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set final_string to ""
set final_string to ""
repeat 5 times
repeat 5 times
set final_string to final_string & str
set final_string to final_string & str
end repeat</lang>
end repeat</lang>


For larger numbers of repetitions, however, it proves significantly faster to progressively double a copy of the original string (concatenating it with itself). Intermediate stages of doubling are appended to an accumulator wherever required for binary composition of the target number.

See the technique of 'Egyptian Multiplication' described in the Rhind Mathematical Papyrus at the British Museum.

{{trans|JavaScript}}
<lang AppleScript>on run
nreps("ha", 50000)
end run


-- String -> Int -> String
on nreps(s, n)
set o to ""
if n < 1 then return o
repeat while (n > 1)
if (n mod 2) > 0 then set o to o & s
set n to (n div 2)
set s to (s & s)
end repeat
return o & s
end nreps
</lang>


=={{header|Applesoft BASIC}}==
=={{header|Applesoft BASIC}}==

Revision as of 21:15, 9 March 2016

Task
Repeat a string
You are encouraged to solve this task according to the task description, using any language you may know.

Take a string and repeat it some number of times. Example: repeat("ha", 5) => "hahahahaha"

If there is a simpler/more efficient way to repeat a single “character” (i.e. creating a string filled with a certain character), you might want to show that as well (i.e. repeat-char("*", 5) => "*****").

4DOS Batch

<lang 4dos>gosub repeat ha 5 echo %@repeat[*,5] quit

Repeat [String Times]
   do %Times%
       echos %String%
   enddo
   echo.

return</lang> Output shows: 

hahahahaha
*****

8th

<lang forth>"ha" 5 s:* . cr</lang> Output shows: 

hahahahaha

ActionScript

ActionScript does not have a built-in way to repeat a string multiple times, but the addition operator can be used to concatenate strings.

In Flex, there is the method mx.utils.StringUtil.repeat().

Iterative version

<lang ActionScript>function repeatString(string:String, numTimes:uint):String { var output:String = ""; for(var i:uint = 0; i < numTimes; i++) output += string; return output; }</lang>

Recursive version

The following double-and-add method is much faster when repeating a string many times. <lang ActionScript>function repeatRecursive(string:String, numTimes:uint):String { if(numTimes == 0) return ""; if(numTimes & 1) return string + repeatRecursive(string, numTimes - 1); var tmp:String = repeatRecursive(string, numTimes/2); return tmp + tmp; }</lang>

Flex

<lang ActionScript>import mx.utils.StringUtil; trace(StringUtil.repeat("ha", 5)); </lang> Sample Output: 

hahahahaha

Ada

In Ada multiplication of an universal integer to string gives the desired result. Here is an example of use: <lang Ada>with Ada.Strings.Fixed; use Ada.Strings.Fixed; with Ada.Text_IO; use Ada.Text_IO;

procedure String_Multiplication is begin 

  Put_Line (5 * "ha");

end String_Multiplication;</lang> Sample output: 

hahahahaha

ALGOL 68

<lang algol68>print (5 * "ha") </lang>


AppleScript

<lang AppleScript>set str to "ha" set final_string to "" repeat 5 times 

   set final_string to final_string & str

end repeat</lang>


For larger numbers of repetitions, however, it proves significantly faster to progressively double a copy of the original string (concatenating it with itself). Intermediate stages of doubling are appended to an accumulator wherever required for binary composition of the target number.

See the technique of 'Egyptian Multiplication' described in the Rhind Mathematical Papyrus at the British Museum.

Translation of: JavaScript

<lang AppleScript>on run 

   nreps("ha", 50000)

end run


-- String -> Int -> String on nreps(s, n) 

   set o to ""
   if n < 1 then return o
   
   repeat while (n > 1)
       if (n mod 2) > 0 then set o to o & s
       set n to (n div 2)
       set s to (s & s)
   end repeat
   return o & s

end nreps </lang>

Applesoft BASIC

<lang ApplesoftBasic>FOR I = 1 TO 5 : S$ = S$ + "HA" : NEXT

? "X" SPC(20) "X"</lang> Output: 

X                    X

ATS

<lang ATS> // // How to compile: // patscc -DATS_MEMALLOC_LIBC -o string_repeat string_repeat.dats //

  1. include

"share/atspre_staload.hats"

fun string_repeat ( 

 x: string, n: intGte(0)

) : Strptr1 = res where { 

 val xs =
   list_make_elt<string>(n, x)
 val res = stringlst_concat($UNSAFE.list_vt2t(xs))
 val ((*freed*)) = list_vt_free(xs)

} (* end of [string_repeat] *)

(* ****** ****** *)

implement main0 () = let // val ha5 = string_repeat("ha", 5) val ((*void*)) = println! ("ha5 = \"", ha5, "\"") val ((*freed*)) = strptr_free (ha5) // in 

 // nothing

end // end of [main0] </lang>

AutoHotkey

<lang AutoHotkey>MsgBox % Repeat("ha",5)

Repeat(String,Times) { 

 Loop, %Times%
   Output .= String
 Return Output

}</lang>

AutoIt

<lang AutoIt>#include <String.au3>

ConsoleWrite(_StringRepeat("ha", 5) & @CRLF)</lang>

AWK

<lang awk>function repeat( str, n, rep, i ) { 

   for( ; i<n; i++ )
       rep = rep str   
   return rep

}

BEGIN { 

   print repeat( "ha", 5 )

}</lang>

Babel

<lang babel>main: { "ha" 5 print_repeat }

print_repeat!: { <- { dup << } -> times }</lang> Outputs: <lang babel>hahahahaha</lang> The '<<' operator prints, 'dup' duplicates the top-of-stack, 'times' does something x number of times. The arrows mean down (<-) and up (->) respectively - it would require a lengthy description to explain what this means, refer to the doc/babel_ref.txt file in the github repo linked from Babel

Batch File

Commandline implementation <lang dos>@echo off if "%2" equ "" goto fail setlocal enabledelayedexpansion set char=%1 set num=%2 for /l %%i in (1,1,%num%) do set res=!res!%char% echo %res%

fail</lang>

'Function' version <lang dos>@echo off set /p a=Enter string to repeat : set /p b=Enter how many times to repeat : set "c=1" set "d=%b%"

a

echo %a% set "c=%c%+=1" if /i _"%c%"==_"%d%" (exit /b) goto :a</lang>

'Function' version 2 <lang dos>@echo off @FOR /L %%i in (0,1,9) DO @CALL :REPEAT %%i @echo That's it! @FOR /L %%i in (0,1,9) DO @CALL :REPEAT %%i @echo. @echo And that! @GOTO END

REPEAT

@echo|set /p="*" @GOTO:EOF

END</lang>

BBC BASIC

<lang bbcbasic> PRINT STRING$(5, "ha")</lang>

Befunge

<lang Befunge>v> ">:#,_v >29*+00p>~:"0"- #v_v $ 

v      ^p0p00:-1g00< $            >
v    p00&p0-1g00+4*65< >00g1-:00p#^_@</lang>

Input sample: 

ha05

Input string has to be zero terminated and less than 18 characters.

Output sample: 

hahahahaha

Bracmat

The code almost explains itself. The repetions are accumulated in a list rep. The str concatenates all elements into a single string, ignoring the white spaces separating the elements.

<lang bracmat>(repeat= 

 string N rep

. !arg:(?string.?N) 

 & !string:?rep
 &   whl
   ' (!N+-1:>0:?N&!string !rep:?rep)
 & str$!rep

);</lang> 

 repeat$(ha.5)
 hahahahaha

Brainf***

Prints "ha" 10 times. Note that this method only works for a number of repetitions that fit into the cell size. <lang bf>+++++ +++++ init first as 10 counter [-> +++++ +++++<] we add 10 to second each loopround

Now we want to loop 5 times to follow std +++++ [-> ++++ . ----- -- . +++<] print h and a each loop

and a newline because I'm kind and it looks good +++++ +++++ +++ . --- .</lang>

Brat

<lang brat>p "ha" * 5 #Prints "hahahahaha"</lang>

Burlesque

<lang burlesque> blsq ) 'h5?* "hhhhh" blsq ) "ha"5.*\[ "hahahahaha" </lang>

C

<lang c>#include <stdio.h>

  1. include <stdlib.h>
  2. include <string.h>

char * string_repeat( int n, const char * s ) { 

 size_t slen = strlen(s);
 char * dest = malloc(n*slen+1);

 int i; char * p;
 for ( i=0, p = dest; i < n; ++i, p += slen ) {
   memcpy(p, s, slen);
 }
 *p = '\0';
 return dest;

}

int main() { 

 char * result = string_repeat(5, "ha");
 puts(result);
 free(result);
 return 0;

}</lang> A variation. <lang c>... char *string_repeat(const char *str, int n) { 

  char *pa, *pb;
  size_t slen = strlen(str);
  char *dest = malloc(n*slen+1);

  pa = dest + (n-1)*slen;
  strcpy(pa, str);
  pb = --pa + slen; 
  while (pa>=dest) *pa-- = *pb--;
  return dest;

}</lang>

To repeat a single character <lang c>#include <stdio.h>

  1. include <stdlib.h>
  2. include <string.h>

char * char_repeat( int n, char c ) { 

 char * dest = malloc(n+1);
 memset(dest, c, n);
 dest[n] = '\0';
 return dest;

}

int main() { 

 char * result = char_repeat(5, '*');
 puts(result);
 free(result);
 return 0;

}</lang>

If you use GLib, simply use g_strnfill ( gsize length, gchar fill_char ) function.

C#

<lang csharp>string s = "".PadLeft(5, 'X').Replace("X", "ha");</lang> or (with .NET 2+) <lang csharp>string s = new String('X', 5).Replace("X", "ha");</lang> or (with .NET 2+) <lang csharp>string s = String.Join("ha", new string[5 + 1]);</lang> or (with .NET 4+) <lang csharp>string s = String.Concat(Enumerable.Repeat("ha", 5));</lang>

To repeat a single character: <lang csharp>string s = "".PadLeft(5, '*');</lang> or (with .NET 2+) <lang csharp>string s = new String('*', 5);</lang>

C++

<lang cpp>#include <string>

  1. include <iostream>

std::string repeat( const std::string &word, int times ) { 

  std::string result ;
  result.reserve(times*word.length()); // avoid repeated reallocation
  for ( int a = 0 ; a < times ; a++ ) 
     result += word ;
  return result ;

}

int main( ) { 

  std::cout << repeat( "Ha" , 5 ) << std::endl ;
  return 0 ;

}</lang>

To repeat a single character: <lang cpp>#include <string>

  1. include <iostream>

int main( ) { 

  std::cout << std::string( 5, '*' ) << std::endl ;
  return 0 ;

}</lang>

Ceylon

<lang ceylon>shared void repeatAString() { print("ha".repeat(5)); }</lang>

Clipper

Also works with Harbour Project compiler Harbour 3.0.0 (Rev. 16951) <lang visualfoxpro> Replicate( "Ha", 5 )</lang>

Clojure

<lang lisp>(apply str (repeat 5 "ha"))</lang>

ColdFusion

<lang cfm> <cfset word = 'ha'> <Cfset n = 5> <Cfoutput> <Cfloop from="1" to="#n#" index="i">#word#</Cfloop> </Cfoutput> </lang>

Common Lisp

<lang lisp>(defun repeat-string (n string) 

 (with-output-to-string (stream)
   (loop repeat n do (write-string string stream))))</lang>

A version which allocates the result string in one step:

<lang lisp>(defun repeat-string (n string 

                    &aux
                    (len (length string))
                    (result (make-string (* n len)
                                         :element-type (array-element-type string))))
 (loop repeat n
       for i from 0 by len
       do (setf (subseq result i (+ i len)) string))
 result)</lang>


For those who love one-liners, even at the expense of readability: <lang lisp>(defun repeat-string (n string) 

 (format nil "~V@{~a~:*~}" n string))</lang>


<lang lisp>(princ (repeat-string 5 "hi"))</lang>


A single character may be repeated using just the builtin make-string: <lang lisp>(make-string 5 :initial-element #\X)</lang> produces “XXXXX”.

D

Repeating a string: <lang d>import std.stdio, std.array;

void main() { 

   writeln("ha".replicate(5));

}</lang> Repeating a character with vector operations: <lang d>import std.stdio;

void main() { 

   char[] chars;     // create the dynamic array
   chars.length = 5; // set the length
   chars[] = '*';    // set all characters in the string to '*'
   writeln(chars);

}</lang>

DCL

Not exactly what the task asks for but at least it is something; <lang DCL>$ write sys$output f$fao( "!AS!-!AS!-!AS!-!AS!-!AS", "ha" ) $ write sys$output f$fao( "!12*d" )</lang>

Output:
$ @repeat_a_string_and_then_character
hahahahaha
dddddddddddd

Delphi

Repeat a string <lang Delphi> function RepeatString(const s: string; count: cardinal): string; var 

 i: Integer;

begin 

 for i := 1 to count do
   Result := Result + s;

end;

Writeln(RepeatString('ha',5)); </lang>

Repeat a character

<lang Delphi> Writeln( StringOfChar('a',5) ); </lang>

Using recursion

<lang Delphi> function RepeatStr(const s: string; i: Cardinal): string; begin 

 if i = 0 then
   result := 
 else
  result := s + RepeatStr(s, i-1)

end; </lang>

Built in RTL function:

<lang Delphi>StrUtils.DupeString</lang>

Déjà Vu

<lang dejavu>!. concat( rep 5 "ha" )</lang>

Output:
"hahahahaha"

DWScript

Repeat a string

<lang Delphi> PrintLn( StringOfString('abc',5) ); </lang>

Repeat a character

<lang Delphi> PrintLn( StringOfChar('a',5) ); </lang>

E

<lang e>"ha" * 5</lang>

ECL

After version 4.2.2 <lang>IMPORT STD; //Imports the Standard Library

STRING MyBaseString := 'abc'; RepeatedString := STD.Str.Repeat(MyBaseString,3); RepeatedString; //returns 'abcabcabc'</lang>

Before version 4.2.2 <lang>RepeatString(STRING InStr, INTEGER Cnt) := FUNCTION 

 rec := {STRING Str};
 ds  := DATASET(Cnt,TRANSFORM(rec,SELF.Str := InStr));
 res := ITERATE(ds,TRANSFORM(rec,SELF.Str := LEFT.Str + RIGHT.Str));
 RETURN Res[Cnt].Str;

END;

RepeatString('ha',3); RepeatString('Who',2);</lang>

Egison

<lang egison> (S.concat (take 5 (repeat1 "ha"))) </lang>

Eiffel

<lang eiffel> 

repeat_string(a_string: STRING; times: INTEGER): STRING
require
  times_positive: times > 0
do
  Result := a_string.multiply(times)
end

</lang>

Elena

<lang elena>#import system.

  1. import system'routines.
  2. import extensions.
  1. symbol program =

[ 

   #var s := 0 repeat &till:5 &each: n [ "ha" ] summarize:(String new) literal.

]. </lang>

Elixir

<lang elixir> String.duplicate("ha", 5) </lang>

Emacs Lisp

Going via a list to repeat the desired string,

<lang lisp>(apply 'concat (make-list 5 "ha"))</lang>

A single character can be repeated with make-string

<lang lisp>(make-string 5 ?x)</lang>

With cl.el the loop macro can repeat and concatenate,

<lang lisp>(require 'cl) (loop repeat 5 concat "ha")</lang>

Erlang

<lang erlang>repeat(X,N) -> 

   lists:flatten(lists:duplicate(N,X)).</lang>

This will duplicate a string or character N times to produce a new string.

ERRE

<lang ERRE> PROCEDURE REPEAT_STRING(S$,N%->REP$) 

  LOCAL I%
  REP$=""
  FOR I%=1 TO N% DO
      REP$=REP$+S$
  END FOR

END PROCEDURE </lang> Note: If N% is less than 1, the result is the empty string "".If S$ is a one-character string you can use the predefined function STRING$ as REP$=STRING$(S$,N%).

Euphoria

<lang Euphoria>function repeat_string(object x, integer times) 

   sequence out
   if atom(x) then
       return repeat(x,times)
   else
       out = ""
       for n = 1 to times do
           out &= x
       end for
       return out
   end if

end function

puts(1,repeat_string("ha",5) & '\n') -- hahahahaha

puts(1,repeat_string('*',5) & '\n') -- *****</lang> Sample Output: 

hahahahaha
*****

<lang Euphoria>-- Here is an alternative method for "Repeat a string" include std/sequence.e printf(1,"Here is the repeated string: %s\n", {repeat_pattern("ha",5)}) printf(1,"Here is another: %s\n", {repeat_pattern("*",5)}) </lang> Sample Output: 

Here is the repeated string: hahahahaha
Here is another: *****

F#

<lang fsharp>> String.replicate 5 "ha";; val it : string = "hahahahaha"</lang> Or <lang fsharp>> String.Concat( Array.create 5 "ha" );; val it : string = "hahahahaha"</lang>

Factor

<lang factor>: repeat-string ( str n -- str' ) swap <repetition> concat ;

"ha" 5 repeat-string print</lang>

Forth

<lang forth>: place-n { src len dest n -- } 

 0 dest c!
 n 0 ?do src len dest +place loop ;

s" ha" pad 5 place-n pad count type \ hahahahaha</lang> The same code without the use of locals: <lang forth>

place-n ( src len dest n -- )
 swap >r 0 r@ c!
 begin dup while -rot 2dup r@ +place rot 1- repeat
 r> 2drop 2drop ;

s" ha" pad 5 place-n pad count type \ hahahahaha</lang> Filling a string with a single character is supported by ANS-Forth: <lang forth>pad 10 char * fill \ repeat a single character pad 10 type \ **********</lang>

Fortran

Works with: Fortran version 90 and later

<lang fortran>program test_repeat 

 write (*, '(a)') repeat ('ha', 5)

end program test_repeat</lang> Output: 

hahahahaha

Frink

<lang frink> println[repeat["ha", 5]] </lang>

GAP

<lang gap>Concatenation(ListWithIdenticalEntries(10, "BOB ")); "BOB BOB BOB BOB BOB BOB BOB BOB BOB BOB "</lang>

Go

<lang go>fmt.Println(strings.Repeat("ha", 5)) // ==> "hahahahaha"</lang> There is no special way to repeat a single character, other than to convert the character to a string. The following works: <lang go>fmt.Println(strings.Repeat(string('h'), 5)) // prints hhhhh</lang>

Groovy

<lang groovy> println 'ha' * 5</lang>

Harbour

<lang visualfoxpro>? Replicate( "Ha", 5 )</lang>

Haskell

For a string of finite length: <lang haskell>concat $ replicate 5 "ha"</lang>

Or with list-monad (a bit obscure): <lang haskell>[1..5] >> "ha"</lang>

Or with Control.Applicative: <lang haskell>[1..5] *> "ha"</lang>

For an infinitely long string: <lang haskell>cycle "ha"</lang>

To repeat a single character: <lang haskell>replicate 5 '*'</lang>

HicEst

<lang HicEst>CHARACTER out*20

EDIT(Text=out, Insert="ha", DO=5)</lang>

Icon and Unicon

The procedure repl is a supplied function in Icon and Unicon. <lang Icon>procedure main(args) 

   write(repl(integer(!args) | 5))

end</lang> If it weren't, one way to write it is: <lang Icon>procedure repl(s, n) 

   every (ns := "") ||:= |s\(0 <= n)
   return ns

end</lang>

Idris

<lang Idris>strRepeat : Nat -> String -> String strRepeat Z s = "" strRepeat (S n) s = s ++ strRepeat n s

chrRepeat : Nat -> Char -> String chrRepeat Z c = "" chrRepeat (S n) c = strCons c $ chrRepeat n c</lang>

Inform 7

<lang inform7>Home is a room.

To decide which indexed text is (T - indexed text) repeated (N - number) times: let temp be indexed text; repeat with M running from 1 to N: let temp be "[temp][T]"; decide on temp.

When play begins: say "ha" repeated 5 times; end the story.</lang>

J

<lang j> 5 # '*' NB. repeat each item 5 times 

  5 # 'ha'              NB. repeat each item 5 times

hhhhhaaaaa 

  5 ((* #) $ ]) 'ha'    NB. repeat array 5 times

hahahahaha 

  5 ;@# < 'ha'          NB. boxing is used to treat the array as a whole

hahahahaha</lang>

Java

Works with: Java version 1.5+

There's no method or operator to do this in Java, so you have to do it yourself. <lang java5>public static String repeat(String str, int times){ 

  StringBuilder ret = new StringBuilder();
  for(int i = 0;i < times;i++) ret.append(str);
  return ret.toString();

}

public static void main(String[] args){ 

 System.out.println(repeat("ha", 5));

}</lang>

Or even shorter: <lang java5>public static String repeat(String str, int times){ 

  return new String(new char[times]).replace("\0", str);

}</lang>

In Apache Commons Lang, there is a StringUtils.repeat() method.

JavaScript

This solution creates an empty array of length n+1, then uses the array's join method to effectively concatenate the string n times. Note that extending the prototype of built-in objects is not a good idea if the code is to run in a shared workspace. <lang javascript>String.prototype.repeat = function(n) { 

   return new Array(1 + (n || 0)).join(this);

}

console.log("ha".repeat(5)); // hahahahaha</lang>

As of ES6, `repeat` is built in, so this can be written as:

<lang javascript> console.log("ha".repeat(5)); // hahahahaha</lang>

For larger numbers of repetitions, however, it proves significantly faster to progressively double a copy of the original string (concatenating it with itself). Intermediate stages of doubling are appended to an accumulator wherever required for binary composition of the target number.

See the technique of 'Egyptian Multiplication' described in the Rhind Mathematical Papyrus at the British Museum.

<lang javascript>function nreps(s, n) { 

 var o = ;
 if (n < 1) return o;
 while (n > 1) {
   if (n & 1) o += s;
   n >>= 1;
   s += s;
 }
 return o + s;

}

nreps('ha', 50000);</lang>

jq

<lang jq>"a " * 3' # => "a a a "</lang>

Note that if the integer multiplicand is 0, then the result is null.

Julia

<lang julia>"ha"^5 '*'^5

  1. the (^) operator is really just call to the `repeat` function

repeat("ha",5)</lang>

K

<lang k> 

 ,/5#,"ha"

"hahahahaha" 

 5#"*"

"*****" </lang>

Kotlin

<lang Kotlin> fun main(args: Array<String>) { 

   println("ha".repeat(5))

} </lang>

LabVIEW

I don't know if there is a built-in function for this, but it is easily achieved with a For loop and Concatenate Strings.

Lasso

<lang Lasso>'ha'*5 // hahahahaha</lang>

<lang Lasso>loop(5) => {^ 'ha' ^} // hahahahaha</lang>

LFE

<lang lisp> (string:copies '"ha" 5) </lang>

Liberty BASIC

<lang lb>a$ ="ha " print StringRepeat$( a$, 5)

end

function StringRepeat$( in$, n) 

   o$ =""
   for i =1 to n
       o$ =o$ +in$
   next i
   StringRepeat$ =o$

end function</lang>

LiveCode

<lang liveCode>on mouseUp 

   put repeatString("ha", 5)

end mouseUp

function repeatString str n 

   repeat n times
       put str after t
   end repeat
   return t

end repeatString</lang>

<lang logo>to copies :n :thing [:acc "||] 

 if :n = 0 [output :acc]
 output (copies :n-1 :thing combine :acc :thing)

end</lang> or using cascade: <lang logo>show cascade 5 [combine "ha ?] "||  ; hahahahaha</lang>

Lhogho doesn't have cascade (yet), nor does it have the initialise a missing parameter capability demonstrated by the [:acc "||] above.

<lang logo>to copies :n :thing :acc 

 if :n = 0 [output :acc]
 output (copies :n-1 :thing combine :acc :thing)

end

print copies 5 "ha "||</lang>

Lua

<lang lua>function repeats(s, n) return n > 0 and s .. repeats(s, n-1) or "" end</lang>

Or use native string library function <lang lua>string.rep(s,n)</lang>

Maple

There are many ways to do this in Maple. First, the "right" (most efficient) way is to use the supplied procedures for this purpose. <lang Maple> > use StringTools in > Repeat( "abc", 10 ); # repeat an arbitrary string > Fill( "x", 20 ) # repeat a character > end use; 

                   "abcabcabcabcabcabcabcabcabcabc"

                        "xxxxxxxxxxxxxxxxxxxx"

</lang> These next two are essentially the same, but are less efficient (though still linear) because they create a sequence of 10 strings before concatenating them (with the built-in procedure cat) to form the result. <lang Maple> > cat( "abc" $ 10 ); 

                   "abcabcabcabcabcabcabcabcabcabc"

> cat( seq( "abc", i = 1 .. 10 ) ); 

                   "abcabcabcabcabcabcabcabcabcabc"

</lang> You can build up a string in a loop, but this is highly inefficient (quadratic); don't do this. <lang Maple> > s := "": > to 10 do s := cat( s, "abc" ) end: s; 

                   "abcabcabcabcabcabcabcabcabcabc"

</lang> If you need to build up a string incrementally, use a StringBuffer object, which keeps things linear.

Finally, note that strings and characters are not distinct datatypes in Maple; a character is just a string of length one.

Mathematica

<lang Mathematica>(* solution 1 *) rep[n_Integer,s_String]:=Apply[StringJoin,ConstantArray[s,{n}]]

(* solution 2 -- @@ is the infix form of Apply[] *) rep[n_Integer,s_String]:=StringJoin@@Table[s,{n}]

(* solution 3 -- demonstrating another of the large number of looping constructs available *) rep[n_Integer,s_String]:=Nest[StringJoin[s, #] &,s,n-1]</lang>

MATLAB / Octave

<lang MATLAB>function S = repeat(s , n) 

   S = repmat(s , [1,n]) ;

return</lang>

Note 1: The repetition is returned, not displayed.
Note 2: To repeat a string, use single quotes. Example: S=repeat('ha',5)

Maxima

<lang maxima>"$*"(s, n) := apply(sconcat, makelist(s, n))$ infix("$*")$

"abc" $* 5; /* "abcabcabcabcabc" */</lang>

Mercury

Mercury's 'string' module provides an efficient char-repeater. The following uses string.builder to repeat strings.

<lang Mercury>:- module repeat.

- interface.
- import_module string, char, int.
- func repeat_char(char, int) = string.
- func repeat(string, int) = string.
- implementation.
- import_module stream, stream.string_writer, string.builder.

repeat_char(C, N) = string.duplicate_char(C, N).

repeat(String, Count) = Repeated :- 

       S0 = string.builder.init,
       Repeated = string.builder.to_string(S),
       printn(string.builder.handle, Count, String, S0, S).
- pred printn(Stream, int, string, State, State)
              <= (stream.writer(Stream, string, State),
                  stream.writer(Stream, character, State)).
- mode printn(in, in, in, di, uo) is det.

printn(Stream, N, String, !S) :- 

       ( N > 0 ->
               print(Stream, String, !S),
               printn(Stream, N - 1, String, !S)
       ; true ).</lang>

Mirah

<lang mirah>x = StringBuilder.new

5.times do 

   x.append "ha"

end

puts x # ==> "hahahahaha"</lang>

Monte

<lang Monte> var s := "ha " * 5 traceln(s) </lang>

MUMPS

<lang MUMPS>RPTSTR(S,N) 

;Repeat a string S for N times
NEW I
FOR I=1:1:N WRITE S
KILL I
QUIT

RPTSTR1(S,N) ;Functionally equivalent, but denser to read 

F I=1:1:N W S
Q

</lang>


This last example uses the $PIECE function. <lang MUMPS> 

;Even better (more terse)
S x="",$P(x,"-",10)="-"
W x

</lang>

Nemerle

Any of the methods shown in the C# solution would also work for Nemerle, but they're all semantically awkward. This example uses an extension method to wrap one of the awkward techniques in order to clarify the semantics (which is also possible in C#, there's nothing really Nemerle specific here except the syntax). <lang Nemerle>using System; using System.Console;

module StrRep { 

   Repeat(this s : string, n : int) : string
   {
       String('x', n).Replace("x", s)
   }
   
   Main() : void
   {
       WriteLine("ha".Repeat(5));
       WriteLine("*".Repeat(5));
       WriteLine(String('*', 5)); // repeating single char

   }

}</lang>

NetRexx

NetRexx has built in functions to manipulate strings. The most appropriate for this task is the copies() function: <lang NetRexx>/* NetRexx */

ha5 = 'ha'.copies(5) </lang>

There are several other built-in functions that can be used to achieve the same result depending on need:

<lang NetRexx>/* NetRexx */ sampleStr = 'ha' -- string to duplicate say ' COPIES:' sampleStr.copies(5) say 'CHANGESTR:' '.....'.changestr('.', sampleStr)

sampleChr = '*' -- character to duplicate say ' LEFT:' sampleChr.left(5, sampleChr) say ' RIGHT:' sampleChr.right(5, sampleChr) say ' CENTRE:' sampleChr.centre(5, sampleChr) say ' OVERLAY:' sampleChr.overlay(sampleChr, 1, 5, sampleChr) say ' SUBSTR:' .substr(1, 5, sampleChr) say 'TRANSLATE:' '.....'.translate(sampleChr, '.') </lang>

NewLISP

<lang NewLISP>(dup "ha" 5)</lang>

Nim

<lang nim> import strutils repeatStr(5, "ha") </lang>

Objeck

<lang objeck>bundle Default { 

 class Repeat {
   function : Main(args : String[]) ~ Nil {
     Repeat("ha", 5)->PrintLine();
   }
   
   function : Repeat(string : String, max : Int) ~ String {
     repeat : String := String->New();
     for(i := 0; i < max; i += 1;) {
       repeat->Append(string);
     };
     
     return repeat;
   }
 }

}</lang>

Objective-C

Objective-C allows developers to extend existing an existing class by adding additional methods to the class without needing to subclass. These extensions are called categories. Category methods are available to all instances of the class, as well as any instances of its subclasses.

This task provides us with an opportunity to visit this aspect of the language feature.

We will extend NSString, the de facto Objective-C string class in environments that are either compatible with or descend directly from the OPENSTEP specification, such as GNUstep and Mac OS X, respectively, with a method that accomplishes the described task.

<lang objc>@interface NSString (RosettaCodeAddition) - (NSString *) repeatStringByNumberOfTimes: (NSUInteger) times; @end

@implementation NSString (RosettaCodeAddition) - (NSString *) repeatStringByNumberOfTimes: (NSUInteger) times { 

   return [@"" stringByPaddingToLength:[self length]*times withString:self startingAtIndex:0];

} @end</lang>

Now, let's put it to use: <lang objc> // Instantiate an NSString by sending an NSString literal our new 

   // -repeatByNumberOfTimes: selector.
   NSString *aString = [@"ha" repeatStringByNumberOfTimes:5];

   // Display the NSString.
   NSLog(@"%@", aString);</lang>

OCaml

Since Ocaml 4.02 strings are immutable, as is convenient for a functional language. Mutable strings are now implemented in the module Bytes. <lang ocaml>let string_repeat s n = 

 let len = Bytes.length s in
 let res = Bytes.create(n * len) in
 for i = 0 to pred n do
   Bytes.blit s 0 res (i * len) len
 done;
 Bytes.to_string res (* not stricly necessary, the bytes type is equivalent to string except mutability *)
</lang>

which gives the signature<lang ocaml> val string_repeat : bytes -> int -> string = <fun></lang>

testing in the toplevel: <lang ocaml># string_repeat "Hiuoa" 3 ;; - : string = "HiuoaHiuoaHiuoa"</lang>

Alternately create an array initialized to s, and concat: <lang ocaml>let string_repeat s n = 

 String.concat "" (Array.to_list (Array.make n s))
</lang>

Or: <lang ocaml>let string_repeat s n = 

 Array.fold_left (^) "" (Array.make n s)
</lang>

To repeat a single character use: <lang ocaml>String.make 5 '*'</lang>

Oforth

<lang Oforth>StringBuffer new "abcd" <<n(5)</lang>

OpenEdge/Progress

<lang Progress (OpenEdge ABL)>MESSAGE FILL( "ha", 5 ) VIEW-AS ALERT-BOX.</lang>

OxygenBasic

<lang oxygenbasic>

'REPEATING A CHARACTER

print string 10,"A" 'result AAAAAAAAAA

'REPEATING A STRING

function RepeatString(string s,sys n) as string 

 sys i, le=len s
 if le=0 then exit function
 n*=le
 function=nuls n
 '
 for i=1 to n step le
   mid function,i,s
 next

end function

print RepeatString "ABC",3 'result ABCABCABC </lang>

Oz

We have to write a function for this: <lang oz>declare 

 fun {Repeat Xs N}
    if N > 0 then
       {Append Xs {Repeat Xs N-1}}
    else
       nil
    end
 end

in 

 {System.showInfo {Repeat "Ha" 5}}</lang>

PARI/GP

This solution is recursive and unimaginably bad. Slightly less bad versions can be designed, but that's not the point: don't use GP for text processing if you can avoid it. If you really need to, it's easy to create an efficient function in PARI (see C) and pass that to GP. <lang parigp>repeat(s,n)={ 

 if(n, Str(repeat(s, n-1), s), "")

};</lang>

concat() joins together a vector of strings, in this case a single string repeated. <lang parigp>repeat(s,n)=concat(vector(n,i, s));</lang>

This solution is recursive and slightly less bad than the others for large n. <lang parigp>repeat(s,n)={ 

 if(n<4, return(concat(vector(n,i, s))));
 if(n%2,
   Str(repeat(Str(s,s),n\2),s)
 ,
   repeat(Str(s,s),n\2)
 );

}</lang>

Pascal

See Delphi

Perl

<lang perl>"ha" x 5</lang>

Perl 6

<lang perl6>"ha" x 5</lang> (Note that the x operator isn't quite the same as in Perl 5: it now only creates strings. To create lists, use xx.)

Phix

<lang Phix>?repeat('*',5) ?join(repeat("ha",5),"")</lang>

Output:
"*****"
"hahahahaha"

PHP

<lang php>str_repeat("ha", 5)</lang>

PicoLisp

<lang PicoLisp>(pack (need 5 "ha")) -> "hahahahaha"</lang> or: <lang PicoLisp>(pack (make (do 5 (link "ha")))) -> "hahahahaha"</lang>

Pike

<lang pike>"ha"*5;</lang>

PL/I

<lang PL/I>s = copy('ha', 5);

/* To repeat a single character a fixed number of times: */

s = (5)'h'; /* asigns 'hhhhh' to s. */ </lang>

PostScript

<lang PostScript>% the comments show the stack content after the line was executed % where rcount is the repeat count, "o" is for orignal, % "f" is for final, and iter is the for loop variable % % usage: rcount ostring times -> fstring

/times { 

 dup length dup    % rcount ostring olength olength
 4 3 roll          % ostring olength olength rcount
 mul dup string    % ostring olength flength fstring
 4 1 roll          % fstring ostring olength flength
 1 sub 0 3 1 roll  % fstring ostring 0 olength flength_minus_one 
 {                 % fstring ostring iter
   1 index 3 index % fstring ostring iter ostring fstring
   3 1 roll        % fstring ostring fstring iter ostring
   putinterval     % fstring ostring
 } for
 pop               % fstring

} def</lang>

PowerBASIC

<lang powerbasic>MSGBOX REPEAT$(5, "ha")</lang>

PowerShell

<lang powershell>"ha" * 5 # ==> "hahahahaha"</lang>

Prolog

<lang prolog>%repeat(Str,Num,Res). repeat(Str,1,Str). repeat(Str,Num,Res):- 

   Num1 is Num-1,
   repeat(Str,Num1,Res1),
   string_concat(Str, Res1, Res).</lang>

Pure

str_repeat is defined by pattern-matching: repeating any string 0 times results in the empty string; while repeating it more than 0 times results in the concatenation of the string and (n-1) further repeats.

<lang pure>> str_repeat 0 s = ""; > str_repeat n s = s + (str_repeat (n-1) s) if n>0; > str_repeat 5 "ha"; "hahahahaha" ></lang>

PureBasic

<lang PureBasic>Procedure.s RepeatString(count, text$=" ") 

  Protected i, ret$=""

  For i = 1 To count
     ret$ + text$
  Next
  ProcedureReturn ret$

EndProcedure

Debug RepeatString(5, "ha")</lang>

Python

<lang python>"ha" * 5 # ==> "hahahahaha"</lang> "Characters" are just strings of length one.

the other way also works: <lang python>5 * "ha" # ==> "hahahahaha"</lang>

R

<lang ruby>paste(rep("ha",5), collapse=)</lang>

Racket

<lang racket>

  1. lang racket
fast

(define (string-repeat n str) 

 (string-append* (make-list n str)))

(string-repeat 5 "ha") ; => "hahahahaha" </lang>

To repeat a single character: <lang racket> (make-string 5 #\*) => "*****" </lang>

RapidQ

<lang vb> 'For a single char showmessage String$(10, "-")

'For strings with more than one char function Repeat$(Expr as string, Count as integer) as string 

   dim x as integer
   for x = 1 to Count
       Result = Result + Expr
   next

end function

showmessage Repeat$("ha", 5) </lang>

REALbasic

<lang vb>Function Repeat(s As String, count As Integer) As String 

 Dim output As String
 For i As Integer = 0 To count
   output = output + s
 Next
 Return output

End Function </lang>

REBOL

<lang rebol>head insert/dup "" "ha" 5</lang>

Retro

<lang Retro>with strings'

repeatString ( $n-$ )
 1- [ dup ] dip [ over prepend ] times nip ;

"ha" 5 repeatString</lang>

REXX

Since the REXX language only supports the "character" type, it's not surprising that there are so many ways to skin a cat. <lang REXX>/*REXX program to show various ways to repeat a string (or repeat a single char).*/

/*all examples are equivalent, but not created equal.*/ 

                          /*───────────────────────────────────────────*/

y='ha' z=copies(y,5) 

                          /*───────────────────────────────────────────*/

z=copies( 'ha', 5 ) 

                          /*───────────────────────────────────────────*/

y='ha' z=y||y||y||y||y 

                          /*───────────────────────────────────────────*/

y='ha' z=y || y || y || y || y /*same as previous, but the "big sky" version*/ 

                          /*───────────────────────────────────────────*/

y='ha' z= 

      do 5
      z=z||y
      end
                          /*───────────────────────────────────────────*/

y="ha" z= 

      do 5
      z=z||y
      end
                          /*───────────────────────────────────────────*/

y="ha" z= 

      do i=101 to 105
      z=z||y
      end

                          /*───────────────────────────────────────────*/

y='+' z=left(,5,y) 

                          /*───────────────────────────────────────────*/

y='+' z=right(,5,y) 

                          /*───────────────────────────────────────────*/

y='+' z=substr(,1,5,y) 

                          /*───────────────────────────────────────────*/

y='+' z=center(,5,y) 

                          /*───────────────────────────────────────────*/

y='+' z=centre(,5,y) 

                          /*───────────────────────────────────────────*/

y='+' z=space(,5,y) 

                          /*───────────────────────────────────────────*/

y='+' z=translate('@@@@@',y,"@") 

                          /*───────────────────────────────────────────*/

y='abcdef' z=five(y) exit

five: procedure expose y; parse arg g if length(g)>=5*length(y) then return g return five(y||g) 

                          /*───────────────────────────────────────────*/

y='something wicked this way comes.' z=y||y||y||y||y||y||y||y||y||y||y||y|\y||y||y z=left(z,5*length(y)) 

                          /*───────────────────────────────────────────*/

y='+' z=copies(,5,y) 

                          /*───────────────────────────────────────────*/

y='+' z=lower(,1,5,y) 

                          /*───────────────────────────────────────────*/

y='+' z=lower(,,5,y) 

                          /*───────────────────────────────────────────*/

z='+' z=upper(,1,5,y) 

                          /*───────────────────────────────────────────*/

z=upper(,,5,y) 

                          /*───────────────────────────────────────────*/

y='charter bus.' z='*****' z=changestr('*',z,y) 

                          /*───────────────────────────────────────────*/

y='what the hey!' z= 

 do until length(z)==5*length(y)
 z=z||y
 end
                          /*───────────────────────────────────────────*/

y='what the hey!' z= 

 do until length(z)==5*length(y)
 z=insert(z,0,y)
 end
                          /*───────────────────────────────────────────*/

y='yippie ki yay' z= 

  do i=1 by 5 for 5
  z=overlay(y,z,i)
  end
                          /*───────────────────────────────────────────*/

y='+' z=justify(,5,y) 

                          /*───────────────────────────────────────────*/

whatever_this_variable_is_____it_aint_referenced_directly= 'boy oh boy.' z=; signal me; me: 

 do 5
 z=z||strip(subword(sourceline(sigl-1),2),,"'")
 end
                          /*───────────────────────────────────────────*/

y="any more examples & the angry townfolk with pitchforks will burn the castle." parse value y||y||y||y||y with z

exit /*stick a fork in it, we're done.*/</lang> Some older REXXes don't have a changestr bif, so one is included here ──► CHANGESTR.REX.

Ring

<lang ring> Copy("ha" , 5) # ==> "hahahahaha"</lang>


Ruby

<lang ruby>"ha" * 5 # ==> "hahahahaha"</lang>

Run BASIC

<lang runbasic>a$ = "ha " for i = 1 to 5 

 a1$ = a1$ + a$

next i a$ = a1$ print a$</lang>

Rust

<lang rust>std::iter::repeat("ha").take(5).collect::<String>(); // ==> "hahahahaha"</lang>

Scala

<lang scala>"ha" * 5 // ==> "hahahahaha"</lang>

Scheme

<lang scheme>(define (string-repeat n str) 

 (apply string-append (vector->list (make-vector n str))))</lang>

with SRFI 1: <lang scheme>(define (string-repeat n str) (fold string-append "" (make-list n str))) (string-repeat 5 "ha") ==> "hahahahaha"</lang>

To repeat a single character: <lang scheme>(make-string 5 #\*)</lang>

Scratch

This example requires making variables named "String", "Count", and "Repeated" first.

sed

Number of ampersands indicates number of repetitions. <lang sed> $ echo ha | sed 's/.*/&&&&&/' hahahahaha </lang>

Seed7

<lang seed7>$ include "seed7_05.s7i";

const proc: main is func 

 begin
   writeln("ha" mult 5);
 end func;</lang>

Output: 

hahahahaha

Sidef

<lang ruby>'ha' * 5; # ==> 'hahahahaha'</lang>

Smalltalk

If n is a small constant, then simply concatenating n times will do; for example, n=5:: <lang smalltalk>v := 'ha'. v,v,v,v,v</lang>

Works with: Pharo version 1.4
Works with: Smalltalk/X

By creating a collection of n 'ha', and joining them to a string:

<lang smalltalk>((1 to: n) collect: [:x | 'ha']) joinUsing: .</lang>

or:

Works with: Smalltalk/X
Works with: VisualWorks Smalltalk

<lang smalltalk>(Array new:n withAll:'ha') asStringWith:.</lang> By creating a WriteStream, and putting N times the string 'ha' into it:

<lang smalltalk>ws := writeStream. n timesRepeat: [ws nextPutAll: 'ha']. ws contents.</lang> alternatively: <lang smalltalk>(String streamContents:[:ws | n timesRepeat: [ws nextPutAll: 'ha']])</lang>

all evaluate to: 

hahahahaha

A string containing a repeated character is generated with: <lang smalltalk>String new:n withAll:$*</lang>

Works with: VA Smalltalk

<lang smalltalk>(String new:n) atAllPut:$*</lang>

SNOBOL4

<lang snobol4> output = dupl("ha",5) end</lang>

Sparkling

<lang sparkling>spn:3> repeat("na", 8) .. " Batman!" = nananananananana Batman!</lang>

SQL

<lang sql>select rpad(, 10, 'ha')</lang>

Standard ML

<lang sml>fun string_repeat (s, n) = 

 concat (List.tabulate (n, fn _ => s))
</lang>

testing in the interpreter: <lang sml>- string_repeat ("Hiuoa", 3) ; val it = "HiuoaHiuoaHiuoa" : string</lang>

To repeat a single character: <lang sml>fun char_repeat (c, n) = 

 implode (List.tabulate (n, fn _ => c))
</lang>

Suneido

<lang Suneido>'ha'.Repeat(5) --> "hahahahaha" '*'.Repeat(5) --> "*****"</lang>

Swift

Using extensions to do the repetition which makes for an easier syntax when repeating Strings, and using String.extend() to get faster evaluation.

<lang swift>extension String { 

 // Slower version
 func repeatString(n: Int) -> String {
   return Array(count: n, repeatedValue: self).joinWithSeparator("")
 }
 
 // Faster version
 // benchmarked with a 1000 characters and 100 repeats the fast version is approx 500 000 times faster :-)
 func repeatString2(n:Int) -> String {
   var result = self
   for _ in 1 ..< n {
     result.appendContentsOf(self)   // Note that String.appendContentsOf is up to 10 times faster than "result += self"
   }
   return result
 }

}

print( "ha".repeatString(5) ) print( "he".repeatString2(5) )</lang>

Output:
hahahahaha
hehehehehe

To repeat a single character: <lang swift>String(count:5, repeatedValue:"*" as Character) </lang>

Note that using the String version on a string of 1 Character, or the repeat single Character version is timewise close to the same. No point in using the Character version for efficiency (tested with repeating up to 100 000 times).

Bitwise Iterative Version

The following version is an enhanced version of the recursive ActionScript, where we're using bit operation along with iterative doubling of the string to get to the correctly repeated version of the text in the most effective manner without recursion. When benchmarked against the plain iterative version in previous section, this version is marginally better, but only my a very small percentage. The critical factor for making the repeat function effective when using larger strings (1000 characters) and multiple repeats (1000 repeats :-) ) was to to exchange the '+=' with 'String.extend' method.

<lang swift>extension String { 

 func repeatBiterative(count: Int) -> String {
       var reduceCount = count
       var result = ""
       var doubled = self
       while reduceCount != 0 {
           if reduceCount & 1 == 1 {
               result.appendContentsOf(doubled)
           }
           reduceCount >>= 1
           if reduceCount != 0 {
               doubled.appendContentsOf(doubled)
           }
       }
       return result
   }

}

"He".repeatBiterative(5)</lang>

Output:
"HeHeHeHeHe"

Tcl

<lang tcl>string repeat "ha" 5  ;# => hahahahaha</lang>

TUSCRIPT

<lang tuscript> $$ MODE TUSCRIPT repeatstring=REPEAT ("ha",5) </lang>

TorqueScript

--Eepos <lang TorqueScript>function strRep(%str,%int) { for(%i = 0; %i < %int; %i++) { %rstr = %rstr@%str; }

return %rstr; }</lang>

UNIX Shell

Using printf

Works with: bash
Works with: ksh93
Works with: zsh

<lang bash>printf "ha"%.0s {1..5}</lang>

With ksh93 and zsh, the count can vary.

Works with: ksh93
Works with: zsh

<lang bash>i=5 printf "ha"%.0s {1..$i}</lang>

With bash, {1..$i} fails, because brace expansion happens before variable substitution. The fix uses eval.

Works with: bash
Works with: ksh93
Works with: zsh

<lang bash>i=5 eval "printf 'ha'%.0s {1..$i}"</lang>

For the general case, one must escape any % or \ characters in the string, because printf would interpret those characters.

Works with: bash
Works with: ksh93
Works with: zsh

<lang bash>reprint() { 

 typeset e="$(sed -e 's,%,%%,g' -e 's,\\,\\\\,g' <<<"$1")"
 eval 'printf "$e"%.0s '"{1..$2}"

} reprint '% ha \' 5</lang>

Using repeat

Works with: zsh
Works with: csh

<lang bash> len=12; str='=' repeat $len printf "$str" </lang>

Using head -c

head -c is a GNU extension, so it only works with those systems. (Also, this script can only repeat a single character.)

Works with: Bourne Shell

<lang sh>width=72; char='=' head -c ${width} < /dev/zero | tr '\0' "$char"</lang>

Ursala

<lang Ursala>#import nat

repeat = ^|DlSL/~& iota

  1. cast %s

example = repeat('ha',5)</lang> output: 

'hahahahaha'

Vala

Repeat a string 5 times: <lang vala> string s = "ha"; string copy = ""; for (int x = 0; x < 5; x++) copy += s; </lang>

Fill a string with a char N times: <lang vala> string s = string.nfill(5, 'c'); </lang>

VBA

<lang VBA> Public Function RepeatStr(aString As String, aNumber As Integer) As String Dim bString As String

bString = aString If aNumber > 1 Then 

 For i = 2 To aNumber
   bString = bString & aString
 Next i

End If RepeatStr = bString End Function </lang>

Sample output: 

print RepeatSTr( "Hello world!", 3)
Hello world!Hello world!Hello world!

Note: unlike Visual Basic .NET, "String(5, "ha") in VBA produces "hhhhh" (only the first character is repeated)!

An alternative method: <lang vba> Public Function RepeatString(stText As String, iQty As Integer) As String RepeatString = Replace(String(iQty, "x"), "x", stText) End Function</lang>

VBScript

Works with: Windows Script Host version *

<lang VBScript> ' VBScript has a String() function that can repeat a character a given number of times ' but this only works with single characters (or the 1st char of a string): WScript.Echo String(10, "123") ' Displays "1111111111"

' To repeat a string of chars, you can use either of the following "hacks"... WScript.Echo Replace(Space(10), " ", "Ha") WScript.Echo Replace(String(10, "X"), "X", "Ha") </lang>

Vedit macro language

<lang vedit>Ins_Text("ha", COUNT, 5) </lang>

Visual Basic .NET

Use the PadLeft & PadRight commands: <lang vbnet>FileContents = "".PadRight(FileSizeBytes, "X")</lang>

Where:

  • FileContents is the string to populate
  • FileSizeBytes is the number of repetitions

This produces: 

"XXXXXXXXXX...........XXXXXXXXX"

This only works for single character, not for Strings. <lang vbnet>FileContents = "".PadRight(3, "XO-")</lang>

produces 

"XXX" not "XO-XO-XO as expected

Visual FoxPro

Use the built in function REPLICATE(string, number): <lang vfp>? REPLICATE("HO", 3)</lang>

produces 

HOHOHO

Wart

<lang wart>def (s * n) :case (string? s) 

 with outstring
   repeat n
     pr s

("ha" * 5) => "hahahahaha"</lang>

Wortel

<lang wortel>@join "" @rep 5 "ha" ; returns "hahahahaha"</lang> As a function: <lang wortel>^(\@join "" @rep)</lang>

XPL0

<lang XPL0>cod T=12; int I; for I gets 1,5 do T(0,"ha")</lang>

Output:
hahahahaha

Yorick

<lang yorick>array("ha", 5)(sum)</lang>

zkl

Same as Ruby <lang zkl>"ha" * 5 # --> "hahahahaha"</lang>

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