Given 2 inputs, a string and a decimal number, output the string multiplied by that number.

The catch is that the number can be a float or an integer.

You should output the string floor(n) time and then the first floor((n-floor(n))*len(string)) letters again.

Other notes:

  • The input will not always be a float, it may be an int. So 1.5, 1, and 1.0 are all possible. It will always be in base 10 though, and if you wish an exception please comment.
  • The string input may contain whitespace, quotes and other characters. No newlines or control chars though.
  • No built-ins for direct string repeating, even string multiplication like the python 'a'*5 are allowed. However string addition is allowed.

Test cases:

The comma & space separate the inputs.

test case, 1 -> test case

case, 2.5 -> casecaseca

(will add more later), 0.3333 -> (will(space)

cats >= dogs, 0.5 -> cats >

Final Note:

I am seeing a lot of answers that use builtin string multiplication or repeation functions. This is not allowed. @VTC's answer is valid though, because it does not multiply the string, only the float input. So the definitive rule is: If it directly multiplies the string, you can't do it.

  • \$\begingroup\$ The wording was modified repeatedly (I did not see the first revision). I suggest to remove direct string repeating (what does this mean?). But all in all you're right \$\endgroup\$ – edc65 Feb 8 '16 at 16:10
  • 1
    \$\begingroup\$ Related quine version \$\endgroup\$ – Sp3000 Feb 8 '16 at 16:47
  • \$\begingroup\$ @Sp3000 yeah, I know. I think the difference is significant enough. \$\endgroup\$ – Rɪᴋᴇʀ Feb 8 '16 at 17:18
  • \$\begingroup\$ "No built-ins for direct string repeating, even string multiplication like the python 'a'*5 are allowed." You don't explain the difference between these. They sound the same to me. \$\endgroup\$ – msh210 Feb 8 '16 at 20:50
  • \$\begingroup\$ @edc65 In Perl you can do list repetition then concatenate the elements of that list, which isn't direct string repetition. In Perl 5: join "", ("case") x 2 vs "case" x 2, in Perl 6 [~] "case" xx 2 vs the same "case" x 2 \$\endgroup\$ – Brad Gilbert b2gills Feb 8 '16 at 20:59

18 Answers 18


Jelly, 5 bytes


Doesn't use a repetition built-in. Try it online!

How it works

×L}Rị  Main link. Left input: n (multiplier). Right input: S (string)

 L}    Yield the length of S.
×      Multiply it with n.
   R   Range; turn n×len(S) into [1, ... floor(n×len(S))].
    ị  Retrieve the elements of S at those indices.
       Indices are 1-based and modular in Jelly, so this begins with the first and
       jump back after reaching the last.

Java 7, 89

void g(char[]a,float b){for(int i=0,l=a.length;i<(int)(l*b);)System.out.print(a[i++%l]);}

takes char[] and float and outputs to STDOUT. basic looping.

  • 3
    \$\begingroup\$ Good golfing, even for java. :P \$\endgroup\$ – Rɪᴋᴇʀ Feb 8 '16 at 16:33
  • \$\begingroup\$ this was suggested on my other answer too, but i don't think i will do this. it does not seem right to me. \$\endgroup\$ – Marky Markov Feb 8 '16 at 16:44
  • \$\begingroup\$ Eh, fair enough. It is recognized here, but alright. :D \$\endgroup\$ – Addison Crump Feb 8 '16 at 16:44
  • \$\begingroup\$ I recommend declaring your language as Java 7. Then no one can tell you to use lambdas. \$\endgroup\$ – feersum Feb 8 '16 at 16:46

Pyth, 9 8


Saved 1 byte thanks to Pietu1998

This takes floor(n * len(string)) letters from the string, using cyclical indexing. I believe this is always equivalent to the given formula.

Test Suite

  • 1
    \$\begingroup\$ No plz don't take this from me this soon. xD \$\endgroup\$ – Addison Crump Feb 8 '16 at 16:02
  • \$\begingroup\$ @VoteToClose I actually didn't read your answer at all, scouts honour :P I didn't even realise that string repetitions were disallowed, this was just shorter that what I came up with that way... \$\endgroup\$ – FryAmTheEggman Feb 8 '16 at 16:14
  • 1
    \$\begingroup\$ You don't even need the second s. range is funny like that. \$\endgroup\$ – PurkkaKoodari Feb 8 '16 at 19:47
  • 1
    \$\begingroup\$ NOO! cries in a corner Ah, oh well. \$\endgroup\$ – Addison Crump Feb 8 '16 at 19:54

JavaScript (ES6), 50 bytes

Edit 2 bytes more to include definition of function f. 1 byte less using the tip of @manatwork. Note: using ~ we have more iterations than necessary, but this is code golf and even 1 byte counts




 ['test case', 1, 'test case'],
 ['case', 3.5, 'casecasecaseca'],
 ['(will add more later)', 0.3333, '(will '],
 ['cats >= dogs', 0.5, 'cats >']]
  var s=t[0],n=t[1],x=t[2],r=f(s,n);
  console.log("«"+s+"» "+n+' => «'+r+'» '+(x==r?'OK':'FAIL expected '+x));
<pre id=O></pre>

  • \$\begingroup\$ Okay, thanks. So far most of the answers have had no problem, and it is really easy to fix. Thanks for correcting it. \$\endgroup\$ – Rɪᴋᴇʀ Feb 8 '16 at 16:28
  • \$\begingroup\$ Tiny typo: n>0 in the code vs. n>1 in the test case. \$\endgroup\$ – manatwork Feb 8 '16 at 16:38
  • \$\begingroup\$ @manatwork thanks. It should work either way \$\endgroup\$ – edc65 Feb 8 '16 at 17:36
  • \$\begingroup\$ Oh. Indeed. But then why not just ~n? (Really just a question. Tried only the given test cases.) \$\endgroup\$ – manatwork Feb 8 '16 at 17:42
  • 3
    \$\begingroup\$ @edc65 Where is f defined in your solution? Aren't you missing f=? \$\endgroup\$ – andlrc Feb 8 '16 at 18:38

Vitsy, 9 bytes

Expects the word as an argument, and the number to multiply by through STDIN.

z          Grab all string argument input.
 l         Get the length of the stack.
  W        Parse STDIN.
   *       Multiply the top two items (length of string and the number of repetitions)
    \[   ] Do the stuff in the loop.
      DO{  Output one char at a time, making sure to duplicate first.

Try it online!

  • \$\begingroup\$ True to your word, you answered fast. \$\endgroup\$ – Rɪᴋᴇʀ Feb 8 '16 at 15:48
  • \$\begingroup\$ @RikerW Martin out FGITW'd me. \$\endgroup\$ – Addison Crump Feb 8 '16 at 15:51
  • \$\begingroup\$ Why do you Grab all string argument input. and then Parse STDIN. again? \$\endgroup\$ – Rɪᴋᴇʀ Feb 8 '16 at 16:08
  • \$\begingroup\$ @RikerW Arguments that are doubles are automatically parsed, pushing them to the stack immediately. Handling that takes more bytes than it's worth. \$\endgroup\$ – Addison Crump Feb 8 '16 at 16:09
  • \$\begingroup\$ Oh okay. That makes more sense now. \$\endgroup\$ – Rɪᴋᴇʀ Feb 8 '16 at 16:10

CJam, 10 bytes


The string is supplied on the first line of STDIN, the float on the second.

Test it here.


l    e# Read string.
_,   e# Duplicate and get its length.
l~   e# Read second line and evaluate.
*    e# Multiply them. If the result, N, was floored it would give us the number of
     e# characters in the required output.
,    e# Get range [0 1 ... ⌊N⌋-1].
\f=  e# For each character in that range, fetch the corresponding character from the
     e# string using cyclic indexing.

Python 2, 71 bytes

lambda s,x:"".join(s for i in range(int(x)))+s[:int(len(s)*(x-int(x)))]

Try it here!

Creates an unnamed lambda which takes the string as first argument and the float as second. Returns the repeated string.

This could be 46 if string repetition builtins were allowed :(

  • 1
    \$\begingroup\$ Much sad. Such string multiplication rules. +1 A+ for effurt. \$\endgroup\$ – Addison Crump Feb 8 '16 at 16:15

Ruby, 49 48 characters


Sample run:

2.1.5 :001 > ->s,n{(0...(n*l=s.size).to_i).map{|i|s[i%l]}*''}['case', 2.5]
 => "casecaseca" 

Perl 6,  46 41  39 bytes

{([~] $^a xx$^b)~$a.substr(0,$a.chars*($b%1))}    # 46 bytes
{substr ([~] $^a xx$^b+1),0,$a.chars*$^b}         # 41 bytes
{substr ([~] $^a xx$^b+1),0,$a.comb*$b}           # 39 bytes

Perl 6 has both a string repetition operator x and a list repetition operator xx.

Since the rules disallow string repetition, we repeat it as if it was a single element list instead. Then the list gets concatenated together, and a substring of it is returned.


# give it a lexical name
my &code = {substr ([~] $^a xx$^b+1),0,$a.chars*$^b}
#          {substr ($^a x$^b+1),0,$a.chars*$^b}

say code('test case', 1).perl;                  # "test case"
say code('case', 2.5).perl;                     # "casecaseca"
say code('(will add more later)', 0.3333).perl; # "(will "
say code('cats >= dogs', 0.5).perl;             # "cats >"
  • \$\begingroup\$ substr ([~] $^a xx$^b+1),0,$a.comb*$b} saves two chars \$\endgroup\$ – raiph Feb 10 '16 at 4:31

osascript, 173 bytes

Oh my days, this is worse than I thought.

on run a
set x to a's item 1's characters
set y to a's item 2
set o to""
set i to 1
set z to x's items's number
repeat y*z
set o to o&x's item i
set i to i mod z+1

Returns the value of the string, another answer using cyclical indexing. Expects input as "string" "repetitions".

  • \$\begingroup\$ Oh my days, this is worse than I thought. So true, so true. \$\endgroup\$ – Rɪᴋᴇʀ Feb 8 '16 at 16:28
  • \$\begingroup\$ Is there a multiple var set at once command? ie set x,y to a's items? \$\endgroup\$ – Rɪᴋᴇʀ Feb 8 '16 at 16:29
  • \$\begingroup\$ @RikerW I don't think so. If there is, I'm seriously missing out. \$\endgroup\$ – Addison Crump Feb 8 '16 at 16:34

Haskell, 44 bytes

c x=x++c x
s#n=take(floor$n*sum[1|a<-s])$c s

Usage example: "(will add more later)" # 0.3333 -> "(will ".

How it works: c concatenates infinite copies of the string x. It behaves like the built-in cycle. sum[1|a<-s] is a custom length function that works with Haskell's strict type system as it returns a Double (the built-in length returns an Int which cannot be multiplied with n). # takes floor (n * length(s)) characters from the cycled string s.


PHP 5, 96 87

9 bytes saved thanks to @manatwork


Pretty straight forward looping answer.


for($i=0; $i < floor(strlen($a)*$argv[2]); $i++) {
    // if the string offset is not set
    // then reset $z back to 0 so we can
    // echo the beginning of ths string again
    @$a[$z] ?: $z=0;
    echo $a[$z];
  • \$\begingroup\$ Not sure when should that error suppression help, for me seems to work without @ too: <?for($i=$z=0;$i++<floor(strlen($a=$argv[1])*$argv[2]);$z++)echo$a[$z]?:$a[$z=0]; \$\endgroup\$ – manatwork Feb 8 '16 at 17:26
  • \$\begingroup\$ I was getting a notice on case #2 which caused the output to render incorrectly, which is when I added in the suppression. (running in CLI mode) \$\endgroup\$ – Samsquanch Feb 8 '16 at 17:30
  • \$\begingroup\$ “PHP 5.3 or later, the default value is E_ALL & ~E_NOTICE & ~E_STRICT & ~E_DEPRECATED.” – error_reporting So we prefer to base our solutions on default configuration and not to care about notices and other good habits. For example ignoring the initialization of $z and $i. \$\endgroup\$ – manatwork Feb 8 '16 at 17:37
  • \$\begingroup\$ Oh, cool. Thanks for the info! \$\endgroup\$ – Samsquanch Feb 8 '16 at 18:05

R, 59 bytes


As an unnamed function. This uses charToRaw to split the string into a vector of raws. This is filled into an array of length * l, converted back to char and output.
I was going to use strsplit, but it ended up being longer.


> f=
+ function(s,l)cat(rawToChar(array(charToRaw(s),nchar(s)*l)))
> f('test case', 1) # -> test case
test case
> f('case', 2.5) # -> casecaseca
> f('(will add more later)', 0.3333) # -> (will(space)
> f('cats >= dogs', 0.5) # -> cats >
cats >

Perl, 51 + 3 = 54 bytes


Requires: -n, -l and -M5.010 | -E:

 $ perl -nlE'$l=<>*y///c;for$i(1..$l){push@a,/./g}say@a[0..$l-1]' <<< $'test case\n1'
 test case
 $ perl -nlE'$l=<>*y///c;for$i(1..$l){push@a,/./g}say@a[0..$l-1]' <<< $'case\n2.5'
 $ perl -nlE'$l=<>*y///c;for$i(1..$l){push@a,/./g}say@a[0..$l-1]' <<< $'(will add more later)\n0.3333'
 $ perl -nlE'$l=<>*y///c;for$i(1..$l){push@a,/./g}say@a[0..$l-1]' <<< $'cats >= dogs\n0.5'
 cats >


$l=<>*y///c;              # Calculate output length (eg. 2.5 * input length)
for$i(1..$l){push@a,/./g} # Push a lot of chars from input into @a
say@a[0..$l-1]            # Slice @a according to output length

c (preprocessor macro), 71

#define f(s,m) l=strlen(s);for(j=0;j<(int)(l*m);)putchar(s[j++%l])

Not much tricky here. Just need to make sure l*m is cast to an int before comparing to j.

Try it online.


Oracle SQL 11.2, 154 152 bytes



WITH v(s,i) AS
  SELECT :1||s,i+1 FROM v WHERE i<=:2

I went the recursive way, with the initialisation select taking care of the decimal part.

Saved 2 bytes thanks to @MickyT

  • \$\begingroup\$ You can save a couple by removing spaces after the ) in the WITH clause and the final select. \$\endgroup\$ – MickyT Feb 8 '16 at 19:44
  • \$\begingroup\$ Another saving would be to replace FLOOR(FLOOR((:2-FLOOR(:2))*LENGTH(:1))) with MOD(:2,1)*LENGTH(:1) \$\endgroup\$ – MickyT Feb 8 '16 at 20:56
  • \$\begingroup\$ And one last one :), you can use LPAD rather than SUBSTR \$\endgroup\$ – MickyT Feb 8 '16 at 21:07

Seriously, 24 bytes


Try it online!


,╗                        get first input (string) and push it to register 0
  ,mi@≈                   get input 2 (x), push frac(x) (f), int(x) (n)
       ╜n                 push n copies of the string
         ╜l(*≈            push length of string, multiply by f, floor (substring length) (z)
              r`╜E`MΣ     push s[:z]
                     )kΣ  move fractional part of string to bottom, concat entire stack

Pyth, 9 bytes


Basically just doing

             z = input()
V*Elz        for N in range(evaluatedInput()*len(z)):    # flooring is automatic
     p@zN        print(z[N], end="")                     # modular indexing

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