG LONG INT #
OP INTEGERINIT = (LONG LONG INT number)INTEGER:(
[1 + ENTIER (SHORTEN SHORTEN long long log(ABS number) / log(digit base))]DIGIT out;
LONG LONG INT carry := number;
FOR digit out FROM UPB out BY next digit TO LWB out DO
LONG LONG INT prev carry := carry;
carry %:= digit base; # avoid MOD as it doesn't under handle -ve numbers #
out[digit out] := SHORTEN SHORTEN (prev carry - carry * digit base)
OD;
out
);
# initialise/cast a INTEGER from an LONG INT #
OP INTEGERINIT = (LONG INT number)INTEGER: INTEGERINIT LENG number;
# initialise/cast a INTEGER from an INT #
OP INTEGERINIT = (INT number)INTEGER: INTEGERINIT LENG LENG number;
# remove leading zero "digits" #
OP NORMALISE = ([]DIGIT number)INTEGER: (
INT leading zeros := LWB number - 1;
FOR digit number FROM LWB number TO UPB number
WHILE number[digit number] = 0 DO leading zeros := digit number OD;
IF leading zeros = UPB number THEN 0 ELSE number[leading zeros+1:] FI
);
#####################################################################
Define a standard representation for the INTEGER mode. Note: this is
rather crude because for a large "digit base" the number is represented as
blocks of decimals. It works nicely for powers of ten (10,100,1000,...),
but for most larger bases (greater then 35) the repr will be a surprise.
#####################################################################
OP REPR = (DIGIT d)STRING:
IF digit base > UPB hex digit repr THEN
STRING out := whole(ABS d, -digit base digit width);
# Replace spaces with zeros #
FOR digit out FROM LWB out TO UPB out DO
IF out[digit out] = " " THEN out[digit out] := "0" FI
OD;
out
ELSE # small enough to represent as ASCII (hex) characters #
hex digit repr[ABS d]
FI;
OP REPR = (INTEGER number)STRING:(
STRING sep = ( digit base digit width > 1 | "," | "" );
INT width := digit base digit width + UPB sep;
[width * UPB number - UPB sep]CHAR out;
INT leading zeros := LWB out - 1;
FOR digit TO UPB number DO
INT start := digit * width - width + 1;
out[start:start+digit base digit width-1] := REPR number[digit];
IF digit base digit width /= 1 & digit /= UPB number THEN
out[start+digit base digit width] := ","
FI
OD;
# eliminate leading zeros #
FOR digit out FROM LWB out TO UPB out
WHILE out[digit out] = "0" OR out[digit out] = sep
DO leading zeros := digit out OD;
CHAR sign = ( number[1]<0 | "-" | "+" );
# finally return the semi-normalised result #
IF leading zeros = UPB out THEN "0" ELSE sign + out[leading zeros+1:] FI
);
################################################################
# Finally Define the required INTEGER multiplication OPerator. #
################################################################
OP * = (INTEGER a, b)INTEGER:(
# initialise out to all zeros #
[UPB a + UPB b]INT ab; FOR place ab TO UPB ab DO ab[place ab]:=0 OD;
FOR place a FROM UPB a BY next digit TO LWB a DO
DIGIT carry := 0;
# calculate each digit (whilst removing the carry) #
FOR place b FROM UPB b BY next digit TO LWB b DO
# n.b. result may be 2 digits #
INT result := ab[place a + place b] + a[place a]*b[place b] + carry;
carry := result % digit base; # avoid MOD as it doesn't under handle -ve numbers #
ab[place a + place b] := result - carry * digit base
OD;
ab[place a + LWB b + next digit] +:= carry
OD;
NORMALISE ab
);
# The following standard operators could (potentially) also be defined #
OP - = (INTEGER a)INTEGER: raise integer not implemented error("monadic minus"),
ABS = (INTEGER a)INTEGER: raise integer not implemented error("ABS"),
ODD = (INTEGER a)INTEGER: raise integer not implemented error("ODD"),
BIN = (INTEGER a)INTEGER: raise integer not implemented error("BIN");
OP + = (INTEGER a, b)INTEGER: raise integer not implemented error("addition"),
- = (INTEGER a, b)INTEGER: raise integer not implemented error("subtraction"),
/ = (INTEGER a, b)REAL: ( VOID(raise integer not implemented error("floating point division")); SKIP),
% = (INTEGER a, b)INTEGER: raise integer not implemented error("fixed point division"),
%* = (INTEGER a, b)INTEGER: raise integer not implemented error("modulo division"),
** = (INTEGER a, b)INTEGER: raise integer not implemented error("to the power of");
LONG INT default integer width := long long int width - 2;
INT fix w = -1177584, fix h = -3915074; # floating point error, probably GMP/hardware specific #
INTEGER golden ratio w := INTEGERINIT ENTIER ((long long sqrt(5)-1) / 2 * LENG LENG 10 ** default integer width + fix w),
golden ratio h := INTEGERINIT ENTIER ((long long sqrt(5)+1) / 2 * LENG LENG 10 ** default integer width + fix h);
test: (
print((
"The approximate golden ratios, width: ", REPR golden ratio w, new line,
" length: ", REPR golden ratio h, new line,
" product is exactly: ", REPR (golden ratio w * golden ratio h), new line));
INTEGER two to the power of 64 = INTEGERINIT(LONG 2 ** 64);
INTEGER neg two to the power of 64 = INTEGERINIT(-(LONG 2 ** 64));
print(("2 ** 64 * -(2 ** 64) = ", REPR (two to the power of 64 * neg two to the power of 64), new line))
)
Output:
The approximate golden ratios, width: +618,033,988,749,894,848,204,586,834,365,638,117,720,309,179,805,762,862,135,448,622,705,261
length: +1,618,033,988,749,894,848,204,586,834,365,638,117,720,309,179,805,762,862,135,448,622,705,261
product is exactly: +1,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,001,201,173,450,350,400,438,606,015,942,314,498,798,603,569,682,901,026,716,145,698,077,078,121
2 ** 64 * -(2 ** 64) = -340,282,366,920,938,463,463,374,607,431,768,211,456
Other libraries or implementation specific extensions
As of February 2009 no open source libraries to do this task have been located.
ALGOL W
begin
% long multiplication of large integers %
% large integers are represented by arrays of integers whose absolute %
% values are in 0 .. ELEMENT_MAX - 1 %
% negative large integers should have negative values in all non-zero %
% elements %
% the least significant digits of the large integer are in element 1 %
integer ELEMENT_DIGITS; % number of digits in an element of a large %
% integer %
integer ELEMENT_MAX; % max absolute value of an element of a large %
% integer - must be 10^( ELEMENT_DIGITS + 1 ) %
integer ELEMENT_COUNT; % number of elements in each large integer %
% implements long multiplication, c is set to a * b %
% c can be the same array as a or b %
% n is the number of elements in the large integers a, b and c %
procedure longMultiply( integer array a, b, c ( * )
; integer value n
) ;
begin
% multiplies the large integer in b by the integer a, the result %
% is added to c, starting from offset %
% overflow is ignored %
procedure multiplyElement( integer value a
; integer array b, c ( * )
; integer value offset, n
) ;
begin
integer carry, cPos;
carry := 0;
cPos := offset;
for bPos := 1 until highestNonZeroElementPosition( b, ( n + 1 ) - offset ) do begin
integer cElement;
cElement := c( cPos ) + ( a * b( bPos ) ) + carry;
if abs cElement < ELEMENT_MAX then carry := 0
else begin
% have digits to carry %
carry := cElement div ELEMENT_MAX;
cElement := ( abs cElement ) rem ELEMENT_MAX;
if carry < 0 then cElement := - cElement
end if_no_carry_ ;
c( cPos ) := cElement;
cPos := cPos + 1
end for_aPos ;
if cPos <= n then c( cPos ) := carry
end multiplyElement ;
integer array mResult ( 1 :: n );
% the result will be computed in mResult, allowing a or b to be c %
for rPos := 1 until n do mResult( rPos ) := 0;
% multiply and add each element to the result %
for aPos := 1 until highestNonZeroElementPosition( a, n ) do begin
if a( aPos ) not = 0 then multiplyElement( a( aPos ), b, mResult, aPos, n )
end for_aPos ;
% return the result in c %
for rPos := 1 until n do c( rPos ) := mResult( rPos )
end longMultiply ;
% writes the decimal value of a large integer a with n elements %
procedure writeonLargeInteger( integer array a ( * )
; integer value n
) ;
begin
integer aMax;
aMax := highestNonZeroElementPosition( a, n );
if aMax < 1 then writeon( "0" )
else begin
% the large integer is non-zero %
writeon( i_w := 1, s_w := 0, a( aMax ) ); % highest element %
% handle the remaining elements - show leading zeros %
for aPos := aMax - 1 step -1 until 1 do begin
integer v;
integer array digits ( 1 :: ELEMENT_DIGITS );
v := abs a( aPos );
for dPos := ELEMENT_DIGITS step -1 until 1 do begin
digits( dPos ) := v rem 10;
v := v div 10
end for_dPos;
for dPos := 1 until ELEMENT_DIGITS do writeon( i_w := 1, s_w := 0, digits( dPos ) )
end for_aPos
end if_aMax_lt_1_
end writeonLargeInteger ;
% returns the position of the highest non-zero element of the large %
% integer a with n elements %
integer procedure highestNonZeroElementPosition( integer array a ( * )
; integer value n
) ;
begin
integer aMax;
aMax := n;
while aMax > 0 and a( aMax ) = 0 do aMax := aMax - 1;
aMax
end highestNonZeroElementPosition ;
% allow each element to contain 4 decimal digits, so element by element %
% multiplication won't overflow 32-bits %
ELEMENT_DIGITS := 4;
ELEMENT_MAX := 10000;
ELEMENT_COUNT := 12; % allows up to 48 digits - enough for the task %
begin
integer array twoTo64, twoTo128 ( 1 :: ELEMENT_COUNT );
integer pwr;
% construct 2^64 in twoTo64 %
for tPos := 2 until ELEMENT_COUNT do twoTo64( tPos ) := 0;
twoTo64( 1 ) := 2;
pwr := 1;
while pwr < 64 do begin
longMultiply( twoTo64, twoTo64, twoTo64, ELEMENT_COUNT );
pwr := pwr * 2
end while_pwr_lt_64 ;
% construct 2^128 %
longMultiply( twoTo64, twoTo64, twoTo128, ELEMENT_COUNT );
write( "2^128: " );
writeonLargeInteger( twoTo128, ELEMENT_COUNT )
end
end.
{{out}}
2^128: 340282366920938463463374607431768211456
AutoHotkey
ahk [http://www.autohotkey.com/forum/viewtopic.php?t=44657&postdays=0&postorder=asc&start=149 discussion]
MsgBox % x := mul(256,256)
MsgBox % x := mul(x,x)
MsgBox % x := mul(x,x) ; 18446744073709551616
MsgBox % x := mul(x,x) ; 340282366920938463463374607431768211456
mul(b,c) { ; <- b*c
VarSetCapacity(a, n:=StrLen(b)+StrLen(c), 48), NumPut(0,a,n,"char")
Loop % StrLen(c) {
i := StrLen(c)+1-A_Index, cy := 0
Loop % StrLen(b) {
j := StrLen(b)+1-A_Index,
t := SubStr(a,i+j,1) + SubStr(b,j,1) * SubStr(c,i,1) + cy
cy := t // 10
NumPut(mod(t,10)+48,a,i+j-1,"char")
}
NumPut(cy+48,a,i+j-2,"char")
}
Return cy ? a : SubStr(a,2)
}
=={{Header|AWK}}== {{works with|gawk|3.1.7}} {{works with|nawk|20100523}} {{trans|Tcl}}
BEGIN {
DEBUG = 0
n = 2^64
nn = sprintf("%.0f", n)
printf "2^64 * 2^64 = %.0f\n", multiply(nn, nn)
printf "2^64 * 2^64 = %.0f\n", n*n
exit
}
function multiply(x, y, len_x,len_y,ax,ay,j,m,c,i,k,d,v,res,mul,result) {
len_x = split_reverse(x, ax)
len_y = split_reverse(y, ay)
print_array(ax)
print_array(ay)
for (j=1; j<=len_y; j++) {
m = ay[j]
c = 0
i = j - 1
for (k=1; k<=len_x; k++) {
d = ax[k]
i++
v = res[i]
if (v == "") {
append_array(res, 0)
v = 0
}
mul = v + c + d*m
c = int(mul / 10)
v = mul % 10
res[i] = v
}
append_array(res, c)
}
print_array(res)
result = reverse_join(res)
sub(/^0+/, "", result)
return result
}
function split_reverse(x, a, a_x) {
split(x, a_x, "")
return reverse_array(a_x, a)
}
function reverse_array(a,b, len,i) {
len = length_array(a)
for (i in a) {
b[1+len-i] = a[i]
}
return len
}
function length_array(a, len,i) {
len = 0
for (i in a) len++
return len
}
function append_array(a, value, len) {
len = length_array(a)
a[++len] = value
}
function reverse_join(a, len,str,i) {
len = length_array(a)
str = ""
for (i=len; i>=1; i--) {
str = str a[i]
}
return str
}
function print_array(a, len,i) {
if (DEBUG) {
len = length_array(a)
print "length=" len
for (i=1; i<=len; i++) {
printf("%s ", i%10)
}
print ""
for (i=1; i<=len; i++) {
#print i " " a[i]
printf("%s ", a[i])
}
print ""
print "===="
}
}
outputs:
2^64 * 2^64 = 340282366920938463463374607431768211456
2^64 * 2^64 = 340282366920938463463374607431768211456
=={{Header|BASIC}}== {{works with|QBasic|}}
Version 1
'PROGRAM : BIG MULTIPLICATION VER #1
'LRCVS 01.01.2010
'THIS PROGRAM SIMPLY MAKES A MULTIPLICATION
'WITH ALL THE PARTIAL PRODUCTS.
'............................................................
DECLARE SUB A.INICIO (A$, B$)
DECLARE SUB B.STORE (CAD$, N$)
DECLARE SUB C.PIZARRA ()
DECLARE SUB D.ENCABEZADOS (A$, B$)
DECLARE SUB E.MULTIPLICACION (A$, B$)
DECLARE SUB G.SUMA ()
DECLARE FUNCTION F.INVCAD$ (CAD$)
RANDOMIZE TIMER
CALL A.INICIO(A$, B$)
CALL B.STORE(A$, "A")
CALL B.STORE(B$, "B")
CALL C.PIZARRA
CALL D.ENCABEZADOS(A$, B$)
CALL E.MULTIPLICACION(A$, B$)
CALL G.SUMA
SUB A.INICIO (A$, B$)
CLS
'Note: Number of digits > 1000
INPUT "NUMBER OF DIGITS "; S
CLS
A$ = ""
B$ = ""
FOR N = 1 TO S
A$ = A$ + LTRIM$(STR$(INT(RND * 9)))
NEXT N
FOR N = 1 TO S
B$ = B$ + LTRIM$(STR$(INT(RND * 9)))
NEXT N
END SUB
SUB B.STORE (CAD$, N$)
OPEN "O", #1, N$
FOR M = LEN(CAD$) TO 1 STEP -1
WRITE #1, MID$(CAD$, M, 1)
NEXT M
CLOSE (1)
END SUB
SUB C.PIZARRA
OPEN "A", #3, "R"
WRITE #3, ""
CLOSE (3)
KILL "R"
END SUB
SUB D.ENCABEZADOS (A$, B$)
LT = LEN(A$) + LEN(B$) + 1
L$ = STRING$(LT, " ")
OPEN "A", #3, "R"
MID$(L$, LT - LEN(A$) + 1) = A$
WRITE #3, L$
CLOSE (3)
L$ = STRING$(LT, " ")
OPEN "A", #3, "R"
MID$(L$, LT - LEN(B$) - 1) = "X " + B$
WRITE #3, L$
CLOSE (3)
END SUB
SUB E.MULTIPLICACION (A$, B$)
LT = LEN(A$) + LEN(B$) + 1
L$ = STRING$(LT, " ")
C$ = ""
D$ = ""
E$ = ""
CT1 = 1
ACUM = 0
OPEN "I", #2, "B"
WHILE EOF(2) <> -1
INPUT #2, B$
OPEN "I", #1, "A"
WHILE EOF(1) <> -1
INPUT #1, A$
RP = (VAL(A$) * VAL(B$)) + ACUM
C$ = LTRIM$(STR$(RP))
IF EOF(1) <> -1 THEN D$ = D$ + RIGHT$(C$, 1)
IF EOF(1) = -1 THEN D$ = D$ + F.INVCAD$(C$)
E$ = LEFT$(C$, LEN(C$) - 1)
ACUM = VAL(E$)
WEND
CLOSE (1)
MID$(L$, LT - CT1 - LEN(D$) + 2) = F.INVCAD$(D$)
OPEN "A", #3, "R"
WRITE #3, L$
CLOSE (3)
L$ = STRING$(LT, " ")
ACUM = 0
C$ = ""
D$ = ""
E$ = ""
CT1 = CT1 + 1
WEND
CLOSE (2)
END SUB
FUNCTION F.INVCAD$ (CAD$)
LCAD = LEN(CAD$)
CADTEM$ = ""
FOR CAD = LCAD TO 1 STEP -1
CADTEM$ = CADTEM$ + MID$(CAD$, CAD, 1)
NEXT CAD
F.INVCAD$ = CADTEM$
END FUNCTION
SUB G.SUMA
CF = 0
OPEN "I", #3, "R"
WHILE EOF(3) <> -1
INPUT #3, R$
CF = CF + 1
AN = LEN(R$)
WEND
CF = CF - 2
CLOSE (3)
W$ = ""
ST = 0
ACUS = 0
FOR P = 1 TO AN
K = 0
OPEN "I", #3, "R"
WHILE EOF(3) <> -1
INPUT #3, R$
K = K + 1
IF K > 2 THEN ST = ST + VAL(MID$(R$, AN - P + 1, 1))
IF K > 2 THEN M$ = LTRIM$(STR$(ST + ACUS))
WEND
'COLOR 10: LOCATE CF + 3, AN - P + 1: PRINT RIGHT$(M$, 1); : COLOR 7
W$ = W$ + RIGHT$(M$, 1)
ACUS = VAL(LEFT$(M$, LEN(M$) - 1))
CLOSE (3)
ST = 0
NEXT P
OPEN "A", #3, "R"
WRITE #3, " " + RIGHT$(F.INVCAD(W$), AN - 1)
CLOSE (3)
CLS
PRINT "THE SOLUTION IN THE FILE: R"
END SUB
Version 2
'PROGRAM: BIG MULTIPLICATION VER # 2
'LRCVS 01/01/2010
'THIS PROGRAM SIMPLY MAKES A BIG MULTIPLICATION
'WITHOUT THE PARTIAL PRODUCTS.
'HERE SEE ONLY THE SOLUTION.
'...............................................................
CLS
PRINT "WAIT"
NA = 2000 'NUMBER OF ELEMENTS OF THE MULTIPLY.
NB = 2000 'NUMBER OF ELEMENTS OF THE MULTIPLIER.
'Solution = 4000 Exacts digits
'......................................................
OPEN "X" + ".MLT" FOR BINARY AS #1
CLOSE (1)
KILL "*.MLT"
'.....................................................
'CREATING THE MULTIPLY >>> A
'CREATING THE MULTIPLIER >>> B
FOR N = 1 TO 2
IF N = 1 THEN F$ = "A" + ".MLT": NN = NA
IF N = 2 THEN F$ = "B" + ".MLT": NN = NB
OPEN F$ FOR BINARY AS #1
FOR N2 = 1 TO NN
RANDOMIZE TIMER
X$ = LTRIM$(STR$(INT(RND * 10)))
SEEK #1, N2: PUT #1, N2, X$
NEXT N2
SEEK #1, N2
CLOSE (1)
NEXT N
'.....................................................
OPEN "A" + ".MLT" FOR BINARY AS #1
FOR K = 0 TO 9
NUM$ = "": Z$ = "": ACU = 0: GG = NA
C$ = LTRIM$(STR$(K))
OPEN C$ + ".MLT" FOR BINARY AS #2
'OPEN "A" + ".MLT" FOR BINARY AS #1
FOR N = 1 TO NA
SEEK #1, GG: GET #1, GG, X$
NUM$ = X$
Z$ = LTRIM$(STR$(ACU + (VAL(X$) * VAL(C$))))
L = LEN(Z$)
ACU = 0
IF L = 1 THEN NUM$ = Z$: PUT #2, N, NUM$
IF L > 1 THEN ACU = VAL(LEFT$(Z$, LEN(Z$) - 1)): NUM$ = RIGHT$(Z$, 1): PUT #2, N, NUM$
SEEK #2, N: PUT #2, N, NUM$
GG = GG - 1
NEXT N
IF L > 1 THEN ACU = VAL(LEFT$(Z$, LEN(Z$) - 1)): NUM$ = LTRIM$(STR$(ACU)): XX$ = XX$ + NUM$: PUT #2, N, NUM$
'CLOSE (1)
CLOSE (2)
NEXT K
CLOSE (1)
'......................................................
ACU = 0
LT5 = 1
LT6 = LT5
OPEN "B" + ".MLT" FOR BINARY AS #1
OPEN "D" + ".MLT" FOR BINARY AS #3
FOR JB = NB TO 1 STEP -1
SEEK #1, JB
GET #1, JB, X$
OPEN X$ + ".MLT" FOR BINARY AS #2: LF = LOF(2): CLOSE (2)
OPEN X$ + ".MLT" FOR BINARY AS #2
FOR KB = 1 TO LF
SEEK #2, KB
GET #2, , NUM$
SEEK #3, LT5
GET #3, LT5, PR$
T$ = ""
T$ = LTRIM$(STR$(ACU + VAL(NUM$) + VAL(PR$)))
PR$ = RIGHT$(T$, 1)
ACU = 0
IF LEN(T$) > 1 THEN ACU = VAL(LEFT$(T$, LEN(T$) - 1))
SEEK #3, LT5: PUT #3, LT5, PR$
LT5 = LT5 + 1
NEXT KB
IF ACU <> 0 THEN PR$ = LTRIM$(STR$(ACU)): PUT #3, LT5, PR$
CLOSE (2)
LT6 = LT6 + 1
LT5 = LT6
ACU = 0
NEXT JB
CLOSE (3)
CLOSE (1)
OPEN "D" + ".MLT" FOR BINARY AS #3: LD = LOF(3): CLOSE (3)
ER = 1
OPEN "D" + ".MLT" FOR BINARY AS #3
OPEN "R" + ".MLT" FOR BINARY AS #4
FOR N = LD TO 1 STEP -1
SEEK #3, N: GET #3, N, PR$
SEEK #4, ER: PUT #4, ER, PR$
ER = ER + 1
NEXT N
CLOSE (4)
CLOSE (3)
KILL "D.MLT"
FOR N = 0 TO 9
C$ = LTRIM$(STR$(N))
KILL C$ + ".MLT"
NEXT N
PRINT "END"
PRINT "THE SOLUTION IN THE FILE: R.MLT"
==={{Header|Applesoft BASIC}}===
100 A$ = "18446744073709551616"
110 B$ = A$
120 GOSUB 400
130 PRINT E$
140 END
400 REM MULTIPLY A$ * B$
410 C$ = "":D$ = "0"
420 FOR I = LEN (B$) TO 1 STEP - 1
430 C = 0:B = VAL ( MID$ (B$,I,1))
440 FOR J = LEN (A$) TO 1 STEP - 1
450 V = B * VAL ( MID$ (A$,J,1)) + C
460 C = INT (V / 10):V = V - C * 10
470 C$ = STR$ (V) + C$
480 NEXT J
490 IF C THEN C$ = STR$ (C) + C$
510 GOSUB 600"ADD C$ + D$
520 D$ = E$:C$ = "0":J = LEN (B$) - I
530 IF J THEN J = J - 1:C$ = C$ + "0": GOTO 530
550 NEXT I
560 RETURN
600 REM ADD C$ + D$
610 E = LEN (D$):E$ = "":C = 0
620 FOR J = LEN (C$) TO 1 STEP - 1
630 IF E THEN D = VAL ( MID$ (D$,E,1))
640 V = VAL ( MID$ (C$,J,1)) + D + C
650 C = V > 9:V = V - 10 * C
660 E$ = STR$ (V) + E$
670 IF E THEN E = E - 1:D = 0
680 NEXT J
700 IF E THEN V = VAL ( MID$ (D$,E,1)) + C:C = V > 9:V = V - 10 * C:E$ = STR$ (V) + E$:E = E - 1: GOTO 700
720 RETURN
Batch File
Based on the JavaScript iterative code.
::Long Multiplication Task from Rosetta Code
::Batch File Implementation
@echo off
call :longmul 18446744073709551616 18446744073709551616 answer
echo(%answer%
exit /b 0
rem The Hellish Procedure
rem Syntax: call :longmul <n1> <n2> <variable to store product>
:longmul
setlocal enabledelayedexpansion
rem Define variables
set "num1=%1"
set "num2=%2"
set "limit1=-1"
set "limit2=-1"
set "length=0"
set "prod="
rem Reverse the digits of each factor
for %%A in (1,2) do (
for /l %%B in (0,1,9) do set "num%%A=!num%%A:%%B=%%B !"
for %%C in (!num%%A!) do ( set /a limit%%A+=1 & set "rev%%A=%%C!rev%%A!" )
)
rem Do the multiplication
for /l %%A in (0,1,%limit1%) do (
for /l %%B in (0,1,%limit2%) do (
set /a iter=%%A+%%B
set /a iternext=iter+1
set /a iternext2=iter+2
set /a prev=digit!iter!
set /a digit!iter!=!rev1:~%%A,1!*!rev2:~%%B,1!
rem The next line updates the length of "digits"
if !iternext! gtr !length! set length=!iternext!
if !iter! lss !length! set /a digit!iter!+=prev
set /a currdigit=digit!iter!
if !currDigit! gtr 9 (
set /a prev=digit!iternext!
set /a digit!iternext!=currdigit/10
set /a digit!iter!=currdigit%%10
rem The next line updates the length of "digits"
if !iternext2! gtr !length! set length=!iternext2!
if !iternext! lss !length! set /a digit!iternext!+=prev
)
)
)
rem Finalize product reversing the digits
for /l %%F in (0,1,%length%) do set "prod=!digit%%F!!prod!"
endlocal & set "%3=%prod%"
goto :eof
{{Out}}
340282366920938463463374607431768211456
=={{Header|BBC BASIC}}== {{works with|BBC BASIC for Windows}} Library method:
INSTALL @lib$+"BB4WMAPMLIB"
MAPM_DllPath$ = @lib$+"BB4WMAPM.DLL"
PROCMAPM_Init
twoto64$ = "18446744073709551616"
PRINT "2^64 * 2^64 = " ; FNMAPM_Multiply(twoto64$, twoto64$)
Explicit method:
twoto64$ = "18446744073709551616"
PRINT "2^64 * 2^64 = " ; FNlongmult(twoto64$, twoto64$)
END
DEF FNlongmult(num1$, num2$)
LOCAL C%, I%, J%, S%, num1&(), num2&(), num3&()
S% = LEN(num1$)+LEN(num2$)
DIM num1&(S%), num2&(S%), num3&(S%)
IF LEN(num1$) > LEN(num2$) SWAP num1$,num2$
$$^num1&(1) = num1$
num1&() AND= 15
FOR I% = LEN(num1$) TO 1 STEP -1
$$^num2&(I%) = num2$
num2&() AND= 15
num3&() += num2&() * num1&(I%)
IF I% MOD 3 = 1 THEN
C% = 0
FOR J% = S%-1 TO I%-1 STEP -1
C% += num3&(J%)
num3&(J%) = C% MOD 10
C% DIV= 10
NEXT
ENDIF
NEXT I%
num3&() += &30
num3&(S%) = 0
IF num3&(0) = &30 THEN = $$^num3&(1)
= $$^num3&(0)
=={{Header|C}}== Doing it as if by hand.
#include <stdio.h> #include <string.h> /* c = a * b. Caller is responsible for memory. c must not be the same as either a or b. */ void longmulti(const char *a, const char *b, char *c) { int i = 0, j = 0, k = 0, n, carry; int la, lb; /* either is zero, return "0" */ if (!strcmp(a, "0") || !strcmp(b, "0")) { c[0] = '0', c[1] = '\0'; return; } /* see if either a or b is negative */ if (a[0] == '-') { i = 1; k = !k; } if (b[0] == '-') { j = 1; k = !k; } /* if yes, prepend minus sign if needed and skip the sign */ if (i || j) { if (k) c[0] = '-'; longmulti(a + i, b + j, c + k); return; } la = strlen(a); lb = strlen(b); memset(c, '0', la + lb); c[la + lb] = '\0'; # define I(a) (a - '0') for (i = la - 1; i >= 0; i--) { for (j = lb - 1, k = i + j + 1, carry = 0; j >= 0; j--, k--) { n = I(a[i]) * I(b[j]) + I(c[k]) + carry; carry = n / 10; c[k] = (n % 10) + '0'; } c[k] += carry; } # undef I if (c[0] == '0') memmove(c, c + 1, la + lb); return; } int main() { char c[1024]; longmulti("-18446744073709551616", "-18446744073709551616", c); printf("%s\n", c); return 0; }
output
## C++
### Version 1
```cpp
#include <iostream>
#include <sstream>
//--------------------------------------------------------------------------------------------------
typedef long long bigInt;
//--------------------------------------------------------------------------------------------------
using namespace std;
//--------------------------------------------------------------------------------------------------
class number
{
public:
number() { s = "0"; neg = false; }
number( bigInt a ) { set( a ); }
number( string a ) { set( a ); }
void set( bigInt a ) { neg = false; if( a < 0 ) { a = -a; neg = true; } ostringstream o; o << a; s = o.str(); clearStr(); }
void set( string a ) { neg = false; s = a; if( s.length() > 1 && s[0] == '-' ) { neg = true; } clearStr(); }
number operator * ( const number& b ) { return this->mul( b ); }
number& operator *= ( const number& b ) { *this = *this * b; return *this; }
number& operator = ( const number& b ) { s = b.s; return *this; }
friend ostream& operator << ( ostream& out, const number& a ) { if( a.neg ) out << "-"; out << a.s; return out; }
friend istream& operator >> ( istream& in, number& a ){ string b; in >> b; a.set( b ); return in; }
private:
number mul( const number& b )
{
number a; bool neg = false;
string r, bs = b.s; r.resize( 2 * max( b.s.length(), s.length() ), '0' );
int xx, ss, rr, t, c, stp = 0;
string::reverse_iterator xi = bs.rbegin(), si, ri;
for( ; xi != bs.rend(); xi++ )
{
c = 0; ri = r.rbegin() + stp;
for( si = s.rbegin(); si != s.rend(); si++ )
{
xx = ( *xi ) - 48; ss = ( *si ) - 48; rr = ( *ri ) - 48;
ss = ss * xx + rr + c; t = ss % 10; c = ( ss - t ) / 10;
( *ri++ ) = t + 48;
}
if( c > 0 ) ( *ri ) = c + 48;
stp++;
}
trimLeft( r ); t = b.neg ? 1 : 0; t += neg ? 1 : 0;
if( t & 1 ) a.s = "-" + r;
else a.s = r;
return a;
}
void trimLeft( string& r )
{
if( r.length() < 2 ) return;
for( string::iterator x = r.begin(); x != ( r.end() - 1 ); )
{
if( ( *x ) != '0' ) return;
x = r.erase( x );
}
}
void clearStr()
{
for( string::iterator x = s.begin(); x != s.end(); )
{
if( ( *x ) < '0' || ( *x ) > '9' ) x = s.erase( x );
else x++;
}
}
string s;
bool neg;
};
//--------------------------------------------------------------------------------------------------
int main( int argc, char* argv[] )
{
number a, b;
a.set( "18446744073709551616" ); b.set( "18446744073709551616" );
cout << a * b << endl << endl;
cout << "Factor 1 = "; cin >> a;
cout << "Factor 2 = "; cin >> b;
cout << "Product: = " << a * b << endl << endl;
return system( "pause" );
}
//--------------------------------------------------------------------------------------------------
{{out}}
340282366920938463463374607431768211456
Factor 1 = 9876548974569852365985574874787454878778975948
Factor 2 = 8954564845421878741168741154541897945138974567
Product: = 88440198241770705041777453160463400993104404280916080859287340887463980926235972531076714516
Version 2
#include <iostream> #include <vector> using namespace std; typedef unsigned long native_t; struct ZPlus_ // unsigned int, represented as digits base 10 { vector<native_t> digits_; // least significant first; value is sum(digits_[i] * 10^i) ZPlus_(native_t n) : digits_(1, n) { while(Sweep()); } bool Sweep() // clean up digits so they are in [0,9] { bool changed = false; int carry = 0; for (auto pd = digits_.begin(); pd != digits_.end(); ++pd) { *pd += carry; carry = *pd / 10; *pd -= 10 * carry; changed = changed || carry > 0; } if (carry) digits_.push_back(carry); return changed || carry > 9; } }; ZPlus_ operator*(const ZPlus_& lhs, const ZPlus_& rhs) { ZPlus_ retval(0); // hold enough space retval.digits_.resize(lhs.digits_.size() + rhs.digits_.size(), 0ul); // accumulate one-digit multiples for (size_t ir = 0; ir < rhs.digits_.size(); ++ir) for (size_t il = 0; il < lhs.digits_.size(); ++il) retval.digits_[ir + il] += rhs.digits_[ir] * lhs.digits_[il]; // sweep clean and drop zeroes while(retval.Sweep()); while (!retval.digits_.empty() && !retval.digits_.back()) retval.digits_.pop_back(); return retval; } ostream& operator<<(ostream& dst, const ZPlus_& n) { for (auto pd = n.digits_.rbegin(); pd != n.digits_.rend(); ++pd) dst << *pd; return dst; } int main(int argc, char* argv[]) { int p2 = 1; ZPlus_ n(2ul); for (int ii = 0; ii < 7; ++ii) { p2 *= 2; n = n * n; cout << "2^" << p2 << " = " << n << "\n"; } return 0; }
2^2 = 4
2^4 = 16
2^8 = 256
2^16 = 65536
2^32 = 4294967296
2^64 = 18446744073709551616
2^128 = 340282366920938463463374607431768211456
C#
{{incorrect|F#|The problem is to implement long multiplication, not to demonstrate bignum support.}}
System.Numerics.BigInteger was added with C# 4.
{{works with|C sharp|C#|4+}}
using System; using System.Numerics; class Program { static void Main() { BigInteger pow2_64 = BigInteger.Pow(2, 64); BigInteger result = BigInteger.Multiply(pow2_64, pow2_64); Console.WriteLine(result); } }
Output:
340282366920938463463374607431768211456
=={{Header|Ceylon}}==
"run() is the main function of this module."
shared void run() {
function multiply(String|Integer|Integer[] top, String|Integer|Integer[] bottom, Integer base = 10) {
function fromString(String s) =>
s
.filter(not(','.equals))
.map((char) => Integer.parse(char.string))
.narrow<Integer>()
.sequence()
.reversed;
function toString(Integer[] ints) =>
""
.join(ints.interpose(',', 3))
.reversed
.trimLeading((char) => char in "0,");
function fromInteger(Integer int) => fromString(int.string);
function convertArg(String|Integer|Integer[] arg) =>
switch(arg)
case (is String) fromString(arg)
case (is Integer) fromInteger(arg)
case (is Integer[]) arg;
value a = convertArg(top);
value b = convertArg(bottom);
value p = a.size;
value q = b.size;
value product = Array.ofSize(p + q, 0);
for (bIndex->bDigit in b.indexed) {
variable value carry = 0;
for (aIndex->aDigit in a.indexed) {
assert (exists prodDigit = product[aIndex + bIndex]);
value temp = prodDigit + carry + aDigit * bDigit;
carry = temp / base;
product[aIndex + bIndex] = temp % base;
}
assert (exists lastDigit = product[bIndex + p]);
product[bIndex + p] = lastDigit + carry;
}
return toString(product.sequence());
}
value twoToThe64th = "18,446,744,073,709,551,616";
value expectedResult = "340,282,366,920,938,463,463,374,607,431,768,211,456";
value result = multiply(twoToThe64th, twoToThe64th);
print("The expected result is ``expectedResult``");
print("The actual result is ``result``");
print("Do they match? ``expectedResult == result then "Yes!" else "No!"``");
}
=={{Header|COBOL}}==
identification division.
program-id. long-mul.
data division.
replace ==ij-lim== by ==7== ==ir-lim== by ==14==.
working-storage section.
1 input-string pic x(26) value "18,446,744,073,709,551,616".
1 a-table.
2 a pic 999 occurs ij-lim.
1 b-table.
2 b pic 999 occurs ij-lim.
1 ir-table value all "0".
2 occurs ij-lim.
3 ir pic 999 occurs ir-lim.
1 s-table value all "0".
2 s pic 999 occurs ir-lim.
1 display.
2 temp-result pic 9(6) value 0.
2 carry pic 999 value 0.
2 remain pic 999 value 0.
1 binary.
2 i pic 9(4) value 0.
2 j pic 9(4) value 0.
2 k pic 9(4) value 0.
procedure division.
begin.
move 1 to j
perform varying i from 1 by 1 until i > ij-lim
unstring input-string delimited ","
into a (i) with pointer j
end-perform
move a-table to b-table
perform intermediate-calc
perform sum-ir
perform display-result
stop run
.
intermediate-calc.
perform varying i from ij-lim by -1 until i < 1
move 0 to carry
perform varying j from ij-lim by -1 until j < 1
compute temp-result = a (i) * b (j) + carry
divide temp-result by 1000 giving carry
remainder remain
compute k = i + j
move remain to ir (i k)
end-perform
subtract 1 from k
move carry to ir (i k)
end-perform
.
sum-ir.
move 0 to carry
perform varying k from ir-lim by -1 until k < 1
move carry to temp-result
perform varying i from ij-lim by -1 until i < 1
compute temp-result = temp-result + ir (i k)
end-perform
divide temp-result by 1000 giving carry
remainder remain
move remain to s (k)
end-perform
.
display-result.
display " " input-string
display " * " input-string
display " = " with no advancing
perform varying k from 1 by 1
until k > ir-lim or s (k) not = 0
end-perform
if s (k) < 100
move 1 to i
inspect s (k) tallying i for leading "0"
display s (k) (i:) "," with no advancing
add 1 to k
end-if
perform varying k from k by 1 until k > ir-lim
display s (k) with no advancing
if k < ir-lim
display "," with no advancing
end-if
end-perform
display space
.
end program long-mul.
18,446,744,073,709,551,616
* 18,446,744,073,709,551,616
= 340,282,366,920,938,463,463,374,607,431,768,211,456
CoffeeScript
# This very limited BCD-based collection of functions
# allows for long multiplication. It works for positive
# numbers only. The assumed data structure is as follows:
# BcdInteger.from_integer(4321) == [1, 2, 3, 4]
BcdInteger =
from_string: (s) ->
arr = []
for c in s
arr.unshift parseInt(c)
arr
from_integer: (n) ->
result = []
while n > 0
result.push n % 10
n = Math.floor n / 10
result
to_string: (arr) ->
s = ''
for elem in arr
s = elem.toString() + s
s
sum: (arr1, arr2) ->
if arr1.length < arr2.length
return BcdInteger.sum(arr2, arr1)
carry = 0
result= []
for d1, pos in arr1
d = d1 + (arr2[pos] || 0) + carry
result.push d % 10
carry = Math.floor d / 10
if carry
result.push 1
result
multiply_by_power_of_ten: (arr, power_of_ten) ->
result = (0 for i in [0...power_of_ten])
result.concat arr
product_by_integer: (arr, n) ->
result = []
for digit, i in arr
prod = BcdInteger.from_integer n * digit
prod = BcdInteger.multiply_by_power_of_ten prod, i
result = BcdInteger.sum result, prod
result
product: (arr1, arr2) ->
result = []
for digit, i in arr1
prod = BcdInteger.product_by_integer arr2, digit
prod = BcdInteger.multiply_by_power_of_ten prod, i
result = BcdInteger.sum result, prod
result
x = BcdInteger.from_integer 1
for i in [1..64]
x = BcdInteger.product_by_integer x, 2
console.log BcdInteger.to_string x # 18446744073709551616
square = BcdInteger.product x, x
console.log BcdInteger.to_string square # 340282366920938463463374607431768211456
Common Lisp
(defun number->digits (number) (do ((digits '())) ((zerop number) digits) (multiple-value-bind (quotient remainder) (floor number 10) (setf number quotient) (push remainder digits)))) (defun digits->number (digits) (reduce #'(lambda (n d) (+ (* 10 n) d)) digits :initial-value 0)) (defun long-multiply (a b) (labels ((first-digit (list) "0 if list is empty, else first element of list." (if (endp list) 0 (first list))) (long-add (digitses &optional (carry 0) (sum '())) "Do long addition on the list of lists of digits. Each list of digits in digitses should begin with the least significant digit. This is the opposite of the digit list returned by number->digits which places the most significant digit first. The digits returned by long-add do have the most significant bit first." (if (every 'endp digitses) (nconc (number->digits carry) sum) (let ((column-sum (reduce '+ (mapcar #'first-digit digitses) :initial-value carry))) (multiple-value-bind (carry column-digit) (floor column-sum 10) (long-add (mapcar 'rest digitses) carry (list* column-digit sum))))))) ;; get the digits of a and b (least significant bit first), and ;; compute the zero padded rows. Then, add these rows (using ;; long-add) and convert the digits back to a number. (do ((a (nreverse (number->digits a))) (b (nreverse (number->digits b))) (prefix '() (list* 0 prefix)) (rows '())) ((endp b) (digits->number (long-add rows))) (let* ((bi (pop b)) (row (mapcar #'(lambda (ai) (* ai bi)) a))) (push (append prefix row) rows)))))
(long-multiply (expt 2 64) (expt 2 64)) 340282366920938463463374607431768211456
D
Using the standard library:
void main() { import std.stdio, std.bigint; writeln(2.BigInt ^^ 64 * 2.BigInt ^^ 64); }
{{out}}
340282366920938463463374607431768211456
Long multiplication, same output: {{trans|JavaScript}}
import std.stdio, std.algorithm, std.range, std.ascii, std.string; auto longMult(in string x1, in string x2) pure nothrow @safe { auto digits1 = x1.representation.retro.map!q{a - '0'}; immutable digits2 = x2.representation.retro.map!q{a - '0'}.array; uint[] res; foreach (immutable i, immutable d1; digits1.enumerate) { foreach (immutable j, immutable d2; digits2) { immutable k = i + j; if (res.length <= k) res.length++; res[k] += d1 * d2; if (res[k] > 9) { if (res.length <= k + 1) res.length++; res[k + 1] = res[k] / 10 + res[k + 1]; res[k] -= res[k] / 10 * 10; } } } //return res.retro.map!digits; return res.retro.map!(d => digits[d]); } void main() { immutable two64 = "18446744073709551616"; longMult(two64, two64).writeln; }
Dc
{{incorrect|Dc|Code does not explicitly implement long multiplication}} Since Dc has arbitrary precision built-in, the task is no different than a normal multiplication:
2 64^ 2 64^ *p
{{incorrect|Dc|A Dc solution might be: Represent bignums as numerical strings and implement arithmetic functions on them.}}
EchoLisp
We implement long multiplication by multiplying polynomials, knowing that the number 1234 is the polynomial x^3 +2x^2 +3x +4 at x=10. As we assume no bigint library is present, long-mul operates on strings.
(lib 'math) ;; for poly multiplication ;; convert string of decimal digits to polynomial ;; "1234" → x^3 +2x^2 +3x +4 ;; least-significant digit first (define (string->long N) (reverse (map string->number (string->list N)))) ;; convert polynomial to string (define (long->string N) (if (pair? N) (string-append (number->string (first N)) (long->string (rest N))) "")) ;; convert poly coefficients to base 10 (define (poly->10 P (carry 0)) (append (for/list ((coeff P)) (set! coeff (+ carry coeff )) (set! carry (quotient coeff 10)) ;; new carry (modulo coeff 10)) (if(zero? carry) null (list carry)))) ;; remove leading 0 if any ;; long multiplication ;; convert input - strings of decimal digits - to polynomials ;; perform poly multiplication in base 10 ;; convert result to string of decimal digits (define (long-mul A B ) (long->string (reverse (poly->10 (poly-mul (string->long A) (string->long B)))))) (define two-64 "18446744073709551616") (long-mul two-64 two-64) → "340282366920938463463374607431768211456" ;; check it (lib 'bigint) Lib: bigint.lib loaded. (expt 2 128) → 340282366920938463463374607431768211456
Euphoria
constant base = 1000000000
function atom_to_long(atom a)
sequence s
s = {}
while a>0 do
s = append(s,remainder(a,base))
a = floor(a/base)
end while
return s
end function
function long_mult(object a, object b)
sequence c
if atom(a) then
a = atom_to_long(a)
end if
if atom(b) then
b = atom_to_long(b)
end if
c = repeat(0,length(a)+length(b))
for i = 1 to length(a) do
c[i .. i+length(b)-1] += a[i]*b
end for
for i = 1 to length(c) do
if c[i] > base then
c[i+1] += floor(c[i]/base) -- carry
c[i] = remainder(c[i],base)
end if
end for
if c[$] = 0 then
c = c[1..$-1]
end if
return c
end function
function long_to_str(sequence a)
sequence s
s = sprintf("%d",a[$])
for i = length(a)-1 to 1 by -1 do
s &= sprintf("%09d",a[i])
end for
return s
end function
sequence a, b, c
a = atom_to_long(power(2,32))
printf(1,"a is %s\n",{long_to_str(a)})
b = long_mult(a,a)
printf(1,"a*a is %s\n",{long_to_str(b)})
c = long_mult(b,b)
printf(1,"a*a*a*a is %s\n",{long_to_str(c)})
Output:
a is 4294967296
a*a is 18446744073709551616
a*a*a*a is 340282366920938463488374607424768211456
=={{header|F Sharp|F#}}==
{{incorrect|F#|The problem is to implement long multiplication, not to demonstrate bignum support.}}
let X = 2I ** 64 * 2I ** 64 ;;
val X : System.Numerics.BigInteger = 340282366920938463463374607431768211456
Factor
USING: kernel math sequences ;
: longmult-seq ( xs ys -- zs )
[ * ] cartesian-map
dup length iota [ 0 <repetition> ] map
[ prepend ] 2map
[ ] [ [ 0 suffix ] dip [ + ] 2map ] map-reduce ;
: integer->digits ( x -- xs ) { } swap [ dup 0 > ] [ 10 /mod swap [ prefix ] dip ] while drop ;
: digits->integer ( xs -- x ) 0 [ swap 10 * + ] reduce ;
: longmult ( x y -- z ) [ integer->digits ] bi@ longmult-seq digits->integer ;
( scratchpad ) 2 64 ^ dup longmult .
340282366920938463463374607431768211456
( scratchpad ) 2 64 ^ dup * .
340282366920938463463374607431768211456
Fortran
{{works with|Fortran|95 and later}}
module LongMoltiplication
implicit none
type longnum
integer, dimension(:), pointer :: num
end type longnum
interface operator (*)
module procedure longmolt_ll
end interface
contains
subroutine longmolt_s2l(istring, num)
character(len=*), intent(in) :: istring
type(longnum), intent(out) :: num
integer :: i, l
l = len(istring)
allocate(num%num(l))
forall(i=1:l) num%num(l-i+1) = iachar(istring(i:i)) - 48
end subroutine longmolt_s2l
! this one performs the moltiplication
function longmolt_ll(a, b) result(c)
type(longnum) :: c
type(longnum), intent(in) :: a, b
integer, dimension(:,:), allocatable :: t
integer :: ntlen, i, j
ntlen = size(a%num) + size(b%num) + 1
allocate(c%num(ntlen))
c%num = 0
allocate(t(size(b%num), ntlen))
t = 0
forall(i=1:size(b%num), j=1:size(a%num)) t(i, j+i-1) = b%num(i) * a%num(j)
do j=2, ntlen
forall(i=1:size(b%num)) t(i, j) = t(i, j) + t(i, j-1)/10
end do
forall(j=1:ntlen) c%num(j) = sum(mod(t(:,j), 10))
do j=2, ntlen
c%num(j) = c%num(j) + c%num(j-1)/10
end do
c%num = mod(c%num, 10)
deallocate(t)
end function longmolt_ll
subroutine longmolt_print(num)
type(longnum), intent(in) :: num
integer :: i, j
do j=size(num%num), 2, -1
if ( num%num(j) /= 0 ) exit
end do
do i=j, 1, -1
write(*,"(I1)", advance="no") num%num(i)
end do
end subroutine longmolt_print
end module LongMoltiplication
program Test
use LongMoltiplication
type(longnum) :: a, b, r
call longmolt_s2l("18446744073709551616", a)
call longmolt_s2l("18446744073709551616", b)
r = a * b
call longmolt_print(r)
write(*,*)
end program Test
FreeBASIC
' version 08-01-2017
' compile with: fbc -s console
Const As UInteger base_ = 1000000000 ' base 1,000,000,000
Function multiply(a1 As String, b1 As String) As String
Dim As String a = a1, b = b1
Trim(a) : Trim(b) ' remove spaces
If Len(a) = 0 Or Len(b) = 0 Then Return "0"
If Len(a) + Len(b) > 10000 Then
Print "number(s) are to big"
Sleep 5000,1
Return ""
End If
If Len(a) < Len(b) Then
Swap a, b
End If
Dim As ULongInt product
Dim As UInteger carry, i, m, shift
Dim As UInteger la = Len(a), lb = Len(b)
Dim As UInteger la9 = la \ 9 + IIf((la Mod 9) = 0, 0, 1)
Dim As UInteger lb9 = lb \ 9 + IIf((lb Mod 9) = 0, 0, 1)
Dim As UInteger arr_a(la9), answer((la9 + lb9) + 2)
Dim As Integer last = la9
' make length a, b a multipy of 9
a = Right((String(9, "0") + a), la9 * 9)
b = Right((String(9, "0") + b), lb9 * 9)
For i = 1 To la9
arr_a(la9 - i +1) = Val(Mid(a, i * 9 -8, 9))
Next
Do
carry = 0
m = Val(Mid(b, lb9 * 9 -8, 9))
For i = 1 To la9
product = CULngInt(arr_a(i)) * m + answer(i + shift) + carry
carry = product \ base_
answer(i + shift) = product - carry * base_
Next
If carry <> 0 Then
last = la9 + shift +1
answer(last) = carry
End If
lb9 = lb9 -1
shift = shift +1
Loop Until lb9 = 0
Dim As String tmp = Str(answer(last))
last = last -1
While last > 0
tmp = tmp + Right(String(9,"0") + Str(answer(last)), 9)
last = last -1
Wend
Return tmp
End Function
' ------=< MAIN >=------
Dim As String a = "2", b = "2", answer
Dim As UInteger i = 1, j
For j = 1 To 7
answer = multiply(a, b)
a = answer
b = answer
i = i + i
Print using "2 ^ ### = "; i;
Print answer
Next
Print
Print "-------------------------------------------------"
Print
a = "2" : b = "1" : answer = ""
For j = 1 To 128
answer = multiply(a, b)
b = answer
Next
Print "2 ^ 128 = "; answer
' empty keyboard buffer
While InKey <> "" : Wend
Print : Print "hit any key to end program"
Sleep
End
{{out}}
2 ^ 2 = 4
2 ^ 4 = 16
2 ^ 8 = 256
2 ^ 16 = 65536
2 ^ 32 = 4294967296
2 ^ 64 = 18446744073709551616
2 ^ 128 = 340282366920938463463374607431768211456
-------------------------------------------------
2 ^ 128 = 340282366920938463463374607431768211456
Go
// Long multiplication per WP article referenced by task description. // That is, multiplicand is multiplied by single digits of multiplier // to form intermediate results. Intermediate results are accumulated // for the product. Used here is the abacus method mentioned by the // article, of summing intermediate results as they are produced, // rather than all at once at the end. // // Limitations: Negative numbers not supported, superfluous leading zeros // not generally removed. package main import "fmt" // argument validation func d(b byte) byte { if b < '0' || b > '9' { panic("digit 0-9 expected") } return b - '0' } // add two numbers as strings func add(x, y string) string { if len(y) > len(x) { x, y = y, x } b := make([]byte, len(x)+1) var c byte for i := 1; i <= len(x); i++ { if i <= len(y) { c += d(y[len(y)-i]) } s := d(x[len(x)-i]) + c c = s / 10 b[len(b)-i] = (s % 10) + '0' } if c == 0 { return string(b[1:]) } b[0] = c + '0' return string(b) } // multipy a number by a single digit func mulDigit(x string, y byte) string { if y == '0' { return "0" } y = d(y) b := make([]byte, len(x)+1) var c byte for i := 1; i <= len(x); i++ { s := d(x[len(x)-i])*y + c c = s / 10 b[len(b)-i] = (s % 10) + '0' } if c == 0 { return string(b[1:]) } b[0] = c + '0' return string(b) } // multiply two numbers as strings func mul(x, y string) string { result := mulDigit(x, y[len(y)-1]) for i, zeros := 2, ""; i <= len(y); i++ { zeros += "0" result = add(result, mulDigit(x, y[len(y)-i])+zeros) } return result } // requested output const n = "18446744073709551616" func main() { fmt.Println(mul(n, n)) }
Output:
340282366920938463463374607431768211456
Haskell
import Data.List (transpose, inits) import Data.Char (digitToInt) longmult :: Integer -> Integer -> Integer longmult x y = foldl1 ((+) . (10 *)) (polymul (digits x) (digits y)) polymul :: [Integer] -> [Integer] -> [Integer] polymul xs ys = sum <$> transpose (zipWith (++) (inits $ repeat 0) ((\f x -> fmap $ flip fmap x . f) (*) xs ys)) digits :: Integer -> [Integer] digits = fmap (fromIntegral . digitToInt) . show main :: IO () main = print $ (2 ^ 64) `longmult` (2 ^ 64)
{{out}}
340282366920938463463374607431768211456
== {{header|Icon}} and {{header|Unicon}} == Large integers are native to Icon and Unicon. Neither libraries nor special programming is required.
procedure main()
write(2^64*2^64)
end
{{output}}
340282366920938463463374607431768211456
J
'''Solution:'''
digits =: ,.&.":
polymult =: +//.@(*/)
buildDecimal=: 10x&#.
longmult=: buildDecimal@polymult&digits
'''Example:'''
longmult~ 2x^64
340282366920938463463374607431768211456
'''Alternatives:'''
longmult could have been defined concisely:
longmult=: 10x&#.@(+//.@(*/)&(,.&.":))
Or, of course, the task may be accomplished without the verb definitions:
10x&#.@(+//.@(*/)&(,.&.":))~2x^64
340282366920938463463374607431768211456
Or using the code (+ 10x&*)/@|. instead of #.:
(+ 10x&*)/@|.@(+//.@(*/)&(,.&.":))~2x^64
340282366920938463463374607431768211456
Or you could use the built-in language support for arbitrary precision multiplication:
(2x^64)*(2x^64)
340282366920938463463374607431768211456
'''Explaining the component verbs:'''
digitstranslates a number to a corresponding list of digits;
,.&.": 123
1 2 3
polymult(multiplies polynomials): '''ref.''' [http://www.jsoftware.com/help/dictionary/samp23.htm]
1 2 3 (+//.@(*/)) 1 2 3
1 4 10 12 9
buildDecimal(translates a list of decimal digits - possibly including "carry" - to the corresponding extended precision number):
(+ 10x&*)/|. 1 4 10 12 9
15129
Java
Decimal version
This version of the code keeps the data in base ten. By doing this, we can avoid converting the whole number to binary and we can keep things simple, but the runtime will be suboptimal.
public class LongMult { private static byte[] stringToDigits(String num) { byte[] result = new byte[num.length()]; for (int i = 0; i < num.length(); i++) { char c = num.charAt(i); if (c < '0' || c > '9') { throw new IllegalArgumentException("Invalid digit " + c + " found at position " + i); } result[num.length() - 1 - i] = (byte) (c - '0'); } return result; } public static String longMult(String num1, String num2) { byte[] left = stringToDigits(num1); byte[] right = stringToDigits(num2); byte[] result = new byte[left.length + right.length]; for (int rightPos = 0; rightPos < right.length; rightPos++) { byte rightDigit = right[rightPos]; byte temp = 0; for (int leftPos = 0; leftPos < left.length; leftPos++) { temp += result[leftPos + rightPos]; temp += rightDigit * left[leftPos]; result[leftPos + rightPos] = (byte) (temp % 10); temp /= 10; } int destPos = rightPos + left.length; while (temp != 0) { temp += result[destPos] & 0xFFFFFFFFL; result[destPos] = (byte) (temp % 10); temp /= 10; destPos++; } } StringBuilder stringResultBuilder = new StringBuilder(result.length); for (int i = result.length - 1; i >= 0; i--) { byte digit = result[i]; if (digit != 0 || stringResultBuilder.length() > 0) { stringResultBuilder.append((char) (digit + '0')); } } return stringResultBuilder.toString(); } public static void main(String[] args) { System.out.println(longMult("18446744073709551616", "18446744073709551616")); } }
Binary version
This version tries to be as efficient as possible, so it converts numbers into binary before doing any calculations. The complexity is higher because of the need to convert to and from base ten, which requires the implementation of some additional arithmetic operations beyond long multiplication itself.
import java.util.Arrays; public class LongMultBinary { /** * A very basic arbitrary-precision integer class. It only handles * non-negative numbers and doesn't implement any arithmetic not necessary * for the task at hand. */ public static class MyLongNum implements Cloneable { /* * The actual bits of the integer, with the least significant place * first. The biggest native integer type of Java is the 64-bit long, * but since we need to be able to store the result of two digits * multiplied, we have to use the second biggest native type, the 32-bit * int. All numeric types are signed in Java, but we don't want to waste * the sign bit, so we need to take extra care while doing arithmetic to * ensure unsigned semantics. */ private int[] digits; /* * The number of digits actually used in the digits array. Since arrays * cannot be resized in Java, we are better off remembering the logical * size ourselves, instead of reallocating and copying every time we need to shrink. */ private int digitsUsed; @Override public MyLongNum clone() { try { MyLongNum clone = (MyLongNum) super.clone(); clone.digits = clone.digits.clone(); return clone; } catch (CloneNotSupportedException e) { throw new Error("Object.clone() threw exception", e); } } private void resize(int newLength) { if (digits.length < newLength) { digits = Arrays.copyOf(digits, newLength); } } private void adjustDigitsUsed() { while (digitsUsed > 0 && digits[digitsUsed - 1] == 0) { digitsUsed--; } } /** * "Short" multiplication by one digit. Used to convert strings to long numbers. */ public void multiply(int multiplier) { if (multiplier < 0) { throw new IllegalArgumentException( "Signed arithmetic isn't supported"); } resize(digitsUsed + 1); long temp = 0; for (int i = 0; i < digitsUsed; i++) { temp += (digits[i] & 0xFFFFFFFFL) * multiplier; digits[i] = (int) temp; // store the low 32 bits temp >>>= 32; } digits[digitsUsed] = (int) temp; digitsUsed++; adjustDigitsUsed(); } /** * "Short" addition (adding a one-digit number). Used to convert strings to long numbers. */ public void add(int addend) { if (addend < 0) { throw new IllegalArgumentException( "Signed arithmetic isn't supported"); } long temp = addend; for (int i = 0; i < digitsUsed && temp != 0; i++) { temp += (digits[i] & 0xFFFFFFFFL); digits[i] = (int) temp; // store the low 32 bits temp >>>= 32; } if (temp != 0) { resize(digitsUsed + 1); digits[digitsUsed] = (int) temp; digitsUsed++; } } /** * "Short" division (dividing by a one-digit number). Used to convert numbers to strings. * @param divisor The digit to divide by. * @return The remainder of the division. */ public int divide(int divisor) { if (divisor < 0) { throw new IllegalArgumentException( "Signed arithmetic isn't supported"); } int remainder = 0; for (int i = digitsUsed - 1; i >= 0; i--) { long twoDigits = (((long) remainder << 32) | (digits[i] & 0xFFFFFFFFL)); remainder = (int) (twoDigits % divisor); digits[i] = (int) (twoDigits / divisor); } adjustDigitsUsed(); return remainder; } public MyLongNum(String value) { // each of our 32-bit digits can store at least 9 decimal digit's worth this.digits = new int[value.length() / 9 + 1]; this.digitsUsed = 0; // To lower the number of bignum operations, handle nine digits at a time. for (int i = 0; i < value.length(); i+=9) { String chunk = value.substring(i, Math.min(i+9, value.length())); int multiplier = 1; int addend = 0; for (int j=0; j<chunk.length(); j++) { char c = chunk.charAt(j); if (c < '0' || c > '9') { throw new IllegalArgumentException("Invalid digit " + c + " found in input"); } multiplier *= 10; addend *= 10; addend += c - '0'; } multiply(multiplier); add(addend); } } @Override public String toString() { if (digitsUsed == 0) { return "0"; } MyLongNum dummy = this.clone(); StringBuilder resultBuilder = new StringBuilder(digitsUsed * 9); while (dummy.digitsUsed > 0) { // To limit the number of bignum divisions, handle nine digits at a time. int decimalDigits = dummy.divide(1000000000); for (int i=0; i<9; i++) { resultBuilder.append((char) (decimalDigits % 10 + '0')); decimalDigits /= 10; } } // Trim any leading zeros we may have created. while (resultBuilder.charAt(resultBuilder.length()-1) == '0') { resultBuilder.deleteCharAt(resultBuilder.length()-1); } return resultBuilder.reverse().toString(); } /** * Long multiplication. */ public void multiply(MyLongNum multiplier) { MyLongNum left, right; // Make sure the shorter number is on the right-hand side to make things a bit more efficient. if (this.digitsUsed > multiplier.digitsUsed) { left = this; right = multiplier; } else { left = multiplier; right = this; } int[] newDigits = new int[left.digitsUsed + right.digitsUsed]; for (int rightPos = 0; rightPos < right.digitsUsed; rightPos++) { long rightDigit = right.digits[rightPos] & 0xFFFFFFFFL; long temp = 0; for (int leftPos = 0; leftPos < left.digitsUsed; leftPos++) { temp += (newDigits[leftPos + rightPos] & 0xFFFFFFFFL); temp += rightDigit * (left.digits[leftPos] & 0xFFFFFFFFL); newDigits[leftPos + rightPos] = (int) temp; temp >>>= 32; } // Roll forward any carry we may have. int destPos = rightPos + digitsUsed; while (temp != 0) { temp += (newDigits[destPos] & 0xFFFFFFFFL); newDigits[destPos] = (int) temp; temp >>>= 32; destPos++; } } this.digits = newDigits; this.digitsUsed = newDigits.length; adjustDigitsUsed(); } } public static void main(String[] args) { MyLongNum one = new MyLongNum("18446744073709551616"); MyLongNum two = one.clone(); one.multiply(two); System.out.println(one); } }
JavaScript
Iterative
With integer expression inputs at this scale, JavaScript still gives a slightly lossy result, despite the subsequent digit by digit string concatenation approach.
The problem is that the JavaScript Math.pow expressions become lossy at around 2^54, and Math.pow(2, 64) evaluates to a rounded:
18446744073709552000 rather than the full 18446744073709551616
This means that to handle larger inputs, the multiplication function needs to have string parameters:
function mult(strNum1,strNum2){ var a1 = strNum1.split("").reverse(); var a2 = strNum2.toString().split("").reverse(); var aResult = new Array; for ( var iterNum1 = 0; iterNum1 < a1.length; iterNum1++ ) { for ( var iterNum2 = 0; iterNum2 < a2.length; iterNum2++ ) { var idxIter = iterNum1 + iterNum2; // Get the current array position. aResult[idxIter] = a1[iterNum1] * a2[iterNum2] + ( idxIter >= aResult.length ? 0 : aResult[idxIter] ); if ( aResult[idxIter] > 9 ) { // Carrying aResult[idxIter + 1] = Math.floor( aResult[idxIter] / 10 ) + ( idxIter + 1 >= aResult.length ? 0 : aResult[idxIter + 1] ); aResult[idxIter] %= 10; } } } return aResult.reverse().join(""); } mult('18446744073709551616', '18446744073709551616')
{{Out}}
340282366920938463463374607431768211456
===Functional (ES 5)===
The function below accepts integer string or native integer arguments, but as JavaScript (unlike Haskell and Python, for example), lacks an arbitrary precision integer type, larger inputs to this function (beyond the scale of c. 2^54) need to take the form of integer strings, to avoid rounding.
For the same reason, the output always takes the form of an arbitrary precision integer string, rather than a native integer data type. (See the '''largeIntegerString()''' helper function below)
(function () { 'use strict'; // Javascript lacks an unbounded integer type // so this multiplication function takes and returns // long integer strings rather than any kind of native integer // longMult :: (String | Integer) -> (String | Integer) -> String function longMult(num1, num2) { return largeIntegerString( digitProducts(digits(num1), digits(num2)) ); } // digitProducts :: [Int] -> [Int] -> [Int] function digitProducts(xs, ys) { return multTable(xs, ys) .map(function (zs, i) { return Array.apply(null, Array(i)) .map(function () { return 0; }) .concat(zs); }) .reduce(function (a, x) { if (a) { var lng = a.length; return x.map(function (y, i) { return y + (i < lng ? a[i] : 0); }) } else return x; }) } // largeIntegerString :: [Int] -> String function largeIntegerString(lstColumnValues) { var dctProduct = lstColumnValues .reduceRight(function (a, x) { var intSum = x + a.carried, intDigit = intSum % 10; return { digits: intDigit .toString() + a.digits, carried: (intSum - intDigit) / 10 }; }, { digits: '', carried: 0 }); return (dctProduct.carried > 0 ? ( dctProduct.carried.toString() ) : '') + dctProduct.digits; } // multTables :: [Int] -> [Int] -> [[Int]] function multTable(xs, ys) { return ys.map(function (y) { return xs.map(function (x) { return x * y; }) }); } // digits :: (Integer | String) -> [Integer] function digits(n) { return (typeof n === 'string' ? n : n.toString()) .split('') .map(function (x) { return parseInt(x, 10); }); } // TEST showing that larged bounded integer inputs give only rounded results // whereas integer string inputs allow for full precision on this scale (2^128) return { fromIntegerStrings: longMult( '18446744073709551616', '18446744073709551616' ), fromBoundedIntegers: longMult( 18446744073709551616, 18446744073709551616 ) }; })();
{{Out}}
{"fromIntegerStrings":"340282366920938463463374607431768211456",
"fromBoundedIntegers":"340282366920938477630474056040704000000"}
jq
{{Works with|jq|1.4}} Since the task description mentions 2^64, the following includes "long_power(i)" for computing n^i.
# multiply two decimal strings, which may be signed (+ or -)
def long_multiply(num1; num2):
def stripsign:
.[0:1] as $a
| if $a == "-" then [ -1, .[1:]]
elif $a == "+" then [ 1, .[1:]]
else [1, .]
end;
def adjustsign(sign):
if sign == 1 then . else "-" + . end;
# mult/2 assumes neither argument has a sign
def mult(num1;num2):
(num1 | explode | map(.-48) | reverse) as $a1
| (num2 | explode | map(.-48) | reverse) as $a2
| reduce range(0; num1|length) as $i1
([]; # result
reduce range(0; num2|length) as $i2 (.;
($i1 + $i2) as $ix
| ( $a1[$i1] * $a2[$i2] +
(if $ix >= length then 0
else .[$ix]
end) ) as $r
| if $r > 9 # carrying
then
.[$ix + 1] = ($r / 10 | floor) +
(if $ix + 1 >= length then 0
else .[$ix + 1]
end)
| .[$ix] = $r - ( $r / 10 | floor ) * 10
else
.[$ix] = $r
end
)
)
| reverse | map(.+48) | implode;
(num1|stripsign) as $a1
| (num2|stripsign) as $a2
| if $a1[1] == "0" or $a2[1] == "0" then "0"
elif $a1[1] == "1" then $a2[1]|adjustsign( $a1[0] * $a2[0] )
elif $a2[1] == "1" then $a1[1]|adjustsign( $a1[0] * $a2[0] )
else mult($a1[1]; $a2[1]) | adjustsign( $a1[0] * $a2[0] )
end;
# Emit (input)^i where input and i are non-negative decimal integers,
# represented as numbers and/or strings.
def long_power(i):
def power(i):
tostring as $self
| (i|tostring) as $i
| if $i == "0" then "1"
elif $i == "1" then $self
elif $self == "0" then "0"
else reduce range(1;i) as $_ ( $self; long_multiply(.; $self) )
end;
(i|tonumber) as $i
| if $i < 4 then power($i)
else ($i|sqrt|floor) as $j
| ($i - $j*$j) as $k
| long_multiply( power($j) | power($j) ; power($k) )
end ;
'''Example''':
2 | long_power(64) | long_multiply(.;.)
{{Out}} $ jq -n -f Long_multiplication.jq "340282366920938463463374607431768211456"
Julia
{{works with|Julia|0.6}} {{trans|Python}}
'''Module''':
module LongMultiplication using Compat function addwithcarry!(r, addend, addendpos) while true pad = max(0, addendpos - lastindex(r)) append!(r, fill(0, pad)) addendrst = addend + r[addendpos] addend, r[addendpos] = divrem(addendrst, 10) iszero(addend) && break addendpos += 1 end return r end function longmult(mult1::AbstractVector{T}, mult2::AbstractVector{T}) where T <: Integer r = T[] for (offset1, digit1) in enumerate(mult1), (offset2, digit2) in zip(eachindex(mult2) + offset1 - 1, mult2) single_multrst = digits(digit1 * digit2) for (addoffset, rstdigit) in zip(eachindex(single_multrst) + offset2 - 1, single_multrst) addwithcarry!(r, rstdigit, addoffset) end end return r end function longmult(a::T, b::T)::T where T <: Integer mult1 = digits(a) mult2 = digits(b) r = longmult(mult1, mult2) return sum(d * T(10) ^ (e - 1) for (e, d) in enumerate(r)) end function longmult(a::AbstractString, b::AbstractString) if !ismatch(r"^\d+", a) || !ismatch(r"^\d+", b) throw(ArgumentError("string must contain only digits")) end mult1 = reverse(collect(Char, a) .- '0') mult2 = reverse(collect(Char, b) .- '0') r = longmult(mult1, mult2) return reverse(join(r)) end end # module LongMultiplication
'''Main''':
@show LongMultiplication.longmult(big(2) ^ 64, big(2) ^ 64) @show LongMultiplication.longmult("18446744073709551616", "18446744073709551616")
{{out}}
LongMultiplication.longmult(big(2) ^ 64, big(2) ^ 64) = 340282366920938463463374607431768211456
LongMultiplication.longmult("18446744073709551616", "18446744073709551616") = "340282366920938463463374607431768211456"
Kotlin
{{trans|Java}}
fun String.toDigits() = mapIndexed { i, c -> if (!c.isDigit()) throw IllegalArgumentException("Invalid digit $c found at position $i") c - '0' }.reversed() operator fun String.times(n: String): String { val left = toDigits() val right = n.toDigits() val result = IntArray(left.size + right.size) right.mapIndexed { rightPos, rightDigit -> var tmp = 0 left.indices.forEach { leftPos -> tmp += result[leftPos + rightPos] + rightDigit * left[leftPos] result[leftPos + rightPos] = tmp % 10 tmp /= 10 } var destPos = rightPos + left.size while (tmp != 0) { tmp += (result[destPos].toLong() and 0xFFFFFFFFL).toInt() result[destPos] = tmp % 10 tmp /= 10 destPos++ } } return result.foldRight(StringBuilder(result.size), { digit, sb -> if (digit != 0 || sb.length > 0) sb.append('0' + digit) sb }).toString() } fun main(args: Array<out String>) { println("18446744073709551616" * "18446744073709551616") }
Lambdatalk
Natural positive numbers are defined as strings, for instance 123 -> "123".
{lambda talk} has a small set of primitives working on strings, [equal?, empty?, chars, charAt, substring]
1) helper functions
{def lastchar
{lambda {:w}
{charAt {- {chars :w} 1} :w}
}}
{def butlast
{lambda {:w}
{substring 0 {- {chars :w} 1} :w}
}}
{def zeros
{lambda {:n}
{if {< :n 1}
then
else 0{zeros {- :n 1}}
}}}
2) add function
{def add
{def add.r
{lambda {:a :b :c :d}
{if {equal? :a #}
then {if {equal? :d 1} then 1 else}{butlast :c}
else {let { {:a :a} {:b :b} {:c :c}
{:d {+ :d {lastchar :a} {lastchar :b} }} }
{add.r {butlast :a} {butlast :b} {lastchar :d}:c
{if {equal? {chars :d} 1} then 0 else 1}}
}}}}
{lambda {:a :b}
{{lambda {:a :b :n}
{add.r #{zeros {- :n {chars :a}}}:a
#{zeros {- :n {chars :b}}}:b # 0}
} :a :b {max {chars :a} {chars :b}}}
}}
3) mul function
{def mul
{def muln
{lambda {:a :b :n}
{if {< :n 1}
then :b
else {muln :a {add :a :b} {- :n 1}}
}}}
{def mul.r
{lambda {:a :b :c :n}
{if {equal? :b #}
then :c
else {mul.r :a {butlast :b}
{add {muln :a 0 {lastchar :b}}{zeros :n} :c} {+ :n 1}}
}}}
{lambda {:a :b}
{mul.r :a #:b 0 0}
}}
4) applying to the task
Due to JS numbers limits, we compute first 2^32 using the JS pow function, then 2^64 and 2^128 using the mul function.
2^32 = '{def p32 {pow 2 32}} -> '{p32} = 4294967296
2^64 = '{def p64 {mul {p32} {p32}}} -> '{p64} = 18446744073709551616
2^128 = '{def p128 {mul {p64} {p64}}} -> '{p128} = 340282366920938463463374607431768211456
This can be tested in http://lambdaway.free.fr/lambdaspeech/?view=numbers8
Liberty BASIC
'[RC] long multiplication
'now, count 2^64
print "2^64"
a$="1"
for i = 1 to 64
a$ = multByD$(a$, 2)
next
print a$
print "(check with native LB)"
print 2^64
print "(looks OK)"
'now let's do b$*a$ stuff
print
print "2^64*2^64"
print longMult$(a$, a$)
print "(check with native LB)"
print 2^64*2^64
print "(looks OK)"
end
'---------------------------------------
function longMult$(a$, b$)
signA = 1
if left$(a$,1) = "-" then a$ = mid$(a$,2): signA = -1
signB = 1
if left$(b$,1) = "-" then b$ = mid$(b$,2): signB = -1
c$ = ""
t$ = ""
shift$ = ""
for i = len(a$) to 1 step -1
d = val(mid$(a$,i,1))
t$ = multByD$(b$, d)
c$ = addLong$(c$, t$+shift$)
shift$ = shift$ +"0"
'print d, t$, c$
next
if signA*signB<0 then c$ = "-" + c$
'print c$
longMult$ = c$
end function
function multByD$(a$, d)
'multiply a$ by digit d
c$ = ""
carry = 0
for i = len(a$) to 1 step -1
a = val(mid$(a$,i,1))
c = a*d+carry
carry = int(c/10)
c = c mod 10
'print a, c
c$ = str$(c)+c$
next
if carry>0 then c$ = str$(carry)+c$
'print c$
multByD$ = c$
end function
function addLong$(a$, b$)
'add a$ + b$, for now only positive
l = max(len(a$), len(b$))
a$=pad$(a$,l)
b$=pad$(b$,l)
c$ = "" 'result
carry = 0
for i = l to 1 step -1
a = val(mid$(a$,i,1))
b = val(mid$(b$,i,1))
c = a+b+carry
carry = int(c/10)
c = c mod 10
'print a, b, c
c$ = str$(c)+c$
next
if carry>0 then c$ = str$(carry)+c$
'print c$
addLong$ = c$
end function
function pad$(a$,n) 'pad from right with 0 to length n
pad$ = a$
while len(pad$)<n
pad$ = "0"+pad$
wend
end function
Maple
longmult := proc(a::integer,b::integer)
local A,B,m,n,i,j;
# Note, return a*b; works in Maple for any sized integer
A := convert(a,base,10);
B := convert(b,base,10);
m := numelems(A);
n := numelems(B);
add( add( A[i]*B[j]*10^(j-1), j=1..n )*10^(i-1), i=1..m );
end;
> longmult( 2^64, 2^64 );
340282366920938463463374607431768211456
Mathematica
We define the long multiplication function:
LongMultiplication[a_,b_]:=Module[{d1,d2},
d1=IntegerDigits[a]//Reverse;
d2=IntegerDigits[b]//Reverse;
Sum[d1[[i]]d2[[j]]*10^(i+j-2),{i,1,Length[d1]},{j,1,Length[d2]}]
]
Example:
n1 = 2^64;
n2 = 2^64;
LongMultiplication[n1, n2]
gives back:
To check the speed difference between built-in multiplication (which is already arbitrary precision) we multiply two big numbers (2^8000 has '''2409''' digits!) and divide their timings:
```Mathematica
n1=2^8000;
n2=2^8000;
Timing[LongMultiplication[n1,n2]][[1]]
Timing[n1 n2][[1]]
Floor[%%/%]
gives back:
72.9686
7.*10^-6
10424088
So our custom function takes about 73 second, the built-in function a couple of millionths of a second, so the long multiplication is about 10.5 million times slower! Mathematica uses Karatsuba multiplication for large integers, which is several magnitudes faster for really big numbers. Making it able to multiply in about a second; the final result has 9542426 digits; result omitted for obvious reasons.
NetRexx
{{trans|REXX}} A reworking of the example at Rexx Version 2.
/* NetRexx */
options replace format comments java crossref symbols nobinary
numeric digits 100
runSample(arg)
return
-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
method multiply(multiplier, multiplicand) public static
result = ''
mpa = s2a(multiplier)
mpb = s2a(multiplicand)
r_ = 0
rim = 1
loop bi = 1 to mpb[0]
loop ai = 1 to mpa[0]
ri = ai + bi -1
p_ = mpa[ai] * mpb[bi]
loop i_ = ri by 1 until p_ = 0
s_ = r_[i_] + p_
r_[i_] = s_ // 10
p_ = s_ % 10
end i_
rim = rim.max(i_)
end ai
end bi
r_[0] = rim
result = a2s(r_)
result = result.strip('l', 0)
if result = '' then result = 0
return result
-- .............................................................................
-- copy characters of a numeric string into a corresponding array
-- digits are numbered 1 to n from right to left
method s2a(numbr) private static
result = 0
lstr = numbr.length()
loop z_ = 1 to lstr
result[z_] = numbr.substr(lstr - z_ + 1, 1)
end z_
result[0] = lstr
return result
-- .............................................................................
-- turn the array of digits into a numeric string
method a2s(numbr) private static
result = ''
loop z_ = numbr[0] to 1 by -1
result = result || numbr[z_]
end z_
return result
-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
method runSample(arg) private static
mms = [ -
123', '123, -
012', '12, -
123456789012' , '44444444444, -
2 ** 64' , '2**64, -
0' ,0 ' -
]
ok = 0
errors = 0
loop mm over mms
parse mm multiplier . ',' multiplicand .
builtIn = multiplier * multiplicand
calculated = multiply(multiplier, multiplicand)
say 'Calculate' multiplier + 0 'x' multiplicand + 0
say 'Built in:' builtIn
say 'Derived: ' calculated
say
if builtIn = calculated then ok = ok + 1
else errors = errors + 1
end mm
say ok 'ok'
say errors 'not ok'
return
{{out}}
Calculate 123 x 123
Built in: 15129
Derived: 15129
Calculate 12 x 12
Built in: 144
Derived: 144
Calculate 123456789012 x 44444444444
Built in: 5486968400478463649328
Derived: 5486968400478463649328
Calculate 18446744073709551616 x 18446744073709551616
Built in: 340282366920938463463374607431768211456
Derived: 340282366920938463463374607431768211456
Calculate 0 x 0
Built in: 0
Derived: 0
5 ok
0 not ok
Nim
{{trans|C}}
import strutils proc ti(a): int = ord(a) - ord('0') proc longmulti(a, b: string): string = var i, j, n, carry, la, lb = 0 k = false # either is zero, return "0" if a == "0" or b == "0": return "0" # see if either a or b is negative if a[0] == '-': i = 1; k = not k if b[0] == '-': j = 1; k = not k # if yes, prepend minus sign if needed and skip the sign if i > 0 or j > 0: result = if k: "-" else: "" result.add longmulti(a[i..a.high], b[j..b.high]) return result = repeatChar(a.len + b.len, '0') for i in countdown(a.high, 0): var carry = 0 var k = i + b.len for j in countdown(b.high, 0): let n = ti(a[i]) * ti(b[j]) + ti(result[k]) + carry carry = n div 10 result[k] = chr(n mod 10 + ord('0')) dec k result[k] = chr(ord(result[k]) + carry) if result[0] == '0': result[0..result.high-1] = result[1..result.high] echo longmulti("-18446744073709551616", "-18446744073709551616")
Output:
3402823669209384634633746074317682114566
Oforth
Oforth handles arbitrary precision integers, so there is no need to implement long multiplication :
{{out}}
2 64 pow dup * println
340282366920938463463374607431768211456
But, if long multiplication was to be implemented :
A natural is implemented as a list of integers with base 1000000000 (in order to print them easier)
Just multiplication is implemented here.
Number Class new: Natural(v)
Natural method: initialize := v ;
Natural method: _v @v ;
Natural classMethod: newValues super new ;
Natural classMethod: newFrom asList self newValues ;
Natural method: *(n)
| v i j l x k |
n _v ->v
ListBuffer initValue(@v size v size + 1+, 0) ->l
v size loop: i [
i v at dup ->x 0 ifEq: [ continue ]
0 @v size loop: j [
i j + 1- ->k
j @v at x * + l at(k) + 1000000000 /mod k rot l put
]
k 1+ swap l put
]
while(l last 0 == l size 0 <> and) [ l removeLast drop ]
l dup freeze Natural newValues ;
Natural method: <<
| i |
@v last <<
@v size 1 - loop: i [ @v at(@v size i -) <<wjp(0, JUSTIFY_RIGHT, 8) ] ;
{{out}}
>Natural newFrom(2 16 pow) .s
[1] (Natural) 65536
ok
>dup * .s
[1] (Natural) 4294967296
ok
>dup * .s
[1] (Natural) 18446744073709551616
ok
>dup * .s
[1] (Natural) 340282366920938463463374607431768211456
ok
>_v .s
[1] (List) [768211456, 374607431, 938463463, 282366920, 340]
ok
PARI/GP
long(a,b)={
a=eval(Vec(a));
b=eval(Vec(b));
my(c=vector(#a+#b),carry=0);
for(i=1,#a,
for(j=1,#b,
c[i+j]+=a[i]*b[j]
)
);
forstep(i=#c,1,-1,
c[i] += carry;
carry = c[i] \ 10;
c[i] = c[i] % 10
);
for(i=1,#c,
if(c[i], return(concat(apply(s->Str(s),vector(#c+1-i,j,c[i+j-1])))))
);
"0"
};
long("18446744073709551616","18446744073709551616")
Output:
%1 = "340282366920938463463374607431768211456"
Pascal
Extracted from a programme to calculate and factor the number (two versions) in Frederick Pohl's book ''The Gold at the Starbow's End'', and compute Godel encodings of text. Compiles with the Free Pascal Compiler. The original would compile with Turbo Pascal (and used pointers to allow access to the "heap" storage scheme) except that does not allow functions to return a "big number" data aggregate, and it is so much nicer to be able to write X:=BigMult(A,B); The original has a special "square" calculation but this task is to exhibit long multiplication. However, raising to a power by iteration is painful, so a special routine for that.
Program TwoUp; Uses DOS, crt; {Concocted by R.N.McLean (whom God preserve), Victoria university, NZ.} Procedure Croak(gasp: string); Begin Writeln; Write(Gasp); HALT; End; const BigBase = 10; {The base of big arithmetic.} const BigEnuff = 333; {The most storage possible is 65532 bytes with Turbo Pascal.} type BigNumberIndexer = word; {To access 0:BigEnuff BigNumberDigit data.} type BigNumberDigit = byte; {The data.} type BigNumberDigit2 = word; {Capable of digit*digit + carry. Like, 255*255 = 65025} type BigNumber = {All sorts of arrangements are possible.} Record {Could include a sign indication.} TopDigit: BigNumberDigit; {Finger the high-order digit.} digit: array[0..BigEnuff] of byte; {The digits: note the "downto" in BigShow.} end; {Could add fractional digits too. Endless, endless.} Procedure BigShow(var a: BigNumber); {Print the number.} var i: integer; {A stepper.} Begin for i:=a.TopDigit downto 0 do {Thus high-order to low, as is the custom.} if BigBase = 10 then write(a.digit[i]) {Constant following by the Turbo Pascal compiler} else if BigBase = 100 then Write(a.digit[i] div 10,a.digit[i] mod 10) {Means that there will be no tests.} else write(a.digit[i],','); {And dead code will be omitted.} End; Procedure BigZero(var A: BigNumber); {A:=0;} Begin; A.TopDigit:=0; A.Digit[0]:=0; End; Procedure BigOne(var A: BigNumber); {A:=1;} Begin; A.TopDigit:=0; A.Digit[0]:=1; End; Function BigInt(n: longint): BigNumber; {A:=N;} var l: BigNumberIndexer; Begin l:=0; if n < 0 then croak('Negative integers are not yet considered.'); repeat {At least one digit is to be placed.} if l > BigEnuff then Croak('BigInt overflowed!'); {Oh dear.} BigInt.Digit[l]:=N mod BigBase; {The low-order digit.} n:=n div BigBase; {Shift down a digit.} l:=l + 1; {Count in anticipation.} until N = 0; {Still some number left?} BigInt.TopDigit:=l - 1; {Went one too far.} End; Function BigMult(a,b: BigNumber): BigNumber; {x:=BigMult(a,b);} {Suppose the digits of A are a5,a4,a3,a2,a1,a0... To multiply A and B. a5 a4 a3 a2 a1 a0: six digits, d1 x b4 b3 b2 b1 b0: five digits, d2 --------------------------- a5b0 a4b0 a3b0 a2b0 a1b0 a0b0 a5b1 a4b1 a3b1 a2b1 a1b1 a0b1 a5b2 a4b2 a3b2 a2b2 a1b2 a0b2 a5b3 a4b3 a3b3 a2b3 a1b3 a0b3 a5b4 a4b4 a3b4 a2b4 a1b4 a0b4 ------------------------------------------------------- carry 9 8 7 6 5 4 3 2 1 0: at least nine digits, ------------------------------------------------------- = d1 + d2 - 1 But the indices are also the powers, so the highest power is 9 = 5 + 4, and a possible tenth for any carry.} var X: BigNumber; {Scratchpad, so b:=BigMult(a,b); doesn't overwrite b as it goes...} var d: BigNumberDigit; {A digit.} var c: BigNumberDigit; {A carry.} var dd: BigNumberDigit2; {A digit product.} var i,j,l: BigNumberIndexer; {Steppers.} Begin if ((A.TopDigit = 0) and (A.Digit[0] = 0)) or((B.TopDigit = 0) and (B.Digit[0] = 0)) then begin BigZero(BigMult); exit; end; l:=A.TopDigit + B.TopDigit; {Minimal digit requirement. (Counting is from zero)} if l > BigEnuff then Croak('BigMult will overflow.'); for i:=l downto 0 do X.Digit[i]:=0; {Clear for action.} for i:=0 to A.TopDigit do {Arbitrarily, choose A on the one hand.} begin {Though there could be a better choice.} d:=A.Digit[i]; {Select the digit.} if d <> 0 then {What the hell. One in BigBase chance.} begin {But not this time.} l:=i; {Locate the power of BigBase.} c:=0; {Start this digit's multiply pass.} for j:=0 to B.TopDigit do {Stepping along B's digits.} begin {One by one.} dd:=BigNumberDigit2(B.Digit[j])*d + X.Digit[l] + c; {The deed.} X.Digit[l]:=dd mod BigBase; {Place the new digit.} c:=dd div BigBase; {And extract the carry.} l:=l + 1; {Ready for the next power up.} end; {Advance to it.} if c > 0 then {The multiply done, place the carry.} begin {Ah. We *will* use the next power up.} if l > BigEnuff then Croak('BigMultX has overflowed.'); {Oh dear.} X.Digit[l]:=c; {Thus as if BigMult..Digit[l] was zeroed.} l:=l + 1; {Preserve the one-too-far for the last case} end; {So much for a carry at the end of a pass.} end; {So much for a non-zero digit.} end; {On to another digit to multiply with.} X.TopDigit:=l - 1; {Remember the one-too-far.} BigMult:=X; {Deliver, possibly scragging A or B, or, both!} End; {of BigMult.} Procedure BigPower(var X: BigNumber; P: longint); {Replaces X by X**P} var A,W: BigNumber; {Scratchpads} label up; Begin {Each squaring doubles the power, melding nicely with binary reduction.} if P <= 0 then Croak('Negative powers are not accommodated!'); BigOne(A); {x**0 = 1} W:=X; {Holds X**1, 2, 4, 8, etc.} up:if P mod 2 = 1 then A:=BigMult(A,W); {Bit on, so include this order.} P:=P div 2; {Halve the power contrariwise to W's doubling.} if P > 0 then {Still some power to come?} begin {Yes.} W:=BigMult(W,W); {Step up to the next bit's power.} goto up; {And see if it is "on".} end; {Odd layout avoids multiply testing P > 0.} X:=A; {The result.} End; var X: BigNumber; var p: longint; BEGIN ClrScr; WriteLn('To calculate x = 2**64, then x*x via multi-digit long multiplication.'); p:=64; {As per the specification.} X:=BigInt(2); {Start with 2.} BigPower(X,p); {First stage: 2**64} Write ('x = 2**',p,' = '); BigShow(X); WriteLn; X:=BigMult(X,X); {Second stage.} Write ('x*x = ');BigShow(X); {Can't have Write('x*x = ',BigShow(BigMult(X,X))), after all. Oh well.} END.
Output: To calculate x = 264, then x*x via multi-digit long multiplication. x = 264 = 18446744073709551616 x*x = 340282366920938463463374607431768211456
Perl
#!/usr/bin/perl -w use strict; # This should probably be done in a loop rather than be recursive. sub add_with_carry { my $resultref = shift; my $addend = shift; my $addendpos = shift; push @$resultref, (0) while (scalar @$resultref < $addendpos + 1); my $addend_result = $addend + $resultref->[$addendpos]; my @addend_digits = reverse split //, $addend_result; $resultref->[$addendpos] = shift @addend_digits; my $carry_digit = shift @addend_digits; &add_with_carry($resultref, $carry_digit, $addendpos + 1) if( defined $carry_digit ) } sub longhand_multiplication { my @multiplicand = reverse split //, shift; my @multiplier = reverse split //, shift; my @result = (); my $multiplicand_offset = 0; foreach my $multiplicand_digit (@multiplicand) { my $multiplier_offset = $multiplicand_offset; foreach my $multiplier_digit (@multiplier) { my $multiplication_result = $multiplicand_digit * $multiplier_digit; my @result_digit_addend_list = reverse split //, $multiplication_result; my $addend_offset = $multiplier_offset; foreach my $result_digit_addend (@result_digit_addend_list) { &add_with_carry(\@result, $result_digit_addend, $addend_offset++) } ++$multiplier_offset; } ++$multiplicand_offset; } @result = reverse @result; return join '', @result; } my $sixtyfour = "18446744073709551616"; my $onetwentyeight = &longhand_multiplication($sixtyfour, $sixtyfour); print "$onetwentyeight\n";
Perl 6
{{works with|rakudo|2015-09-17}} For efficiency (and novelty), this program explicitly implements long multiplication, but in base 10000. That base was chosen because multiplying two 5-digit numbers can overflow a 32-bit integer, but two 4-digit numbers cannot.
sub num_to_groups ( $num ) { $num.flip.comb(/.**1..4/)».flip };
sub groups_to_num ( @g ) { [~] flat @g.pop, @g.reverse».fmt('%04d') };
sub long_multiply ( Str $x, Str $y ) {
my @group_sums;
for flat num_to_groups($x).pairs X num_to_groups($y).pairs -> $xp, $yp {
@group_sums[ $xp.key + $yp.key ] += $xp.value * $yp.value;
}
for @group_sums.keys -> $k {
next if @group_sums[$k] < 10000;
@group_sums[$k+1] += @group_sums[$k].Int div 10000;
@group_sums[$k] %= 10000;
}
return groups_to_num @group_sums;
}
my $str = '18446744073709551616';
long_multiply( $str, $str ).say;
# cross-check with native implementation
say +$str * +$str;
{{out}}
340282366920938463463374607431768211456
340282366920938463463374607431768211456
Phix
{{Trans|Euphoria}} Simple longhand multiplication. To keep things as simple as possible, this does not handle negative numbers.
If bcd1 is a number split into digits 0..9, bcd9 is a number split into "digits" 000,000,000..999,999,999, which fit in an integer.
They are held lsb-style mainly so that trimming a trailing 0 does not alter their value.
constant base = 1_000_000_000
function bcd9_mult(sequence a, sequence b)
sequence c
integer j
atom ci
c = repeat(0,length(a)+length(b))
for i=1 to length(a) do
j = i+length(b)-1
c[i..j] = sq_add(c[i..j],sq_mul(a[i],b))
end for
for i=1 to length(c) do
ci = c[i]
if ci>base then
c[i+1] += floor(ci/base) -- carry
c[i] = remainder(ci,base)
end if
end for
if c[$]=0 then
c = c[1..$-1]
end if
return c
end function
function atom_to_bcd9(atom a)
sequence s = {}
while a>0 do
s = append(s,remainder(a,base))
a = floor(a/base)
end while
return s
end function
function bcd9_to_str(sequence a)
string s = sprintf("%d",a[$])
for i=length(a)-1 to 1 by -1 do
s &= sprintf("%09d",a[i])
end for
-- (might want to trim leading 0s here)
return s
end function
sequence a, b, c
a = atom_to_bcd9(power(2,32))
printf(1,"a is %s\n",{bcd9_to_str(a)})
b = bcd9_mult(a,a)
printf(1,"a*a is %s\n",{bcd9_to_str(b)})
c = bcd9_mult(b,b)
printf(1,"a*a*a*a is %s\n",{bcd9_to_str(c)})
{{out}}
a is 4294967296
a*a is 18446744073709551616
a*a*a*a is 340282366920938463488374607488768211456
PHP
<?php function longMult($a, $b) { $as = (string) $a; $bs = (string) $b; for($pi = 0, $ai = strlen($as) - 1; $ai >= 0; $pi++, $ai--) { for($p = 0; $p < $pi; $p++) { $regi[$ai][] = 0; } for($bi = strlen($bs) - 1; $bi >= 0; $bi--) { $regi[$ai][] = $as[$ai] * $bs[$bi]; } } return $regi; } function longAdd($arr) { $outer = count($arr); $inner = count($arr[$outer-1]) + $outer; for($i = 0; $i <= $inner; $i++) { for($o = 0; $o < $outer; $o++) { $val = isset($arr[$o][$i]) ? $arr[$o][$i] : 0; @$sum[$i] += $val; } } return $sum; } function carry($arr) { for($i = 0; $i < count($arr); $i++) { $s = (string) $arr[$i]; switch(strlen($s)) { case 2: $arr[$i] = $s{1}; @$arr[$i+1] += $s{0}; break; case 3: $arr[$i] = $s{2}; @$arr[$i+1] += $s{0}.$s{1}; break; } } return ltrim(implode('',array_reverse($arr)),'0'); } function lm($a,$b) { return carry(longAdd(longMult($a,$b))); } if(lm('18446744073709551616','18446744073709551616') == '340282366920938463463374607431768211456') { echo 'pass!'; }; // 2^64 * 2^64 </Lang> ## PicoLisp ```PicoLisp (de multi (A B) (setq A (format A) B (reverse (chop B))) (let Result 0 (for (I . X) B (setq Result (+ Result (* (format X) A (** 10 (dec I)))))) ) )
PL/I
/* Multiply a by b, giving c. */
multiply: procedure (a, b, c);
declare (a, b, c) (*) fixed decimal (1);
declare (d, e, f) (hbound(a,1)) fixed decimal (1);
declare pr (-hbound(a,1) : hbound(a,1)) fixed decimal (1);
declare p fixed decimal (2), (carry, s) fixed decimal (1);
declare neg bit (1) aligned;
declare (i, j, n, offset) fixed binary (31);
n = hbound(a,1);
d = a;
e = b;
s = a(1) + b(1);
neg = (s = 9);
if a(1) = 9 then call complement (d);
if b(1) = 9 then call complement (e);
pr = 0;
offset = 0; carry = 0;
do i = n to 1 by -1;
do j = n to 1 by -1;
p = d(i) * e(j) + pr(j-offset) + carry;
if p > 9 then do; carry = p/10; p = mod(p, 10); end; else carry = 0;
pr(j-offset) = p;
end;
offset = offset + 1;
end;
do i = hbound(a,1) to 1 by -1;
c(i) = pr(i);
end;
do i = -hbound(a,1) to 1;
if pr(i) ^= 0 then signal fixedoverflow;
end;
if neg then call complement (c);
end multiply;
complement: procedure (a);
declare a(*) fixed decimal (1);
declare i fixed binary (31), carry fixed decimal (1);
declare s fixed decimal (2);
carry = 1;
do i = hbound(a,1) to 1 by -1;
s = 9 - a(i) + carry;
if s > 9 then do; s = s - 10; carry = 1; end; else carry = 0;
a(i) = s;
end;
end complement;
Calling sequence:
a = 0; b = 0; c = 0;
a(60) = 1;
do i = 1 to 64; /* Generate 2**64 */
call add (a, a, b);
put skip;
call output (b);
a = b;
end;
call multiply (a, b, c);
put skip;
call output (c);
Final output:
18446744073709551616
340282366920938463463374607431768211456
PowerShell
Implementation
# LongAddition only supports Unsigned Integers represented as Strings/Character Arrays Function LongAddition ( [Char[]] $lhs, [Char[]] $rhs ) { $lhsl = $lhs.length $rhsl = $rhs.length if(($lhsl -gt 0) -and ($rhsl -gt 0)) { $maxplace = [Math]::Max($rhsl,$lhsl)+1 1..$maxplace | ForEach-Object { $carry = 0 $result = "" } { $add1 = 0 $add2 = 0 if( $_ -le $lhsl ) { $add1 = [int]$lhs[ -$_ ] - 48 } if( $_ -le $rhsl ) { $add2 = [int]$rhs[ -$_ ] - 48 } $iresult = $add1 + $add2 + $carry if( ( $_ -lt $maxplace ) -or ( $iresult -gt 0 ) ) { $result = "{0}{1}" -f ( $iresult % 10 ),$result } $carry = [Math]::Floor( $iresult / 10 ) } { $result } } elseif($lhsl -gt 0) { [String]::Join( '', $lhs ) } elseif($rhsl -gt 0) { [String]::Join( '', $rhs ) } else { "0" } } # LongMultiplication only supports Unsigned Integers represented as Strings/Character Arrays Function LongMultiplication ( [Char[]] $lhs, [Char[]] $rhs ) { $lhsl = $lhs.length $rhsl = $rhs.length if(($lhsl -gt 0) -and ($rhsl -gt 0)) { 1..$lhsl | ForEach-Object { $carry0 = "" $result0 = "" } { $i = -$_ $add1 = ( 1..$rhsl | ForEach-Object { $carry1 = 0 $result1 = "" } { $j = -$_ $mult1 = [int]$lhs[ $i ] - 48 $mult2 = [int]$rhs[ $j ] - 48 $iresult1 = $mult1 * $mult2 + $carry1 $result1 = "{0}{1}" -f ( $iresult1 % 10 ), $result1 $carry1 = [Math]::Floor( $iresult1 / 10 ) } { if( $carry1 -gt 0 ) { $result1 = "{0}{1}" -f $carry1, $result1 } $result1 } ) $iresult0 = ( LongAddition $add1 $carry0 ) $iresultl = $iresult0.length $result0 = "{0}{1}" -f $iresult0[-1],$result0 if( $iresultl -gt 1 ) { $carry0 = [String]::Join( '', $iresult0[ -$iresultl..-2 ] ) } else { $carry0 = "" } } { if( $carry0 -ne "" ) { $result0 = "{0}{1}" -f $carry0, $result0 } $result0 } } else { "0" } } LongMultiplication "18446744073709551616" "18446744073709551616"
Library Method
{{works with|PowerShell|4.0}}
[BigInt]$n = [Math]::Pow(2,64) [BigInt]::Multiply($n,$n)
Output:
340282366920938463463374607431768211456
Prolog
Arbitrary precision arithmetic is native in most Prolog implementations.
?- X is 2**64 * 2**64.
X = 340282366920938463463374607431768211456.
PureBasic
Explicit Implementation
Structure decDigitFmt ;decimal digit format
Array Digit.b(0) ;contains each digit of number, right-most digit is index 0
digitCount.i ;zero based
sign.i ; {x < 0} = -1, {x = 0} = 0, {x > 0} = 1
EndStructure
Global zero_decDigitFmt.decDigitFmt ;represents zero in the decimal digit format
;converts string representation of integer into the digit format, number can include signus but no imbedded spaces
Procedure stringToDecDigitFmt(numString.s, *x.decDigitFmt)
Protected *c.Character, digitIdx, digitCount
If numString And *x
*c.Character = @numString
Repeat
Select *c\c
Case '0' To '9', '-', '+'
*c + SizeOf(Character)
Default
numString = Left(numString, *c - @numString)
Break
EndSelect
ForEver
*c = @numString
Select *c\c
Case '-'
*x\sign = -1
*c + SizeOf(Character)
Case '+'
*x\sign = 1
*c + SizeOf(Character)
Case '0' To '9'
*x\sign = 1
EndSelect
numString = LTrim(PeekS(*c), "0") ;remove leading zeroes
If numString = "" ;is true if equal to zero or if only a signus is present
CopyStructure(@zero_decDigitFmt, *x, decDigitFmt)
ProcedureReturn
EndIf
*c = @numString
digitCount = Len(PeekS(*c)) - 1
Dim *x\Digit(digitCount)
*x\digitCount = digitCount
digitIdx = 0
While *c\c
If *c\c >= '0' And *c\c <= '9'
*x\Digit(digitCount - digitIdx) = *c\c - '0'
digitIdx + 1
*c + SizeOf(Character)
Else
Break
EndIf
Wend
EndIf
EndProcedure
;converts digit format representation of integer into string representation
Procedure.s decDigitFmtToString(*x.decDigitFmt)
Protected i, number.s
If *x
If *x\sign = 0
number = "0"
Else
For i = *x\digitCount To 0 Step -1
number + Str(*x\Digit(i))
Next
number = LTrim(number, "0")
If *x\sign = -1
number = "-" + number
EndIf
EndIf
EndIf
ProcedureReturn number
EndProcedure
;handles only positive numbers and zero, negative numbers left as an exercise for the reader ;)
Procedure add_decDigitFmt(*a.decDigitFmt, *b.decDigitFmt, *sum.decDigitFmt, digitPos = 0) ;*sum contains the result of (*a ) * 10^digitPos + (*b)
Protected carry, i, newDigitCount, workingSum, a_dup.decDigitFmt
If *a And *b And *sum
If *a = *sum: CopyStructure(*a, @a_dup, decDigitFmt): *a = @a_dup: EndIf ;handle special case of *sum + *b = *sum
If *b <> *sum: CopyStructure(*b, *sum, decDigitFmt): EndIf ;handle general case of *a + *b = *sum and special case of *a + *sum = *sum
;calculate number of digits needed for sum and resize array of digits if necessary
newDigitCount = *a\digitCount + digitPos
If newDigitCount >= *sum\digitCount
If *sum\digitCount = newDigitCount And *sum\Digit(*sum\digitCount) <> 0
newDigitCount + 1
EndIf
If *sum\digitCount <> newDigitCount
*sum\digitCount = newDigitCount
Redim *sum\Digit(*sum\digitCount)
EndIf
EndIf
i = 0
Repeat
If i <= *a\digitCount
workingSum = *a\Digit(i) + *sum\Digit(digitPos) + carry
Else
workingSum = *sum\Digit(digitPos) + carry
EndIf
If workingSum > 9
carry = 1
workingSum - 10
Else
carry = 0
EndIf
*sum\Digit(digitPos) = workingSum
digitPos + 1
i + 1
Until i > *a\digitCount And carry = 0
If *a\sign <> 0 Or *sum\sign <> 0
*sum\sign = 1 ;only handle positive numbers and zero for now
EndIf
EndIf
EndProcedure
Procedure multiply_decDigitFmt(*a.decDigitFmt, *b.decDigitFmt, *product.decDigitFmt) ;*product contains the result of (*a) * (*b)
Protected i, digitPos, productSignus
Protected Dim multTable.decDigitFmt(9)
Protected NewList digitProduct.decDigitFmt()
If *a And *b And *product
If *a\sign = 0 Or *b\sign = 0
CopyStructure(zero_decDigitFmt, *product, decDigitFmt)
ProcedureReturn
EndIf
If *b\digitCount > *a\digitCount: Swap *a, *b: EndIf
;build multiplication table
CopyStructure(*a, @multTable(1), decDigitFmt): multTable(1)\sign = 1 ;always positive
For i = 2 To 9
add_decDigitFmt(*a, multTable(i - 1), multTable(i))
Next
;collect individual digit products for later summation; these could also be added as we go along
For i = 0 To *b\digitCount
AddElement(digitProduct())
digitProduct() = multTable(*b\Digit(i))
Next
;determine sign of product
If *a\sign <> *b\sign
productSignus = -1
Else
productSignus = 1
EndIf
digitPos = 0
CopyStructure(zero_decDigitFmt, *product, decDigitFmt)
ForEach digitProduct()
add_decDigitFmt(digitProduct(), *product, *product, digitPos)
digitPos + 1
Next
*product\sign = productSignus ;set sign of product
EndIf
EndProcedure
;handles only positive integer exponents or an exponent of zero, does not raise an error for 0^0
Procedure exponent_decDigitFmt(*a.decDigitFmt, exponent, *product.decDigitFmt)
Protected i, a_dup.decDigitFmt
If *a And *product And exponent >= 0
If *a = *product: CopyStructure(*a, @a_dup, decDigitFmt): *a = @a_dup: EndIf
stringToDecDigitFmt("1", *product)
For i = 1 To exponent: multiply_decDigitFmt(*product, *a, *product): Next
EndIf
EndProcedure
If OpenConsole()
Define a.decDigitFmt, product.decDigitFmt
stringToDecDigitFmt("2", a)
exponent_decDigitFmt(a, 64, a) ;2^64
multiply_decDigitFmt(a, a, product)
PrintN("The result of 2^64 * 2^64 is " + decDigitFmtToString(product))
Print(#crlf$ + #crlf$ + "Press ENTER to exit"): Input()
CloseConsole()
EndIf
Output:
The result of 2^64 * 2^64 is 340282366920938463463374607431768211456
Library Method
{{works with|PureBasic|4.41}}
Using [http://www.purebasic.fr/english/viewtopic.php?p=309763#p309763 Decimal.pbi] by Stargåte allows for calculation with long numbers, this is useful since version 4.41 of PureBasic mostly only supporter data types native to x86/x64/PPC etc processors.
XIncludeFile "decimal.pbi"
Define.Decimal *a, *b
*a=PowerDecimal(IntegerToDecimal(2),IntegerToDecimal(64))
*b=TimesDecimal(*a,*a,#NoDecimal)
Print("2^64*2^64 = "+DecimalToString(*b))
'''Outputs 2^64*2^64 = 340282366920938463463374607431768211456
Python
(Note that Python comes with arbitrary length integers).
#!/usr/bin/env python print 2**64*2**64
{{works with|Python|3.0}} {{trans|Perl}}
#!/usr/bin/env python def add_with_carry(result, addend, addendpos): while True: while len(result) < addendpos + 1: result.append(0) addend_result = str(int(addend) + int(result[addendpos])) addend_digits = list(addend_result) result[addendpos] = addend_digits.pop() if not addend_digits: break addend = addend_digits.pop() addendpos += 1 def longhand_multiplication(multiplicand, multiplier): result = [] for multiplicand_offset, multiplicand_digit in enumerate(reversed(multiplicand)): for multiplier_offset, multiplier_digit in enumerate(reversed(multiplier), start=multiplicand_offset): multiplication_result = str(int(multiplicand_digit) * int(multiplier_digit)) for addend_offset, result_digit_addend in enumerate(reversed(multiplication_result), start=multiplier_offset): add_with_carry(result, result_digit_addend, addend_offset) result.reverse() return ''.join(result) if __name__ == "__main__": sixtyfour = "18446744073709551616" onetwentyeight = longhand_multiplication(sixtyfour, sixtyfour) print(onetwentyeight)
Shorter version: {{trans|Haskell}} {{Works with|Python|3.7}}
'''Long multiplication''' from functools import reduce def longmult(x, y): '''Long multiplication.''' return reduce( digitSum, polymul(digits(x), digits(y)), 0 ) def digitSum(a, x): '''Left to right decimal digit summing.''' return a * 10 + x def polymul(xs, ys): '''List of specific products.''' return map( lambda *vs: sum(filter(None, vs)), *[ [0] * i + zs for i, zs in enumerate(mult_table(xs, ys)) ] ) def mult_table(xs, ys): '''Rows of all products.''' return [[x * y for x in xs] for y in ys] def digits(x): '''Digits of x as a list of integers.''' return [int(c) for c in str(x)] if __name__ == '__main__': print( longmult(2 ** 64, 2 ** 64) )
Ol
Ol already supports long numbers "out-of-the-box".
(define x (* 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2)) ; 2^64
(print (* x x))
340282366920938463463374607431768211456
R
Using GMP
{{libheader|gmp}}
library(gmp) a <- as.bigz("18446744073709551616") mul.bigz(a,a)
"340282366920938463463374607431768211456"
A native implementation
This code is more verbose than necessary, for ease of understanding.
longmult <- function(xstr, ystr) { #get the number described in each string getnumeric <- function(xstr) as.numeric(unlist(strsplit(xstr, ""))) x <- getnumeric(xstr) y <- getnumeric(ystr) #multiply each pair of digits together mat <- apply(x %o% y, 1, as.character) #loop over columns, then rows, adding zeroes to end of each number in the matrix to get the correct positioning ncols <- ncol(mat) cols <- seq_len(ncols) for(j in cols) { zeroes <- paste(rep("0", ncols-j), collapse="") mat[,j] <- paste(mat[,j], zeroes, sep="") } nrows <- nrow(mat) rows <- seq_len(nrows) for(i in rows) { zeroes <- paste(rep("0", nrows-i), collapse="") mat[i,] <- paste(mat[i,], zeroes, sep="") } #add zeroes to the start of the each number, so they are all the same length len <- max(nchar(mat)) strcolumns <- formatC(cbind(as.vector(mat)), width=len) strcolumns <- gsub(" ", "0", strcolumns) #line up all the numbers below each other strmat <- matrix(unlist(strsplit(strcolumns, "")), byrow=TRUE, ncol=len) #convert to numeric and add them mat2 <- apply(strmat, 2, as.numeric) sum1 <- colSums(mat2) #repeat the process on each of the totals, until each total is a single digit repeat { ntotals <- length(sum1) totals <- seq_len(ntotals) for(i in totals) { zeroes <- paste(rep("0", ntotals-i), collapse="") sum1[i] <- paste(sum1[i], zeroes, sep="") } len2 <- max(nchar(sum1)) strcolumns2 <- formatC(cbind(as.vector(sum1)), width=len2) strcolumns2 <- gsub(" ", "0", strcolumns2) strmat2 <- matrix(unlist(strsplit(strcolumns2, "")), byrow=TRUE, ncol=len2) mat3 <- apply(strmat2, 2, as.numeric) sum1 <- colSums(mat3) if(all(sum1 < 10)) break } #Concatenate the digits together ans <- paste(sum1, collapse="") ans } a <- "18446744073709551616" longmult(a, a)
"340282366920938463463374607431768211456"
Racket
#lang racket
(define (mult A B)
(define nums
(let loop ([B B] [zeros '()])
(if (null? B)
'()
(cons (append zeros (let loop ([c 0] [A A])
(cond [(pair? A)
(define-values [q r]
(quotient/remainder
(+ c (* (car A) (car B)))
10))
(cons r (loop q (cdr A)))]
[(zero? c) '()]
[else (list c)])))
(loop (cdr B) (cons 0 zeros))))))
(let loop ([c 0] [nums nums])
(if (null? nums)
'()
(let-values ([(q r) (quotient/remainder (apply + c (map car nums)) 10)])
(cons r (loop q (filter pair? (map cdr nums))))))))
(define (number->list n)
(if (zero? n) '()
(let-values ([(q r) (quotient/remainder n 10)])
(cons r (number->list q)))))
(define 2^64 (number->list (expt 2 64)))
(for-each display (reverse (mult 2^64 2^64))) (newline)
;; for comparison
(* (expt 2 64) (expt 2 64))
;; Output:
;; 340282366920938463463374607431768211456
;; 340282366920938463463374607431768211456
REXX
version 1
This REXX version supports: ::* leading signs ::* decimal points ::* automatically adjusting the number of decimal digits needed
Programming note: && is REXX's '''exclusive or''' operand.
/*REXX program performs long multiplication on two numbers (without the "E"). */
numeric digits 300 /*be able to handle gihugeic input #s. */
parse arg x y . /*obtain optional arguments from the CL*/
if x=='' | x=="," then x=2**64 /*Not specified? Then use the default.*/
if y=='' | y=="," then y=x /* " " " " " " */
if x<0 && y<0 then sign= '-' /*there only a single negative number? */
else sign= /*no, then result sign must be positive*/
xx=x; x=strip(x, 'T', .); x1=left(x, 1) /*remove any trailing decimal points. */
yy=y; y=strip(y, 'T', .); y1=left(y, 1) /* " " " " " */
if x1=='-' | x1=="+" then x=substr(x, 2) /*remove a leading ± sign. */
if y1=='-' | y1=="+" then y=substr(y, 2) /* " " " " " */
parse var x '.' xf; parse var y "." yf /*obtain the fractional part of X and Y*/
#=length(xf || yf) /*#: digits past the decimal points (.)*/
x=space( translate( x, , .), 0) /*remove decimal point if there is any.*/
y=space( translate( y, , .), 0) /* " " " " " " " */
Lx=length(x); Ly=length(y) /*get the lengths of the new X and Y. */
numeric digits max(digits(), Lx + Ly) /*use a new decimal digits precision.*/
$=0 /*$: is the product (so far). */
do j=Ly by -1 for Ly /*almost like REXX does it, ··· but no.*/
$=$ + ((x*substr(y, j, 1))copies(0, Ly-j) )
end /*j*/
f=length($) - # /*does product has enough decimal digs?*/
if f<0 then $=copies(0, abs(f) + 1)$ /*Negative? Add leading 0s for INSERT.*/
say 'long mult:' xx "*" yy '──►' sign || strip( insert(., $, length($) - #), 'T', .)
say ' built─in:' xx "*" yy '──►' xx*yy /*stick a fork in it, we're all done. */
'''output''' when using the default inputs:
long mult: 18446744073709551616 * 18446744073709551616 ──► 340282366920938463463374607431768211456
built─in: 18446744073709551616 * 18446744073709551616 ──► 340282366920938463463374607431768211456
'''output''' when using the inputs of: 123 -456789000
long mult: 123 * -456789000 ──► -56185047000
built─in: 123 * -456789000 ──► -56185047000
'''output''' when using the inputs of: -123.678 +456789000
long mult: -123.678 * +456789000 ──► -56494749942.000
built─in: -123.678 * +456789000 ──► -56494749942.000
version 2
/* REXX **************************************************************
* While REXX can multiply arbitrary large integers
* here is the algorithm asked for by the task description
* 13.05.2013 Walter Pachl
*********************************************************************/
cnt.=0
Numeric Digits 100
Call test 123 123
Call test 12 12
Call test 123456789012 44444444444
Call test 2**64 2**64
Call test 0 0
say cnt.0ok 'ok'
say cnt.0nok 'not ok'
Exit
test:
Parse Arg a b
soll=a*b
haben=multiply(a b)
Say 'soll =' soll
Say 'haben=' haben
If haben<>soll Then
cnt.0nok=cnt.0nok+1
Else
cnt.0ok=cnt.0ok+1
Return
multiply: Procedure
/* REXX **************************************************************
* Multiply(a b) -> a*b
*********************************************************************/
Parse Arg a b
Call s2a 'a'
Call s2a 'b'
r.=0
rim=1
r0=0
Do bi=1 To b.0
Do ai=1 To a.0
ri=ai+bi-1
p=a.ai*b.bi
Do i=ri by 1 Until p=0
s=r.i+p
r.i=s//10
p=s%10
End
rim=max(rim,i)
End
End
res=strip(a2s('r'),'L','0')
If res='' Then
res='0'
Return res
s2a:
/**********************************************************************
* copy characters of a string into a corresponding array
* digits are numbered 1 to n fron right to left
**********************************************************************/
Parse arg name
string=value(name)
lstring=length(string)
do z=1 to lstring
Call value name'.'z,substr(string,lstring-z+1,1)
End
Call value name'.0',lstring
Return
a2s:
/**********************************************************************
* turn the array of digits into a string
**********************************************************************/
call trace 'o'
Parse Arg name
ol=''
Do z=rim To 1 By -1
ol=ol||value(name'.z')
End
Return ol
Output:
soll = 15129
haben= 15129
soll = 144
haben= 144
soll = 5486968400478463649328
haben= 5486968400478463649328
soll = 340282366920938463463374607431768211456
haben= 340282366920938463463374607431768211456
soll = 0
haben= 0
5 ok
0 not ok
Ruby
{{trans|Tcl}}
def longmult(x,y) result = [0] j = 0 y.digits.each do |m| c = 0 i = j x.digits.each do |d| v = result[i] result << 0 if v.zero? c, v = (v + c + d*m).divmod(10) result[i] = v i += 1 end result[i] += c j += 1 end # calculate the answer from the result array of digits result.reverse.inject(0) {|sum, n| 10*sum + n} end n=2**64 printf " %d * %d = %d\n", n, n, n*n printf "longmult(%d, %d) = %d\n", n, n, longmult(n,n)
18446744073709551616 * 18446744073709551616 = 340282366920938463463374607431768211456
longmult(18446744073709551616, 18446744073709551616) = 340282366920938463463374607431768211456
Scala
This implementation does not rely on an arbitrary precision numeric type. Instead, only single digits are ever multiplied or added, and all partial results are kept as string.
def addNums(x: String, y: String) = { val padSize = x.length max y.length val paddedX = "0" * (padSize - x.length) + x val paddedY = "0" * (padSize - y.length) + y val (sum, carry) = (paddedX zip paddedY).foldRight(("", 0)) { case ((dx, dy), (acc, carry)) => val sum = dx.asDigit + dy.asDigit + carry ((sum % 10).toString + acc, sum / 10) } if (carry != 0) carry.toString + sum else sum } def multByDigit(num: String, digit: Int) = { val (mult, carry) = num.foldRight(("", 0)) { case (d, (acc, carry)) => val mult = d.asDigit * digit + carry ((mult % 10).toString + acc, mult / 10) } if (carry != 0) carry.toString + mult else mult } def mult(x: String, y: String) = y.foldLeft("")((acc, digit) => addNums(acc + "0", multByDigit(x, digit.asDigit)))
Sample:
scala> mult("18446744073709551616", "18446744073709551616")
res25: java.lang.String = 340282366920938463463374607431768211456
{{works with|Scala|2.8}}
Scala 2.8 introduces scanLeft and scanRight which can be used to simplify this further:
def adjustResult(result: IndexedSeq[Int]) = ( result .map(_ % 10) // remove carry from each digit .tail // drop the seed carry .reverse // put most significant digits on the left .dropWhile(_ == 0) // remove leading zeroes .mkString ) def addNums(x: String, y: String) = { val padSize = (x.length max y.length) + 1 // We want to keep a zero to the left, to catch the carry val paddedX = "0" * (padSize - x.length) + x val paddedY = "0" * (padSize - y.length) + y adjustResult((paddedX zip paddedY).scanRight(0) { case ((dx, dy), last) => dx.asDigit + dy.asDigit + last / 10 }) } def multByDigit(num: String, digit: Int) = adjustResult(("0"+num).scanRight(0)(_.asDigit * digit + _ / 10)) def mult(x: String, y: String) = y.foldLeft("")((acc, digit) => addNums(acc + "0", multByDigit(x, digit.asDigit)))
Scheme
Since Scheme already supports arbitrary precision arithmetic, build it out of church numerals. Don't try converting these to native integers. You will die waiting for the answer.
(define one (lambda (f) (lambda (x) (f x))))
(define (add a b) (lambda (f) (lambda (x) ((a f) ((b f) x)))))
(define (mult a b) (lambda (f) (lambda (x) ((a (b f)) x))))
(define (expo a b) (lambda (f) (lambda (x) (((b a) f) x))))
(define two (add one one))
(define six (add two (add two two)))
(define sixty-four (expo two six))
(display (mult (expo two sixty-four) (expo two sixty-four)))
Seed7
Seed7 supports arbitrary-precision arithmetic. The library [http://seed7.sourceforge.net/libraries/bigint.htm bigint.s7i] defines the type [http://seed7.sourceforge.net/manual/types.htm#bigInteger bigInteger]. A bigInteger is a signed integer number of unlimited size. With library support the task can be solved by using the multiplication operator [http://seed7.sourceforge.net/libraries/bigint.htm#%28in_bigInteger%29*%28in_bigInteger%29 *]:
$ include "seed7_05.s7i";
include "bigint.s7i";
const proc: main is func
begin
writeln(2_**64 * 2_**64);
end func;
Output:
340282366920938463463374607431768211456
This task seems to prefer an inferior implementation of a long multiplication, where long numbers are stored in decimal strings. Besides type safety there are seveal other drawbacks triggered by such a representation. E.g.: In almost all cases a representation with decimal strings leads to significant lower computing speed. The multiplication example below uses the requested inferior implementation:
$ include "seed7_05.s7i";
const func string: (in string: a) * (in string: b) is func
result
var string: product is "";
local
var integer: i is 1;
var integer: j is 1;
var integer: k is 0;
var integer: carry is 0;
begin
if startsWith(a, "-") then
if startsWith(b, "-") then
product := a[2 ..] * b[2 ..];
else
product := "-" & a[2 ..] * b;
end if;
elsif startsWith(b, "-") then
product := "-" & a * b[2 ..];
else
product := "0" mult length(a) + length(b);
for i range length(a) downto 1 do
k := i + length(b);
carry := 0;
for j range length(b) downto 1 do
carry +:= (ord(a[i]) - ord('0')) * (ord(b[j]) - ord('0')) + (ord(product[k]) - ord('0'));
product @:= [k] chr(carry rem 10 + ord('0'));
carry := carry div 10;
decr(k);
end for;
product @:= [k] chr(ord(product[k]) + carry);
end for;
while startsWith(product, "0") and length(product) >= 2 do
product := product[2 ..];
end while;
end if;
end func;
const proc: main is func
begin
writeln("-18446744073709551616" * "-18446744073709551616");
end func;
The output is the same as with the superior solution.
Sidef
(Note that arbitrary precision arithmetic is native in Sidef).
say (2**64 * 2**64);
{{trans|Python}}
func add_with_carry(result, addend, addendpos) { loop { while (result.len < addendpos+1) { result.append(0) } var addend_digits = (addend.to_i + result[addendpos] -> to_s.chars) result[addendpos] = addend_digits.pop addend_digits.len > 0 || break addend = addend_digits.pop addendpos++ } } func longhand_multiplication(multiplicand, multiplier) { var result = [] var multiplicand_offset = 0 multiplicand.reverse.each { |multiplicand_digit| var multiplier_offset = multiplicand_offset multiplier.reverse.each { |multiplier_digit| var multiplication_result = (multiplicand_digit.to_i * multiplier_digit.to_i -> to_s) var addend_offset = multiplier_offset multiplication_result.reverse.each { |result_digit_addend| add_with_carry(result, result_digit_addend, addend_offset) addend_offset++ } multiplier_offset++ } multiplicand_offset++ } return result.join.reverse } say longhand_multiplication('18446744073709551616', '18446744073709551616')
{{out}}
340282366920938463463374607431768211456
Slate
{{incorrect|Slate|Code does not explicitly implement long multiplication}}
(2 raisedTo: 64) * (2 raisedTo: 64).
Smalltalk
(Note that arbitrary precision arithmetic is native in Smalltalk).
(2 raisedTo: 64) * (2 raisedTo: 64).
Tcl
{{works with|Tcl|8.5}} Tcl 8.5 supports arbitrary-precision integers, which improves math operations on large integers. It is easy to define our own by following rules for long multiplication; we can then check this against the built-in's result:
package require Tcl 8.5 proc longmult {x y} { set digits [lreverse [split $x ""]] set result {0} set j -2 foreach m [lreverse [split $y ""]] { set c 0 set i [incr j] foreach d $digits { set v [lindex $result [incr i]] if {$v eq ""} { lappend result 0 set v 0 } regexp (.)(.)$ 0[expr {$v + $c + $d*$m}] -> c v lset result $i $v } lappend result $c } # Reconvert digit list into a decimal number set result [string trimleft [join [lreverse $result] ""] 0] if {$result == ""} then {return 0} else {return $result} } puts [set n [expr {2**64}]] puts [longmult $n $n] puts [expr {$n * $n}]
outputs
18446744073709551616
340282366920938463463374607431768211456
340282366920938463463374607431768211456
UNIX Shell
In real shell scripts, I would use either bc or dc for this:
multiply() { echo "$1 $2 * p" | dc; }
But you can also do it with bash's built-in arithmetic:
add() { # arbitrary-precision addition local a="$1" b="$2" sum= carry=0 if (( ${#a} < ${#b} )); then local t="$a" a="$b" b="$t" fi while (( ${#a} )); do local -i d1="${a##${a%?}}" d2="10#0${b##${b%?}}" s=carry+d1+d2 sum="${s##${s%?}}$sum" carry="10#0${s%?}" a="${a%?}" b="${b%?}" done echo "$sum" } multiply() { # arbitrary-precision multiplication local a="$1" b="$2" product=0 if (( ${#a} < ${#b} )); then local t="$a" a="$b" b="$t" fi local zeroes= while (( ${#b} )); do local m1="$a" local m2="${b##${b%?}}" local partial=$zeroes local -i carry=0 while (( ${#m1} )); do local -i d="${m1##${m1%?}}" m1="${m1%?}" local -i p=d*m2+carry partial="${p##${p%?}}$partial" carry="10#0${p%?}" done partial="${carry#0}$partial" product="$(add "$product" "$partial")" zeroes=0$zeroes b="${b%?}" done echo "$product" }
Output is the same either way:
$ multiply 18446744073709551616 18446744073709551616
340282366920928463463374607431768211456
Ursala
Natural numbers of unlimited size are a built in type, and arithmetic operations on them are available as library functions. However, since the task calls for explicitly implementing long multiplication, here is an implementation using nothing but language primitives. The numbers are represented as lists of booleans, LSB first. The compiler already knows how to parse and display them in decimal.
successor = ~&a^?\1! ~&ah?/~&NfatPRC ~&NNXatPC
sum = ~&B^?a\~&Y@a ~&B?abh/successor@alh2fabt2RC ~&Yabh2Ofabt2RC
product = ~&alrB^& sum@NfalrtPXPRCarh2alPNQX
x = 18446744073709551616
#show+
y = %nP product@iiX x
output:
340282366920938463463374607431768211456
Vedit macro language
This example multiplies the value on current line with the value on next line and stores result on the 3rd line.
BOL
#11 = EOL_Pos-Cur_Pos
#12 = EOL_Pos-1
Line(1)
#21 = EOL_Pos-Cur_Pos
#22 = EOL_Pos-1
EOL Ins_Newline
Ins_Char('0', COUNT, #11+#21)
#32 = Cur_Pos-1
for (#2 = 0; #2 < #21; #2++) {
Goto_Pos(#22-#2) #5 = Cur_Char - '0'
for (#1 = 0; #1 < #11; #1++) {
Goto_Pos(#12-#1) #6 = Cur_Char - '0'
#7 = #5 * #6
#3 = #1 + #2
while (#7 > 0) {
Goto_Pos(#32-#3)
#7 += Cur_Char - '0'
Ins_Char(#7%10 + '0', OVERWRITE)
#3++
#7 = #7/10
}
}
}
Sample input and output:
18446744073709551616
18446744073709551616
0340282366920938463463374607431768211456
XPL0
include c:\cxpl\stdlib;
char Two64, Product(40);
[Two64:= "18446744073709551616";
StrNMul(Two64, Two64, Product, 20);
Product(39):= Product(39)!$80; \terminate string
Text(0, Product+1); \skip leading zero
]
Output:
340282366920938463463374607431768211456
zkl
[gnu] BigNums are supported via an extension library
var BN=Import("zklBigNum");
BN(2).pow(64) * BN(2).pow(64)
340282366920938463463374607431768211456
BN(2).pow(128) : "%,d".fmt(_)
340,282,366,920,938,463,463,374,607,431,768,211,456
//42!, also BN(42).factorial()
[2..42].reduce(fcn(p,n){p*n},BN(1)) : "%,d".fmt(_)
1,405,006,117,752,879,898,543,142,606,244,511,569,936,384,000,000,000
{{omit from|Erlang|Erlang has this built in}}