Repeat a string: Difference between revisions
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(→{{header|Haskell}}: Or unpacking the mechanism of '''replicate''' a little, and using a monoid operation) |
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To repeat a single character:
<lang haskell>replicate 5 '*'</lang>
Or, unpacking the mechanism of '''replicate''' a little, and using a monoid operation:
<lang haskell>repString :: String -> Int -> String
repString s n =
let rep x = xs
where
xs = mappend x xs
in take (n * length s) $ rep s
main :: IO ()
main = print $ repString "ha" 5</lang>
{{Out}}
<pre>"hahahahaha"</pre>
=={{header|HicEst}}==
| |||
Revision as of 20:28, 29 March 2017
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>
APL
Fill up a string of length 10 with 'ha': <lang apl> 10⍴'ha' hahahahaha</lang> Alternatively, define a function: <lang apl> REPEAT←{(⍺×⍴⍵)⍴⍵}
5 REPEAT 'ha'
hahahahaha</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.
<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 //
- 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>
beeswax
<lang beeswax> p <
p0~1<}~< d@<
_VT@1~>yg~9PKd@M'd;</lang>
Example:
julia> beeswax("repeat a string.bswx")
sha
i5
hahahahaha
Program finished!
s tells the user that the program expects a string as input.
i tells the user that the program expects an integer as input.
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>
- include <stdlib.h>
- 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>
- include <stdlib.h>
- 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>
- 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>
- 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>
COBOL
Virtually a one-liner. <lang cobol>IDENTIFICATION DIVISION. PROGRAM-ID. REPEAT-PROGRAM. DATA DIVISION. WORKING-STORAGE SECTION. 77 HAHA PIC A(10). PROCEDURE DIVISION.
MOVE ALL 'ha' TO HAHA. DISPLAY HAHA. STOP RUN.</lang>
- Output:
hahahahaha
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.
- import system'routines.
- import extensions.
- 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
<lang fortran>program test_repeat
write (*, '(a)') repeat ('ha', 5)
end program test_repeat</lang> Output:
hahahahaha
FreeBASIC
<lang freebasic>' FB 1.05.0 Win64
' A character is essentially a string of length 1 in FB though there is a built-in function, String, ' which creates a string by repeating a character a given number of times.
' To avoid repeated concatenation (a slow operation) when the string to be repeated has a length ' greater than one, we instead create a buffer of the required size and then fill that.
Function repeat(s As String, n As Integer) As String
If n < 1 Then Return ""
If n = 1 Then Return s
Var size = Len(s)
If size = 0 Then Return s ' empty string
If size = 1 Then Return String(n, s[0]) ' repeated single character
Var buffer = Space(size * n) 'create buffer for size > 1
For i As Integer = 0 To n - 1
For j As Integer = 0 To size - 1
buffer[i * size + j] = s[j]
Next j
Next i
Return buffer
End Function
Print repeat("rosetta", 1) Print repeat("ha", 5) Print repeat("*", 5) Print Print "Press any key to quit program" Sleep</lang>
- Output:
rosetta hahahahaha *****
Frink
<lang frink> println[repeat["ha", 5]] </lang>
Gambas
<lang>Public Sub Form_Open()
Print String$(5, "ha")
End</lang>
Output = hahahahaha
GAP
<lang gap>Concatenation(ListWithIdenticalEntries(10, "BOB ")); "BOB BOB BOB BOB BOB BOB BOB BOB BOB BOB "</lang>
Glee
<lang glee>'*' %% 5</lang>
<lang glee>'ha' => Str; Str# => Len; 1..Len %% (Len * 5) => Idx; Str [Idx] $;</lang>
<lang glee>'ha'=>S[1..(S#)%%(S# *5)]</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>
Or, unpacking the mechanism of replicate a little, and using a monoid operation: <lang haskell>repString :: String -> Int -> String repString s n =
let rep x = xs
where
xs = mappend x xs
in take (n * length s) $ rep s
main :: IO () main = print $ repString "ha" 5</lang>
- Output:
"hahahahaha"
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
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
- 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>
Lingo
- Take a string and repeat it some number of times.
<lang lingo>on rep (str, n)
res = "" repeat with i = 1 to n put str after res end repeat return res
end</lang> <lang lingo>put rep("ha", 5) -- "hahahahaha"</lang>
- If there is a simpler/more efficient way to repeat a single “character”...
<lang lingo>put bytearray(5, chartonum("*")).readRawString(5) -- "*****"</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>
Logo
<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
Version #1. Based on recursion.
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>
Version #2. Simple loop based.
Basic set of string functions is very handy for presentation purposes. At the same time, it is true that PARI/GP is not an appropriate tool for the heavy text processing.
<lang parigp> \\ Repeat a string str the specified number of times ntimes and return composed string. \\ 3/3/2016 aev srepeat(str,ntimes)={ my(srez=str,nt=ntimes-1); if(ntimes<1||#str==0,return("")); if(ntimes==1,return(str)); for(i=1,nt, srez=concat(srez,str)); return(srez); }
{ \\ TESTS print(" *** Testing srepeat:"); print("1.",srepeat("a",5)); print("2.",srepeat("ab",5)); print("3.",srepeat("c",1)); print("4.|",srepeat("d",0),"|"); print("5.|",srepeat("",5),"|"); print1("6."); for(i=1,10000000, srepeat("e",10)); } </lang>
- Output:
*** Testing srepeat: 1.aaaaa 2.ababababab 3.c 4.|| 5.|| 6. (16:00) gp > ## *** last result computed in 1min, 2,939 ms.
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>
- 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>
Since 1.16: <lang rust>"ha".repeat(5); // ==> "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>
By creating a collection of n 'ha', and joining them to a string:
<lang smalltalk>((1 to: n) collect: [:x | 'ha']) joinUsing: .</lang>
or:
<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:$*</lang>
<lang smalltalk>(String new:n) atAllPut:$*</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
<lang bash>printf "ha"%.0s {1..5}</lang>
With ksh93 and zsh, the count can vary.
<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.
<lang bash>i=5 eval "printf 'ha'%.0s {1..$i}"</lang>
For the general case, one must escape any % or \ characters in the string, because printf would interpret those characters.
<lang bash>reprint() {
typeset e="$(sed -e 's,%,%%,g' -e 's,\\,\\\\,g' <<<"$1")"
eval 'printf "$e"%.0s '"{1..$2}"
} reprint '% ha \' 5</lang>
Using repeat
<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.)
<lang sh>width=72; char='=' head -c ${width} < /dev/zero | tr '\0' "$char"</lang>
Ursala
<lang Ursala>#import nat
repeat = ^|DlSL/~& iota
- 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
<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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