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 |,円s,円|)\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.

• edc65: 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
• Sp3000: Related quine version
• Riker: @Sp3000 yeah, I know. I think the difference is significant enough.
• msh210: "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.
• Brad Gilbert b2gills: @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

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.

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

Answer 2

Pyth, (削除) 9 (削除ここまで) 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

• Addison Crump: No plz don't take this from me this soon. xD
• FryAmTheEggman: @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...
• PurkkaKoodari: You don't even need the second s. range is funny like that.
• Addison Crump: NOO! cries in a corner Ah, oh well.

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>

• Riker: Okay, thanks. So far most of the answers have had no problem, and it is really easy to fix. Thanks for correcting it.
• manatwork: Tiny typo: n>0 in the code vs. n>1 in the test case.
• manatwork: Oh. Indeed. But then why not just ~n? (Really just a question. Tried only the given test cases.)
• Andreas Louv: @edc65 Where is f defined in your solution? Aren't you missing f=?
• edc65: @dev-null oh oh oh you'r perfectly right, this is recursive. My fault. Thanks for pointing it out

Answer 4

Jelly, 5 bytes

×ばつL}Rị

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

How it works

×ばつL}Rị Main link. Left input: n (multiplier). Right input: S (string)
 L} Yield the length of ×ばつ Multiply it with n.
 R Range; turn n×ばつlen(S) into [1, ... floor(n×ばつ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!

• Riker: True to your word, you answered fast.
• Addison Crump: @RikerW Martin out FGITW'd me.
• Riker: Why do you Grab all string argument input. and then Parse STDIN. again?
• Addison Crump: @RikerW Arguments that are doubles are automatically parsed, pushing them to the stack immediately. Handling that takes more bytes than it's worth.
• Riker: Oh okay. That makes more sense now.

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 :(

• Addison Crump: Much sad. Such string multiplication rules. +1 A+ for effurt.

Answer 8

Ruby, (削除) 49 (削除ここまで) 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, (削除) 46 41 (削除ここまで) 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 >"

• raiph: substr ([~] $^a xx$^b+1),0,$a.comb*$b} saves two chars

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".

• Riker: Oh my days, this is worse than I thought. So true, so true.
• Riker: Is there a multiple var set at once command? ie set x,y to a's items?
• Addison Crump: @RikerW I don't think so. If there is, I'm seriously missing out.

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, (削除) 96 (削除ここまで) 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++;
}

• manatwork: 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];
• Samsquanch: 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)
• manatwork: "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.
• Samsquanch: Oh, cool. Thanks for the info!

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, (削除) 154 (削除ここまで) 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

• MickyT: You can save a couple by removing spaces after the ) in the WITH clause and the final select.
• MickyT: Another saving would be to replace FLOOR(FLOOR((:2-FLOOR(:2))*LENGTH(:1))) with MOD(:2,1)*LENGTH(:1)
• MickyT: And one last one :), you can use LPAD rather than SUBSTR

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

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 20

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 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))
 }
}

Answer 22

SAKO, 211 bytes

PODPROGRAM:F(X,*W)
CALKOWITE:*W,I,J
STRUKTURA(9):W
J=-1
2)J=J+1
GDYW(J)=58:1,INACZEJ2
1)GDYENT(X)=0:3,INACZEJ4
*4)DRUKUJWIERSZ:W
POWTORZ:I=1(1)ENT(X)
*3)W(I)=0
POWTORZ:I=ENT((X-ENT(X))×ばつJ)(1)9
DRUKUJWIERSZ:W
WROC

Explanation:

1. We get our S (here W) and N (here X)
2. We count how many characters there are in X and store it in J.
3. We check if \$\lfloor X \rfloor\$ is zero and if yes skip the next loop.
4. We loop for \$\lfloor X \rfloor\$ times and print the string.
5. We check how many is \$\lfloor (X - \lfloor X \rfloor) \cdot |,円J,円|)\rfloor\$.
6. We replace all characters after that with zeroes.
7. We print our shortened string.

Note: This will work only in KW6 encoding system, for other adjust the value in the first if statement. We also guess that the string is 9 characters long + string ending for lessening the bytes. It can be increased indefinitely according to specific needs.

Full programme version, 227 bytes

CALKOWITE:*W,I,J
BLOK(9):W
W(=I)=0
CZYTAJ:X
CZYTAJWIERSZ:W
J=-1
2)J=J+1
GDYW(J)=58:1,INACZEJ2
1)GDYENT(X)=0:3,INACZEJ4
*4)DRUKUJWIERSZ:W
POWTORZ:I=1(1)ENT(X)
*3)W(I)=0
POWTORZ:I=ENT((X-ENT(X))×ばつJ)(1)9
DRUKUJWIERSZ:W
STOP1
KONIEC

Answer 23

Perl 5 -lF, 43 + 2 = 45 bytes

$b=@F*<>;@F=(@F,@F)for 0..$b;say@F[0..$b-1]

Try it online!