Decimal Multiplication of Strings

Question

Given two inputs, a string \$s\$ and a decimal number \$n\$, 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 \$\lfloor n \rfloor\$ times and then the first \$\lfloor (n - \lfloor n \rfloor) \cdot |\,s\,|)\rfloor\$ characters again.

Other notes:

• The input will not always be a float, it may be an integer. 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 white space except newlines, quotes and other characters. Control characters are excluded, too, if this helps.
• No built-ins for direct string repetition are allowed. That means for instance Python’s 'a'*5 is forbidden. Nevertheless string concatenation (frequently denoted by the + sign) is allowed.

Test cases

\$s\$	\$n\$	result
test case	1	test case
case	2.5	casecaseca
(will add more later)	0.3333	(will
cats >= dogs	0.5	cats >

Final Note

I am seeing a lot of answers that use builtin string multiplication or repetition functions. This is not allowed. The rule is: If it directly multiplies the string, you cannot do it.

Answer 1

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.

Answer 2

Pyth, 9 8

s@Lz*lzQ

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

Answer 3

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

f=(s,n,l=s.length*n)=>~n?f(s+s,n-1,l):s.slice(0,l)

TEST

f=(s,n,l=s.length*n)=>~n?f(s+s,n-1,l):s.slice(0,l)

//TEST
console.log=x=>O.textContent+=x+'\n'
;[
 ['test case', 1, 'test case'],
 ['case', 3.5, 'casecasecaseca'],
 ['(will add more later)', 0.3333, '(will '],
 ['cats >= dogs', 0.5, 'cats >']]
.forEach(t=>{
  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>

Answer 4

Jelly, 5 bytes

×L}Rị

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.

Answer 5

Vitsy, 9 bytes

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

zlW*\[DO{]
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!

Answer 6

CJam, 10 bytes

l_,l~*,\f=

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

Test it here.

Explanation

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.

Answer 7

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 :(

Answer 8

Ruby, 49 48 characters

->s,n{(0...(n*l=s.size).to_i).map{|i|s[i%l]}*''}

Sample run:

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

Answer 9

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.

Usage:

# 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 >"

Answer 10

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
end
o
end

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

Answer 11

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.

Answer 12

PHP 5, 96 87

9 bytes saved thanks to @manatwork

<?for($i=$z=0;$i++<floor(strlen($a=$argv[1])*$argv[2]);$z++)echo$a[$z]?:$a[$z=0‌​];

Pretty straight forward looping answer.

Ungolfed

<?
$a=$argv[1];
$z=0;
for($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];
    $z++;
}

Answer 13

R, 59 bytes

function(s,l)cat(rawToChar(array(charToRaw(s),nchar(s)*l)))

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.

Test

> 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
casecaseca
> f('(will add more later)', 0.3333) # -> (will(space)
(will 
> f('cats >= dogs', 0.5) # -> cats >
cats >
> 

Answer 14

Perl, 51 + 3 = 54 bytes

$l=<>*y///c;for$i(1..$l){push@a,/./g}say@a[0..$l-1]

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

Explanation:

$l=<>*y///c;              # Calculate output length (eg. 2.5 * input length)
for$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

Answer 15

c (preprocessor macro), 71

j,l;
#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.

Answer 16

Oracle SQL 11.2, 154 152 bytes

WITH v(s,i)AS(SELECT SUBSTR(:1,1,FLOOR(FLOOR((:2-FLOOR(:2))*LENGTH(:1)))),1 FROM DUAL UNION ALL SELECT :1||s,i+1 FROM v WHERE i<=:2)SELECT MAX(s)FROM v;

Un-golfed

WITH v(s,i) AS
(
  SELECT SUBSTR(:1,1,FLOOR(FLOOR((:2-FLOOR(:2))*LENGTH(:1)))),1 FROM DUAL 
  UNION ALL 
  SELECT :1||s,i+1 FROM v WHERE i<=:2
)
SELECT MAX(s) FROM v;

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

Saved 2 bytes thanks to @MickyT

Answer 17

Seriously, 24 bytes

,╗,mi@≈╜n╜l(*≈r`╜E`MΣ)kΣ

Try it online!

Explanation:

,╗,mi@≈╜n╜l(*≈r`╜E`MΣ)kΣ
,╗                        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

Answer 18

Pyth, 9 bytes

V*Elzp@zN

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

Answer 19

Pascal, 138 B

This full program requires a processor compliant with ISO standard 10206 “Extended Pascal”. Trivial cases have been excluded: It is presumed that the input string s is non-empty and n is positive. NB: The golfed version caps s at a reasonable length of 999.

program p(input,output);var s:string(999);n:real;begin
read(s,n);while n>1 do begin n:=n-1;write(s)end;write(s[1..trunc(length(s)*n)])end.

Ungolfed:

program decimalMultiplicationOfStrings(input, output);
    var
        { `string` is a schema data type defined by Extended Pascal.
          The `string(maxInt)` discriminates the schema data type.
          Henceforth the specific `string` has a capacity of `maxInt`. }
        s: string(maxInt);
        n: real;
    begin
        { You can specify `real` as well as `integer` literals for `n`.
          An `integer` value is automatically promoted to `real`. }
        read(s, n);
        { The `string` subrange notation used below in the last line
          may not specify an empty range (e. g. `1‥0` is illegal).
          Therefore the `while` loop’s condition is `n > 1`.
          If `n = 1`, the one copy is printed via the final line. }
        while n > 1 do
        begin
            n ≔ n − 1;
            write(s);
        end;
        { The subrange notation is defined by Extended Pascal.
          It cannot be used for strings that are `bindable`. }
        write(s[1‥trunc(length(s) * n)]);
    end.

Input is end-of-line separated. s can otherwise contain any character from chr(0) to (the implementation-defined) maxChar range.

Answer 20

Nim, 75 bytes

func r[S,F](s:S,n:F):S=
 for i in 0..<int n*s.len.float:result&=s[i%%s.len]

Attempt This Online!

Answer 21

Scala, 68 bytes

Golfed version. Try it online!

(s,n)=>{val l=s.length;(0 to (n*l).toInt-1).map(i=>s(i%l)).mkString}

Ungolfed version. Try it online!

object Main {
  def repeatString(s: String, n: Double): String = {
    val len = s.length
    val repeatLen = (n * len).toInt
    (0 until repeatLen).map(i => s(i % len)).mkString
  }

  def main(args: Array[String]): Unit = {
    println(repeatString("case", 2.5))
  }
}