dim as Str15 s, c s = "Hello!" c = s print c

Output:

```txt

"Hello!"

Gambas

Note that the DIM statement is required in Gambas.

'''[https://gambas-playground.proko.eu/?gist=b88224f45b9b5be09eafdf069b059076 Click this link to run this code]'''

Public Sub main()
Dim src As String
Dim dst As String

src = "Hello"
dst = src

Print src
Print dst
End

GAP

#In GAP strings are lists of characters. An affectation simply copy references
a := "more";
b := a;
b{[1..4]} := "less";
a;
# "less"

# Here is a true copy
a := "more";
b := ShallowCopy(a);
b{[1..4]} := "less";
a;
# "more"

GML

src = "string";
dest = src;

Go

Just use assignment:

src := "Hello"
dst := src

Strings in Go are immutable. Because of this, there is no need to distinguish between copying the contents and making an additional reference. Technically, Go strings are immutable byte slices. A slice is an object that contains a reference to an underlying array. In the assignment shown above, a new slice object is created for dst. Its internal reference is likely to point to the same underlying array as src, but the language does not specify this behavior or make any guarantees about it.

Groovy

The dynamics of references and object creation are very much the same as in [[#Java|Java]]. However, the meaning of the equality (==) operator is different in Groovy, so we show those differences here, even though they are not relevant to the actual copying.

Example and counter-example:

def string = 'Scooby-doo-bee-doo'    // assigns string object to a variable reference
def stringRef = string               // assigns another variable reference to the same object
def stringCopy = new String(string)  // copies string value into a new object, and assigns to a third variable reference

Test Program:

assert string == stringRef           // they have equal values (like Java equals(), not like Java ==)
assert string.is(stringRef)          // they are references to the same objext (like Java ==)
assert string == stringCopy          // they have equal values
assert ! string.is(stringCopy)       // they are references to different objects (like Java !=)

'''Caveat Lector''': Strings are immutable objects in Groovy, so it is wasteful and utterly unnecessary to ever make copies of them within a Groovy program.

GUISS

Start.Programs,Accessories,Notepad,
Type:Hello world[pling],Highlight:Hello world[pling],
Menu,Edit,Copy,Menu,Edit,Paste

Haskell

In Haskell, every value is immutable, including ''String''s. So one never needs to copy them; references are shared.

HicEst

src = "Hello World"
dst = src

i

//Strings are immutable in 'i'.
software {
	a = "Hello World"
	b = a //This copies the string.

	a += "s"

	print(a)
	print(b)
}

=={{header|Icon}} and {{header|Unicon}}== Strings in Icon are immutable.

procedure main()
    a := "qwerty"
    b := a
    b[2+:4] := "uarterl"
    write(a," -> ",b)
end

Under the covers 'b' is created as a reference to the same string as 'a'; the sub-string assignment creates a new copy of the string. However, there is no way to tell this in the language. While most of the time this is transparent, programs that create very long strings through repeated concatenation need to avoid generating intermediate strings. Instead using a list and concatenating at the last minute can perform much better.

Note that strings are indicated using double quotes. However, single quotes are another type called character sets or csets.

J

src  =: 'hello'
dest =: src

J has copy-on-write semantics. So both src and dest are references to the same memory, until src changes, at which time dest retains a copy of the original value of src.

Java

In Java, Strings are immutable, so it doesn't make that much difference to copy it.

String src = "Hello";
String newAlias = src;
String strCopy = new String(src);

//"newAlias == src" is true
//"strCopy == src" is false
//"strCopy.equals(src)" is true

Instead, maybe you want to create a StringBuffer (mutable string) from an existing String or StringBuffer:

StringBuffer srcCopy = new StringBuffer("Hello");

JavaScript

Objects can be copied in JavaScript via simple reassignment. Changes to the properties of one will be reflected in the other:

var container = {myString: "Hello"};
var containerCopy = container; // Now both identifiers refer to the same object

containerCopy.myString = "Goodbye"; // container.myString will also return "Goodbye"

If you copy property values with reassignment, such as properties of the global object (window in browsers), only the value will be copied and not the reference

var a = "Hello";
var b = a; // Same as saying window.b = window.a

b = "Goodbye" // b contains a copy of a's value and a will still return "Hello"

jq

jq is a functional language and all data types, including strings, are immutable. If a string were to be copied (e.g. by exploding and imploding it), the resultant string would be equal in all respects to the original, and from the jq programmer's perspective, the two would be identical.

jq does however have a type of variable, though their values actually don't change -- they are just context-dependent. For example, consider the sequence of steps in the following function:

def demo:
  "abc" as $s    # assignment of a string to a variable
  | $s as $t     # $t points to the same string as $s
  | "def" as $s  # This $s shadows the previous $s
  | $t           # $t still points to "abc"
;

demo

{{Out}} "abc"

Joy

"hello" dup

Strings are immutable.

Julia

Strings are immutable in Julia. Assignment of one string valued variable to another is effectively a copy, as subsequent changes to either variable have no effect on the other.

s = "Rosetta Code"
t = s

println("s = \"", s, "\" and, after \"t = s\", t = \"", t, "\"")

s = "Julia at "*s

println("s = \"", s, "\" and, after this change, t = \"", t, "\"")

{{out}}

s = "Rosetta Code" and, after "t = s", t = "Rosetta Code"
s = "Julia at Rosetta Code" and, after this change, t = "Rosetta Code"

KonsolScript

Var:String str1 = "Hello";
Var:String str2 = str1;

Kotlin

val s = "Hello"
val alias = s      // alias === s
val copy = "" + s  // copy !== s

LabVIEW

In LabVIEW, one can simply wire an input to more than one output.
{{VI snippet}}
[[File:LabVIEW_Copy_a_string.png]]

Lang5

'hello dup

Lasso

While other datatypes like arrays require ->asCopy & ->asCopyDeep methods, assigning strings creates a copy, not a reference, as is seen below.

local(x = 'I saw a rhino!')
local(y = #x)

#x //I saw a rhino!
'\r'
#y //I saw a rhino!

'\r\r'
#x = 'I saw one too'
#x //I saw one too
'\r'
#y //I saw a rhino!

'\r\r'
#y = 'it was grey.'
#x //I saw one too
'\r'
#y //it was grey.

LC3 Assembly

Copying a string is the same as copying any other zero-terminated array. This program copies the string at SRC to COPY, then prints the copy to show it has worked.

        .ORIG      0x3000

        LEA        R1,SRC
        LEA        R2,COPY

LOOP    LDR        R3,R1,0
        STR        R3,R2,0
        BRZ        DONE
        ADD        R1,R1,1
        ADD        R2,R2,1
        BRNZP      LOOP

DONE    LEA        R0,COPY
        PUTS

        HALT

SRC     .STRINGZ   "What, has this thing appeared again tonight?"

COPY    .BLKW      128

        .END

{{out}}

What, has this thing appeared again tonight?

LFE

(let* ((a '"data assigned to a")
       (b a))
  (: io format '"Contents of 'b': ~s~n" (list b)))

{{out}}

Contents of 'b': data assigned to a

One can also use set to copy a sting when one is in the LFE REPL:

 (set a '"data")
"data"
> a
"data"
> (set b a)
"data"
> b
"data"

Liberty BASIC

src$ = "Hello"
dest$ = src$
print src$
print dest$

Lingo

str = "Hello world!"
str2 = str

Syntax-wise strings are not immuatable in Lingo. You can alter an existing string without new assignment:

put "X" before str
put "X" after str
put "X" into char 6 of str
put str
-- "XHellX world!X"

But memory-wise they are immutable: Lingo internally stores references to strings, and as soon as a string is altered, a new copy is created on the fly, so other references to the original string are not affected by the change.

Lisaac

+ scon : STRING_CONSTANT;
+ svar : STRING;

scon := "sample";
svar := STRING.create 20;
svar.copy scon;
svar.append "!\n";

svar.print;

STRING_CONSTANT is immutable, STRING is not.

Little

string a = "A string";
string b = a;
a =~ s/$/\./;
puts(a);
puts(b);

LiveCode

put "foo" into bar
put bar into baz
answer bar && baz

Copies are nearly always made, on function calls parameters may be passed by reference (pointer) by prepending @ to a parameter in the function definition, however this is the only case where it is usually performed.

Logo

As a functional language, words are normally treated as symbols and cannot be modified. The EQUAL? predicate compares contents instead of identity. In [[UCB Logo]] the .EQ predicate tests for "thing" identity.

make "a "foo
make "b "foo
print .eq :a :b   ; true, identical symbols are reused

make "c :a
print .eq :a :c   ; true, copy a reference

make "c word :b "||  ; force a copy of the contents of a word by appending the empty word
print equal? :b :c   ; true
print .eq :b :c     ; false

Lua

Lua strings are immutable, so only one reference to each string exists.

a = "string"
b = a
print(a == b) -->true
print(b) -->string

Maple

In Maple, you cannot really copy a string in the sense that there can be two copies of the string in memory. As soon as you create a second copy of a string that already exists, it get turned into a reference to the first copy. However, you can copy a reference to a string by a simple assignment statement.

> s := "some string";
                           s := "some string"

> t := "some string";
                           t := "some string"

> evalb( s = t ); # they are equal
                                  true

> addressof( s ) = addressof( t ); # not just equal data, but the same address in memory
                        3078334210 = 3078334210

> u := t: # copy reference

=={{header|Mathematica}} / {{header|Wolfram Language}}==

a="Hello World"
b=a

MATLAB

string1 = 'Hello';
string2 = string1;

Maxima

/* It's possible in Maxima to access individual characters by subscripts, but it's not the usual way.
Also, the result is "Lisp character", which cannot be used by other Maxima functions except cunlisp. The usual
way to access characters is charat, returning a "Maxima character" (actually a one characte string). With the latter,
it's impossible to modify a string in place, thus scopy is of little use. */

a: "loners"$
b: scopy(a)$
c: a$

c[2]: c[5]$

a;
"losers"

b;
"loners"

c;
"losers"

MAXScript

str1 = "Hello"
str2 = copy str1

Metafont

Metafont will always copy a string (does not make references).

string s, a;
s := "hello";
a := s;
s := s & " world";
message s;  % writes "hello world"
message a;  % writes "hello"
end

MiniScript

phrase = "hi"
copy = phrase
print phrase
print copy

MIPS Assembly

This does a full copy of the string, not just copying the pointer to the string's contents.

.data
	ex_msg_og: .asciiz "Original string:\n"
	ex_msg_cpy: .asciiz "\nCopied string:\n"
	string: .asciiz "Nice string you got there!\n"

.text
	main:
		la $v1,string #load addr of string into $v0
		la $t1,($v1)  #copy addr into $t0 for later access
		lb $a1,($v1)  #load byte from string addr
	strlen_loop:
		beqz $a1,alloc_mem
		addi $a0,$a0,1 #increment strlen_counter
		addi $v1,$v1,1 #increment ptr
		lb $a1,($v1)   #load the byte
		j strlen_loop

	alloc_mem:
		li $v0,9 #alloc memory, $a0 is arg for how many bytes to allocate
		         #result is stored in $v0
		syscall
		la $t0,($v0) #$v0 is static, $t0 is the moving ptr
		la $v1,($t1) #get a copy we can increment
	copy_str:
		lb $a1,($t1) #copy first byte from source

	strcopy_loop:
		beqz $a1,exit_procedure #check if current byte is NULL
		sb $a1,($t0)            #store the byte at the target pointer
		addi $t0,$t0,1          #increment source ptr
		addi $t1,$t1,1          #decrement source ptr
		lb $a1,($t1)            #load next byte from source ptr
		j strcopy_loop


	exit_procedure:
		la $a1,($v0) #store our string at $v0 so it doesn't get overwritten
		li $v0,4 #set syscall to PRINT

		la $a0,ex_msg_og  #PRINT("original string:")
		syscall

		la $a0,($v1)      #PRINT(original string)
		syscall

		la $a0,ex_msg_cpy #PRINT("copied string:")
		syscall

		la $a0,($a1)      #PRINT(strcopy)
		syscall

		li $v0,10         #EXIT(0)
		syscall

Mirah

src = "Hello"
new_alias = src

puts 'interned strings are equal' if src == new_alias

str_copy = String.new(src)
puts 'non-interned strings are not equal' if str_copy != src
puts 'compare strings with equals()' if str_copy.equals(src)

=={{header|Modula-3}}== Strings in Modula-3 have the type TEXT.

VAR src: TEXT := "Foo";
VAR dst: TEXT := src;

MUMPS

SET S1="Greetings, Planet" SET S2=S1

## Nanoquery


```nanoquery
$a = "Hello"
$b = $a

Neko

var src = "Hello"
var dst = src

Nemerle

Nemerle gives you the option of declaring a variable - even a string - as mutable, so the caveats of languages with only immutable strings don't necessarily apply. However, Nemerle binds the value of the string to the new name when copying; to sort of emulate copying a reference you can use lazy evaluation.

using System;
using System.Console;
using Nemerle;

module StrCopy
{
    Main() : void
    {
        mutable str1 = "I am not changed";      // str1 is bound to literal
        def str2 = lazy(str1);                  // str2 will be bound when evaluated
        def str3 = str1;                        // str3 is bound to value of str1
        str1 = "I am changed";                  // str1 is bound to new literal
        Write($"$(str1)\n$(str2)\n$(str3)\n");  // str2 is bound to value of str1
        // Output: I am changed
        //         I am changed
        //         I am not changed
    }
}

NetRexx

In addition to the string capabilities provided by the Java String libraries (see [[#Java|Java]] for some examples) NetRexx provides comprehensive string capabilities through the built-in Rexx type. Rexx strings can be copied by simple assignment; as follows:

/* NetRexx */
options replace format comments java crossref symbols nobinary

s1 = 'This is a Rexx string'
s2 = s1

s2 = s2.changestr(' ', '_')

say s1
say s2

In this example a string is created, the string is copied then the copy is modified with the changestr built-in function. Finally both strings are displayed to confirm that the original string wasn't modified by the call to changestr.

{{out}}

This is a Rexx string
This_is_a_Rexx_string

NewLISP

(define (assert f msg) (if (not f) (println msg)))

(setq  s "Greetings!"  c (copy s))
(reverse c) ; Modifies c in place.

(assert (= s c) "Strings not equal.")

; another way
; Nehal-Singhal 2018-05-25

> (setq a "abcd")
"abcd"
> (setq b a)
"abcd"
> b
"abcd"
> (= a b)
true

Nim

var
  c = "This is a string"
  d = c # Copy c into a new string

=={{header|Oberon-2}}==

MODULE CopyString;
TYPE
	String = ARRAY 128 OF CHAR;
VAR
	a,b: String;

BEGIN
	a := "plain string";
	COPY(a,b);
END CopyString.

Objeck

a := "GoodBye!";
b := a;

=={{header|Objective-C}}== Immutable strings - since they are immutable, you may get the same instance with its references count increased. Or, you can get a copy which is mutable if you use mutableCopy. Remember that both copy and mutableCopy return a retained instance. You can also get a copy by doing [NSString stringWithString:] or [[NSString alloc] initWithString:].

Note that both copy and initWithString:/stringWithString: are optimized to return the original string object (possibly retained) if it is immutable.

NSString *original = @"Literal String";
NSString *new = [original copy];
NSString *anotherNew = [NSString stringWithString:original];
NSString *newMutable = [original mutableCopy];

Mutable strings - you can get either new mutable (if you use mutableCopy) or immutable (if you use copy) string:

NSMutableString *original = [NSMutableString stringWithString:@"Literal String"];
NSString *immutable = [original copy];
NSString *anotherImmutable = [NSString stringWithString:original];
NSMutableString *mutable = [original mutableCopy];

Copying a CString into an NSString:

const char *cstring = "I'm a plain C string";
NSString *string = [NSString stringWithUTF8String:cstring];

Copying from data, possibly not null terminated:

char bytes[] = "some data";
NSString *string = [[NSString alloc] initWithBytes:bytes length:9 encoding:NSASCIIStringEncoding];

And of course, if a C string is needed, you can use standard functions like strcpy.

OCaml

## Octave


```octave>str2 = str1</lang



## Oforth

To make a copy of the reference, just dup the string

```Oforth
"abcde" dup

There is no need to copy a string content as strings are immutable. If really needed :

StringBuffer new "abcde" <<

OxygenBasic

string s, t="hello"
s=t

ooRexx

/* Rexx ***************************************************************
* 16.05.2013 Walter Pachl
**********************************************************************/

s1 = 'This is a Rexx string'
s2 = s1 /* does not copy the string */

Say 's1='s1
Say 's2='s2
i1=s1~identityhash; Say 's1~identityhash='i1
i2=s2~identityhash; Say 's2~identityhash='i2

s2 = s2~changestr('*', '*') /* creates a modified copy */

Say 's1='s1
Say 's2='s2
i1=s1~identityhash; Say 's1~identityhash='i1
i2=s2~identityhash; Say 's2~identityhash='i2

{{out}}

s1=This is a Rexx string
s2=This is a Rexx string
s1~identityhash=17587366586244
s2~identityhash=17587366586244
s1=This is a Rexx string
s2=This is a Rexx string
s1~identityhash=17587366586244
s2~identityhash=17587366588032

PARI/GP

Assignment in GP always copies.

In PARI, strings can be copied and references can be made.

```C
GEN string_copy = gcopy(string);
GEN string_ref = string;

Pascal

program in,out;

type

   pString = ^string;

var

   s1,s2 : string ;
   pStr  : pString ;

begin

   /* direct copy */
   s1 := 'Now is the time for all good men to come to the aid of their party.'
   s2 := s1 ;

   writeln(s1);
   writeln(s2);

   /* By Reference */
   pStr := @s1 ;
   writeln(pStr^);

   pStr := @s2 ;
   writeln(pStr^);

end;

Perl

To copy a string, just use ordinary assignment:

my $original = 'Hello.';
my $new = $original;
$new = 'Goodbye.';
print "$original\n";   # prints "Hello."

To create a reference to an existing string, so that modifying the referent changes the original string, use a backslash:

my $original = 'Hello.';
my $ref = \$original;
$$ref = 'Goodbye.';
print "$original\n";   # prints "Goodbye."

If you want a new name for the same string, so that you can modify it without dereferencing a reference, assign a reference to a typeglob:

my $original = 'Hello.';
our $alias;
local *alias = \$original;
$alias = 'Good evening.';
print "$original\n";   # prints "Good evening."

Note that our $alias, though in most cases a no-op, is necessary under stricture. Beware that local binds dynamically, so any subroutines called in this scope will see (and possibly modify!) the value of $alias assigned here.

To make a lexical variable that is an alias of some other variable, the [http://search.cpan.org/perldoc?Lexical::Alias Lexical::Alias] module can be used:

use Lexical::Alias;
my $original = 'Hello.';
my $alias;
alias $alias, $original;
$alias = 'Good evening.';
print "$original\n";   # prints "Good evening."

Perl 6

There is no special handling needed to copy a string; just assign it to a new variable:

my $original = 'Hello.';
my $copy = $original;
say $copy;            # prints "Hello."
$copy = 'Goodbye.';
say $copy;            # prints "Goodbye."
say $original;        # prints "Hello."

You can also bind a new variable to an existing one so that each refers to, and can modify the same string.

my $original = 'Hello.';
my $bound := $original;
say $bound;           # prints "Hello."
$bound = 'Goodbye.';
say $bound;           # prints "Goodbye."
say $original;        # prints "Goodbye."

Phix

Use of strings is utterly intuitive with no unexpected side effects. For example

string this = "feed"
string that = this     -- (that becomes "feed", this remains "feed")
that[2..3] = "oo"      -- (that becomes "food", this remains "feed")
this[1] = 'n'          -- (that remains "food", this becomes "need")
?{this,that}

{{out}}

{"need","food"}

Phix variables are reference counted (except for integers). When a simple copy is made, it increases the reference count and shares the data, making it very fast on large sequences and long strings. Attempts to modify any data with a reference count greater than one cause a copy to be made, and all other variables are left unchanged. Strings can be modified "in situ", no problem.

PHP

$src = "Hello";
$dst = $src;

PicoLisp

(setq Str1 "abcdef")
(setq Str2 Str1)                       # Create a reference to that symbol
(setq Str3 (name Str1))                # Create new symbol with name "abcdef"

Pike

int main(){
   string hi = "Hello World.";
   string ih = hi;
}

PL/I

   declare (s1, s2) character (20) varying;
   s1 = 'now is the time';
   s2 = s1;

Pop11

In Pop11 normal data are represented by references, so plain assignment will copy references. To copy data one has to use copy procedure:

vars src, dst;
'Hello' -> src;
copy(src) -> dst;

One can also combine assignment (initialization) with variable declarations:

vars src='Hello';
vars dst=copy(src);

PostScript

In PostScript,

(hello) dup length string copy

Prolog

Values in Prolog are immutable so unifying with a variable that already has the value of a string will effectively copy that string. You cannot reassign a value once it has been unified, it is not logical to have a value equal more than one thing.

?- A = "A test string", A = B.
A = B, B = "A test string".

PowerShell

Since PowerShell uses .NET behind the scenes and .NET strings are immutable you can simply assign the same string to another variable without breaking anything:

$str = "foo"
$dup = $str

To actually create a copy the Clone() method can be used:

$dup = $str.Clone()

PureBasic

src$ = "Hello"
dst$ = src$

ProDOS

editvar /newvar /value=a /userinput=1 /title=Enter a string to be copied:
editvar /newvar /value=b /userinput=1 /title=Enter current directory of the string:
editvar /newvar /value=c /userinput=1 /title=Enter file to copy to:
copy -a- from -b- to -c-

Python

{{works with|Python|2.3, 2.4, and 2.5}} Since strings are immutable, all copy operations return the same string. Probably the reference is increased.

 src = "hello"
>>> a = src
>>> b = src[:]
>>> import copy
>>> c = copy.copy(src)
>>> d = copy.deepcopy(src)
>>> src is a is b is c is d
True

To actually copy a string:

 a = 'hello'
>>> b = ''.join(a)
>>> a == b
True
>>> b is a  ### Might be True ... depends on "interning" implementation details!
False

As a result of object "interning" some strings such as the empty string and single character strings like 'a' may be references to the same object regardless of copying. This can potentially happen with any Python immutable object and should be of no consequence to any proper code.

Be careful with ''is'' - use it only when you want to compare the identity of the object. To compare string values, use the ''=='' operator. For numbers and strings any given Python interpreter's implementation of "interning" may cause the object identities to coincide. Thus any number of names to identical numbers or strings might become references to the same objects regardless of how those objects were derived (even if the contents were properly "copied" around). The fact that these are immutable objects makes this a reasonable behavior.

R

Copy a string by value:

str1 <- "abc"
str2 <- str1

Racket

#lang racket

(let* ([s1 "Hey"]
       [s2 s1]
       [s3 (string-copy s1)]
       [s4 s3])
  (printf "s1 and s2 refer to ~a strings\n"
          (if (eq? s1 s2) "the same" "different")) ; same
  (printf "s1 and s3 refer to ~a strings\n"
          (if (eq? s1 s3) "the same" "different")) ; different
  (printf "s3 and s4 refer to ~a strings\n"
          (if (eq? s3 s4) "the same" "different")) ; same
  (string-fill! s3 #\!)
  (printf "~a~a~a~a\n" s1 s2 s3 s4)) ; outputs "HeyHey!!!!!!"

Raven

Copy a string by reference:

'abc' as a
a as b

Copy a string by value:

'abc' as a
a copy as b

REBOL

REBOL [
    Title: "String Copy"
    URL: http://rosettacode.org/wiki/Copy_a_string
]

x: y: "Testing."
y/2: #"X"
print ["Both variables reference same string:" mold x "," mold y]

x: "Slackeriffic!"
print ["Now reference different strings:" mold x "," mold y]

y: copy x        ; String copy here!
y/3: #"X"        ; Modify string.
print ["x copied to y, then modified:" mold x "," mold y]

y: copy/part x 7 ; Copy only the first part of y to x.
print ["Partial copy:" mold x "," mold y]

y: copy/part  skip x 2  3
print ["Partial copy from offset:" mold x "," mold y]

{{out}}

Script: "String Copy" (16-Dec-2009)
Both variables reference same string: "TXsting." , "TXsting."
Now reference different strings: "Slackeriffic!" , "TXsting."
x copied to y, then modified: "Slackeriffic!" , "SlXckeriffic!"
Partial copy: "Slackeriffic!" , "Slacker"
Partial copy from offset: "Slackeriffic!" , "ack"

Red

Red[]
originalString: "hello wordl"
copiedString: originalString
; OR
copiedString2: copy originalString

Retro

"this is a string" dup tempString

REXX

The example shows how to copy the contents of one string into another string.

Note that delimiters for literal strings, REXX accepts either of: ::* ''' ' ''' (an apostrophe) ::* ''' " ''' (a double quote)

Also note that ''all'' REXX values (variables) are stored as (varying length) ''character strings''.

src = "this is a string"
dst = src

RLaB

 s1 = "A string"
A string
>> s2 = s1
A string

Ring

cStr1 = "Hello!"   # create original string
cStr2 = cStr1      # make new string from original

Robotic

set "$string1" to "This is a string"
set "$string2" to "$string1"
* "&$string2&"

Ruby

In Ruby, String are mutable.

original = "hello"
reference = original          # copies reference
copy1 = original.dup          # instance of original.class
copy2 = String.new(original)  # instance of String

original << " world!"         # append
p reference                   #=> "hello world!"
p copy1                       #=> "hello"
p copy2                       #=> "hello"

There is a method of Object#clone, too, in the copy of the object.

original = "hello".freeze     # prevents further modifications
copy1 = original.dup          # copies contents (without status)
copy2 = original.clone        # copies contents (with status)
p copy1.frozen?               #=> false
p copy1 << " world!"          #=> "hello world!"
p copy2.frozen?               #=> true
p copy2 << " world!"          #=> can't modify frozen String (RuntimeError)

Run BASIC

origString$ = "Hello!"     ' create original string
newString$  = origString$  ' make new strig from original

Rust

fn main() {
    let s1 = "A String";
    let mut s2 = s1;

    s2 = "Another String";

    println!("s1 = {}, s2 = {}", s1, s2);
}

Output: s1 = A String, s2 = Another String

## Sather


```sather
class MAIN is
  main is
    s  ::= "a string";
    s1 ::= s;
    -- s1 is a copy
  end;
end;

Scala

  val src = "Hello"
  // Its actually not a copy but a reference
  // That is not a problem because String is immutable
  // In fact its a feature
  val des = src
  assert(src eq des) // Proves the same reference is used.
  // To make a real copy makes no sense.
  // Actually its hard to make a copy, the compiler is too smart.
  // mkString, toString makes also not a real copy
  val cop = src.mkString.toString
  assert((src eq cop))                 // Still no copyed image
  val copy = src.reverse.reverse       // Finally double reverse makes a copy
  assert(src == copy && !(src eq copy))// Prove, but it really makes no sense.

Scheme

(define dst (string-copy src))

Seed7

var string: dest is "";

dest := "Hello";

Shiny

src: 'hello'
cpy: src

Sidef

var original = "hello";               # new String object
var reference = original;             # points at the original object
var copy1 = String.new(original);     # creates a new String object
var copy2 = original+'';              # ==//==

Slate

[ | :s | s == s copy] applyTo: {'hello'}. "returns False"

Smalltalk

|s1 s2|
"bind the var s1 to the object string on the right"
s1 := 'i am a string'.
"bind the var s2 to the same object..."
s2 := s1.
"bind s2 to a copy of the object bound to s1"
s2 := (s1 copy).

SNOBOL4

* copy a to b
          b = a = "test"
          output = a
          output = b
* change the copy
          b "t" = "T"
          output = b
end

{{out}}

  test
  test
  Test

Standard ML

In Standard ML, strings are immutable, so you don't copy it.

Instead, maybe you want to create a CharArray.array (mutable string) from an existing string:

val src = "Hello";
val srcCopy = CharArray.array (size src, #"x"); (* 'x' is just dummy character *)
CharArray.copyVec {src = src, dst = srcCopy, di = 0};
src = CharArray.vector srcCopy; (* evaluates to true *)

or from another CharArray.array:

val srcCopy2 = CharArray.array (CharArray.length srcCopy, #"x"); (* 'x' is just dummy character *)
CharArray.copy {src = srcCopy, dst = srcCopy2, di = 0};

Swift

Just use assignment:

var src = "Hello"
var dst = src

Strings in Swift are value types, so assigning copies the string.

Tcl

set src "Rosetta Code"
set dst $src

Tcl copies strings internally when needed. To be exact, it uses a basic value model based on simple objects that are immutable when shared (i.e., when they have more than one effective reference to them); when unshared, they can be changed because the only holder of a reference has to be the code requesting the change. At the script level, this looks like Tcl is making a copy when the variable is assigned as above, but is more efficient in the common case where a value is not actually modified.

=={{header|TI-83 BASIC}}==

:"Rosetta Code"→Str1
:Str1→Str2

=={{header|TI-89 BASIC}}==

:"Rosetta Code"→str1
:str1→str2

Toka

" hello" is-data a
a string.clone is-data b

Trith

Strings are immutable character sequences, so copying a string just means duplicating the reference at the top of the stack:

"Hello" dup

TUSCRIPT

$$ MODE TUSCRIPT
str="Hello"
dst=str

UNIX Shell

foo="Hello"
bar=$foo    # This is a copy of the string

Ursa

decl string a b
set a "hello"
set b a

V

dup really makes a reference, but the language is functional, so the string is immutable.

"hello" dup

VBA

This program copies string in variable a to variable b. Mutating variable a subsequently doesn't alter variable b. Variable b is not a reference.

Sub copystring()
    a = "Hello World!"
    b = a
    a = "I'm gone"
    Debug.Print b
    Debug.Print a
End Sub

{{out}}

Hello World!
I'm gone

Vim Script

let str1 = "original string"
let str2 = str1
let str1 = "new string"

echo "String 1:" str1
echo "String 2:" str2

{{Out}}

String 1: new string
String 2: original string

Visual Basic .NET

'''Platform:''' [[.NET]]

{{works with|Visual Basic .NET|9.0+}}

'Immutable Strings
Dim a = "Test string"
Dim b = a 'reference to same string
Dim c = New String(a.ToCharArray) 'new string, normally not used

'Mutable Strings
Dim x As New Text.StringBuilder("Test string")
Dim y = x 'reference
Dim z = New Text.StringBuilder(x.ToString) 'new string

Alternatively, you can use, with all versions of the .NET framework:

Dim a As String = "Test String"
Dim b As String = String.Copy(a) ' New string

XPL0

The default method of terminating strings is to set the most significant bit of the last character. An alternative is to use the 'string 0' command to specify zero-terminated strings. The string copy routine from the standard library is shown.

proc StrCopy(A, B);     \Copy string: A --> B
char A, B;              \Strings: B must already have enough space "Reserved"
int  I;                 \Beware if strings overlap
for I:= 0 to -1>>1-1 do
    [B(I):= A(I);
    if A(I) >= $80 then return
    ];

char S1, S2, S3(13);
[S1:= "Hello, world!";  \S1 now points to the string
S2:= S1;                \S2 now also points to the string
StrCopy(S1, S3);        \S3 points to a separate copy of the string
]

zkl

Strings are immutable so copy is just return the string:

"abc".copy()  // noop

zonnon

module Main;
var
	s,r: string;
	c: array 60 of char;
begin
	s := "plain string";r := s; writeln(s);
	(* copy string to array of char *)
	copy(s,c);c[0] := 'P';
	(* copy array of char to string *)
	copy(c,r);writeln(r);
end Main.

ZX Spectrum Basic

10 LET a$ = "Hello": REM a$ is the original string
20 LET b$ = a$: REM b$ is the copy