# 1, 2, Fizz, 4, Buzz

## Introduction

In our recent effort to collect catalogues of shortest solutions for standard programming exercises, here is PPCG's first ever vanilla FizzBuzz challenge. If you wish to see other catalogue challenges, there is "Hello World!" and "Is this number a prime?".

## Challenge

Write a program that prints the decimal numbers from 1 to 100 inclusive. But for multiples of three print “Fizz” instead of the number and for the multiples of five print “Buzz”. For numbers which are multiples of both three and five print “FizzBuzz”.

## Output

The output will be a list of numbers (and Fizzes, Buzzes and FizzBuzzes) separated by a newline (either \n or \r\n). A trailing newline is acceptable, but a leading newline is not. Apart from your choice of newline, the output should look exactly like this:

1
2
Fizz
4
Buzz
Fizz
7
8
Fizz
Buzz
11
Fizz
13
14
FizzBuzz
16
17
Fizz
19
Buzz
Fizz
22
23
Fizz
Buzz
26
Fizz
28
29
FizzBuzz
31
32
Fizz
34
Buzz
Fizz
37
38
Fizz
Buzz
41
Fizz
43
44
FizzBuzz
46
47
Fizz
49
Buzz
Fizz
52
53
Fizz
Buzz
56
Fizz
58
59
FizzBuzz
61
62
Fizz
64
Buzz
Fizz
67
68
Fizz
Buzz
71
Fizz
73
74
FizzBuzz
76
77
Fizz
79
Buzz
Fizz
82
83
Fizz
Buzz
86
Fizz
88
89
FizzBuzz
91
92
Fizz
94
Buzz
Fizz
97
98
Fizz
Buzz


The only exception to this rule is constant output of your language's interpreter that cannot be suppressed, such as a greeting, ANSI color codes or indentation.

## Further Rules

• This is not about finding the language with the shortest approach for playing FizzBuzz, this is about finding the shortest approach in every language. Therefore, no answer will be marked as accepted.

• Submissions are scored in bytes in an appropriate preexisting encoding, usually (but not necessarily) UTF-8. Some languages, like Folders, are a bit tricky to score--if in doubt, please ask on Meta.

• Nothing can be printed to STDERR.

• Feel free to use a language (or language version) even if it's newer than this challenge. If anyone wants to abuse this by creating a language where the empty program generates FizzBuzz output, then congrats for paving the way for a very boring answer.

Note that there must be an interpreter so the submission can be tested. It is allowed (and even encouraged) to write this interpreter yourself for a previously unimplemented language.

• If your language of choice is a trivial variant of another (potentially more popular) language which already has an answer (think BASIC or SQL dialects, Unix shells or trivial Brainfuck derivatives like Alphuck and ???), consider adding a note to the existing answer that the same or a very similar solution is also the shortest in the other language.

• Because the output is fixed, you may hardcode the output (but this may not be the shortest option).

• You may use preexisting solutions, as long as you credit the original author of the program.

• Standard loopholes are otherwise disallowed.

As a side note, please don't downvote boring (but valid) answers in languages where there is not much to golf; these are still useful to this question as it tries to compile a catalogue as complete as possible. However, do primarily upvote answers in languages where the authors actually had to put effort into golfing the code.

## Catalogue

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• Nothing can be printed to STDERR. Is this true only when running, or also when compiling (assuming that is a separate step?) – AShelly Sep 24 '15 at 20:47
• @AShelly Only when running – Beta Decay Sep 24 '15 at 20:48
• I’m not sure I like the fact that you hardcoded the 100 into the challenge. That way, a program that just generates the expected output is a valid entry, but is not interesting for this challenge. I think the challenge should expect the program to input the number of items to output. – Timwi Sep 24 '15 at 23:28
• @Timwi While I agree that it would make it (only slightly) more interesting, I've very often seen FizzBuzz as strictly 1 to 100 (on Wikipedia and Rosetta Code, for example). If the goal is to have a "canonical" FB challenge, it makes sense. – Geobits Sep 25 '15 at 0:50
• A "vanilla fizzbuzz" sounds delicious. – Reinstate Monica -- notmaynard Sep 25 '15 at 15:12

# Forth (gforth), 95 bytes

: f 'e 1 DO 1 i 3 mod 0= if ." Fizz"0 then i 5 mod 0= if ." Buzz"0 then if i . then cr LOOP ; f


Try it online!

With help from @Bubbler, it's shorter than the other forth answer. Explanation coming soon.

# !@#$%^&*()_+, 378 bytes No TIO link, since the language has been updated to include the = command. ^!!($!!!!(!_++_^^^_$_^_$!!^^_+=!!_+^$!!_+$(+$_^_$!!_+)+!_++!!^^_+=!!_+^$!!_+$(+$_^_$!!_+)+!_+++!!_+^_$+_)+!!_+^$!!_+$(+$_^_$!!_+)+!_++$!_++(zziF@@@@ &!_+%!_+)+!!(!_++_^^^^^_$_^_$!!^^_+=!!_+^$!!_+$(+$_^_$!!_+)+!_++!!^^_+=!!_+^$!!_+$(+$_^_$!!_+)+!_+++!!_+^_$+_)+!!_+^$!!_+$(+$_^_$!!_+)+!_++$!_++(zzuB@@@@ &!_+%!_+)+ &($!#$!_+)
@^%$!_++!d_+!!_+^$!!_+$(+$_^_$!!_+)+!_++!!_+^_$+_$^$)


There are unprintables in this code. Here's an xxd dump:

00000000: 5e21 2128 2421 2121 2128 215f 2b2b 5f5e  ^!!($!!!!(!_++_^ 00000010: 5e5e 5f24 5f5e 5f24 2121 5e5e 5f2b 3d21 ^^_$_^_$!!^^_+=! 00000020: 215f 2b5e 2421 215f 2b24 282b 245f 5e5f !_+^$!!_+$(+$_^_
00000030: 2421 215f 2b29 2b21 5f2b 2b21 215e 5e5f  $!!_+)+!_++!!^^_ 00000040: 2b3d 2121 5f2b 5e24 2121 5f2b 2428 2b24 +=!!_+^$!!_+$(+$
00000050: 5f5e 5f24 2121 5f2b 292b 215f 2b2b 2b21  _^_$!!_+)+!_+++! 00000060: 215f 2b5e 5f24 2b5f 292b 2121 5f2b 5e24 !_+^_$+_)+!!_+^$00000070: 2121 5f2b 2428 2b24 5f5e 5f24 2121 5f2b !!_+$(+$_^_$!!_+
00000080: 292b 215f 2b2b 2421 5f2b 2b28 7a7a 6946  )+!_++$!_++(zziF 00000090: 4040 4040 0026 215f 2b25 215f 2b29 2b21 @@@@.&!_+%!_+)+! 000000a0: 2128 215f 2b2b 5f5e 5e5e 5e5e 5f24 5f5e !(!_++_^^^^^_$_^
000000b0: 5f24 2121 5e5e 5f2b 3d21 215f 2b5e 2421  _$!!^^_+=!!_+^$!
000000c0: 215f 2b24 282b 245f 5e5f 2421 215f 2b29  !_+$(+$_^_$!!_+) 000000d0: 2b21 5f2b 2b21 215e 5e5f 2b3d 2121 5f2b +!_++!!^^_+=!!_+ 000000e0: 5e24 2121 5f2b 2428 2b24 5f5e 5f24 2121 ^$!!_+$(+$_^_$!! 000000f0: 5f2b 292b 215f 2b2b 2b21 215f 2b5e 5f24 _+)+!_+++!!_+^_$
00000100: 2b5f 292b 2121 5f2b 5e24 2121 5f2b 2428  +_)+!!_+^$!!_+$(
00000110: 2b24 5f5e 5f24 2121 5f2b 292b 215f 2b2b  +$_^_$!!_+)+!_++
00000120: 2421 5f2b 2b28 7a7a 7542 4040 4040 0026  $!_++(zzuB@@@@.& 00000130: 215f 2b25 215f 2b29 2b00 2628 2421 2324 !_+%!_+)+.&($!#$00000140: 215f 2b29 0a40 5e25 2421 5f2b 2b21 645f !_+).@^%$!_++!d_
00000150: 2b21 215f 2b5e 2421 215f 2b24 282b 245f  +!!_+^$!!_+$(+$_ 00000160: 5e5f 2421 215f 2b29 2b21 5f2b 2b21 215f ^_$!!_+)+!_++!!_
00000170: 2b5e 5f24 2b5f 245e 2429                 +^_$+_$^$)  Alternatively, here's a version with no unprintables (387 bytes): ^!!($!!!!(!_++_^^^_$_^_$!!^^_+=!!_+^$!!_+$(+$_^_$!!_+)+!_++!!^^_+=!!_+^$!!_+$(+$_^_$!!_+)+!_+++!!_+^_$+_)+!!_+^$!!_+$(+$_^_$!!_+)+!_++$!_++(zziF@@@@!!_+&!_+%!_+)+!!(!_++_^^^^^_$_^_$!!^^_+=!!_+^$!!_+$(+$_^_$!!_+)+!_++!!^^_+=!!_+^$!!_+$(+$_^_$!!_+)+!_+++!!_+^_$+_)+!!_+^$!!_+$(+$_^_$!!_+)+!_++$!_++(zzuB@@@@!!_+&!_+%!_+)+!!_+&($!#$!_+)
@^%$!_++!d_+!!_+^$!!_+$(+$_^_$!!_+)+!_++!!_+^_$+_$^$)


## Notes

There's likely still room for golfing. I decided to approach this by manually calculating mod 3 and mod 5 for each integer from 1 to 100, whereas an approach by maintaining separate counters for mod 3 and mod 5 might save tons of bytes. This was more of a proof of concept.

I made this snippet using a templating language of my own creation, which I will publish at some point. Here's the metaprogram:

;let ~ !!_+^$!!_+$(+$_^_$!!_+)+!_++
;let DROP !_++
;let omx !!_+^_$+_ ^ !! ($
!!
!!(
DROP
_ 3{^} _
$_ 1{^} _$

-2=~-2=~+ omx
)+~$DROP ( @"Fizz" 0&!_+% !_+ )+ !!( DROP _ 5{^} _$ _ 1{^} _
$-2=~-2=~+ omx )+ ~$DROP
(
@"Buzz"
0&!_+%
!_+
)+

0& ($!#$!_+)

10@

^%
$DROP ! -100 +~omx$^$)  # GNU AWK, 60 bytes + 1 input byte An approach different from Cabbie407's, that golfed 2 bytes off thanks to the more flexible parsing of the GNU's implementation. Still needs one EOF input. END{for(;++n<101;print i?i:n)i=(n%3?e:"Fizz")(n%5?e:"Buzz")}  Try it online! END # starts after commanding the EOF (Ctrl+D). { for(; ++n<101; # the _n_ variable is started here on the conditional, at the value 1. print i?i:n # in the end of each loop, prints _i_ if non-null, or _n_. ) i=(n%3?e:"Fizz")(n%5?e:"Buzz") # two ternary conditionals that concatenates Fizz and Buzz accordingly. # by the way, _e_ is a not assigned variable, which will return a null string (""). # if n != 0 (mod 15), then _i_ will return null at the print statement above. }  For no input, use the BEGIN pattern instead (62 bytes total): BEGIN{for(;++n<101;print i?i:n)i=(n%3?e:"Fizz")(n%5?e:"Buzz")}  # uBASIC, 106 96 bytes Improved syntax, and fixed a bug 0ForI=1To100:S$="":IfIMod3=0ThenS$="Fizz" 1IfIMod5=0ThenS$=S$+"Buzz" 2IfS$=""Then?I,
3?S$4NextI  Try it online! # Canvas, 31 bytes Ａ２＾｛ｗ３％０≡Fizz×ｙ５％０≡Buzz×＋：？Ｏ］ｙＯ  Try it here! Okay seriously guys why is it that both ascii-art focused languages didn't already have a FizzBuzz? This uses my favourite FizzBuzz approach found in the Vyxal answer. # Python 3, 888577 73 bytes i=0 exec("i+=1;print(i%3*i%5and i or(i%3<1)*'Fizz'+(i%5<1)*'Buzz');"*100)  Try it online! # Hassium, 160 Bytes Here's it in Hassium. Surprised there's never been a FizzBuzz challenge before. There's also some lengthier (but more interesting) FizzBuzz examples here func main(){foreach(x in range(1,100)){if(x%15==0){println("fizzbuzz");}else if(x%3==0){println("fizz");}else if(x%5==0){println("buzz");}else println(x);}}  Run online and see expanded version here • @FryAmTheEggman Duly noted and added. Thanks. – Jacob Misirian Sep 25 '15 at 2:19 • 1. The capitalization of Fizz and Buzz is wrong. 2. <1 is shorter than ==0. 3. All curly braces ({}) can be eliminated. – Dennis Sep 25 '15 at 5:29 # Frink, 131 bytes This is still to be golfed, but because the docs are bare bones, it will be golfed through experimentation for x=1 to 100 { if x%15==0 { println["FizzBuzz"] } else { if x%3==0 { println["Fizz"] } else { if x%5==0 { println["Buzz"] } else { println[x] } } } }  • By my reading of the linked page only the for loop's braces are necessary. – Neil Sep 29 '15 at 14:56 • @Neil Then you'd need then which would increase the byte count – Beta Decay Sep 29 '15 at 16:01 • No, that's only if you want the controlled statement on the same line. – Neil Sep 29 '15 at 16:13 • I might have misread so you might still need the braces on the outer else clauses too. – Neil Sep 29 '15 at 16:36 # Groovy, 71 bytes (1..100).each{i->println i%15<1?'FizzBuzz':i%5<1?'Buzz':i%3<1?'Fizz':i}  # Swift, 77 bytes for i in 1...100{print(i%15<1 ?"FizzBuzz":i%3<1 ?"Fizz":i%5<1 ?"Buzz":"\(i)")}  • @RichardG.Nielsen It's considered poor etiquette to edit some else's golfed code. Suggesting it in a comment is fine, or you could post your own answer since you came up with it independently. – feersum Oct 27 '15 at 11:43 ## STATA, 115 bytes qui{ set ob 100 g a="Fizz" if!mod(_n,3) g b="Buzz" if!mod(_n,5) g c=a+b replace c=string(_n) if c=="" } l c,noo noh  qui{} suppresses output for everything in that block. First set the number of observations to be 100. Then generate variable a to be "Fizz" for every observation where its index number is divisible by 3. Then generate variable b to be "Buzz" for every observation where its index number is divisible by 5. Generate variable c to be the concatenation of these two. Then, replace c with the index number (STATA uses 1 indexing) if it is still an empty string. Then list the results of c in a table without observation numbers or headers. Only works in the "real" STATA interpreter. I need to add functions and conditions to the online interpreter for it to work there. A different solution in 116 bytes: forv x=1/100{ if!mod(x',3){ di"Fizz"_c if!mod(x',5) di"Buzz"_c } else if!mod(x',5) di"Buzz"_c else di x' _c di }  This solution goes through a for loop and checks whether the loop variable is divisible by 3 or not. If it is, it prints "Fizz". Then it checks if it is divisible by 5 and prints "Buzz". Otherwise, it checks if it is divisible by 5 and prints "Buzz". If not, it prints the loop variable. # zsh, 65 63 bytes repeat 100 x=|let ++i%3||x=Fizz&&let i%5||x+=Buzz&&<<<${x:-$i}  Changed echo to <<<. It's now 2 bytes shorter, because <<< doesn't need a space. repeat 100 x=|let ++i%3||x=Fizz&&let i%5||x+=Buzz&&echo${x:-$i}  # Python 2, 148 133 Bytes def f(n): if n%3+n%5<1:return"FizzBuzz" if n%5<1:return"Buzz" if n%3<1:return"Fizz" return n for x in map(f,range(1,101)):print x  • it is possible to save some bytes by reducing the amount of indentation. (1 space is sufficient) – Mohammad Sep 27 '15 at 18:12 • @Mhmd it's probably actually tabs, SE converts them to 4 spaces. – undergroundmonorail Sep 27 '15 at 22:18 • if n%3+n%5==0:return"FizzBuzz" -> if n%3+n%5==0:return f(3)+f(5) EDIT: nevermind i miscounted, it's the same – undergroundmonorail Sep 27 '15 at 22:19 • oh, but you can change ==0 to <1 everywhere it appears – undergroundmonorail Sep 27 '15 at 22:58 • This can be shortened significantly by removing the function and return logic: for i in range(100):print(i%3+i%5<1and'FizzBuzz')or(i%5<1and'Buzz')or(i%3<1and'Fizz')or i golfs it down to 89 bytes. – Skyler Oct 26 '15 at 13:47 ## UniBasic, 106 bytes FOR I=1 TO 100;D='';IF MOD(I,3)=0 THEN D='Fizz' IF MOD(I,5)=0 THEN D:='Buzz' IF D='' THEN D=I CRT D;NEXT I  ## C# using LINQ, 168 186 using System.Linq;class A{static void Main(){foreach(var s in Enumerable.Range(1,100).Select(n=>n%3==0?n%5==0?"FizzBuzz":"Fizz":n%5==0?"Buzz":n.ToString()))System.Console.WriteLine(s);}}  • Don't you need to include using System.Linq; so that Enumerable can be referenced? – Ken Gregory Sep 28 '15 at 20:13 • @KenGregory You're right - thank's for pointing that out. I forgot that LINQPad (where I tried it) adds this implicitly. I've edited my anwer accordingly. – Bill Tür Sep 28 '15 at 20:31 # Windows Batch, 172 @setlocal enableDelayedExpansion&for /l %%N in (1 1 100) do @(set v=&set/a1/(%%N%%3^)||set v=Fizz&set/a1/(%%N%%5^)||set v=!v!Buzz&if defined v (echo !v!)else echo %%N)2>nul  ## C#, 155 142 Bytes class a{static void Main(){for(int i=0;i++<100;){var s="";if(i%3<1)s="Fizz";if(i%5<1)s+="Buzz";if(s=="")s=i+"";System.Console.WriteLine(s);}}}  Added as an alternate approach to the example using LINQ Thanks @Riokmij! • You can replace the ==0's with <1, change the declaration of s with var instead of string, and replace the call to .ToString() by +"". Also, initial value for i should be 0. – Najkin Sep 29 '15 at 15:56 # MSX-BASIC, 106 bytes 1FORI=1TO100:IFIMOD3=0ANDIMOD5=0THEN?"FizzBuzz"ELSEIFIMOD3=0THEN?"Fizz"ELSEIFIMOD5=0THEN?"Buzz"ELSE?I 2NEXT  The one-liner version to be executed in direct mode would be 120 bytes because all of the extra NEXTs needed before the ELSEs: FORI=1TO100:IFIMOD3=0ANDIMOD5=0THEN?"FizzBuzz":NEXTELSEIFIMOD3=0THEN?"Fizz":NEXTELSEIFIMOD5=0THEN?"Buzz":NEXTELSE?I:NEXT  • IFIMOD3=0ANDIMOD5=0 -> IFIMOD15=0? I think MSX BASIC could be tokenized too, which would decrease your byte count. – lirtosiast Sep 30 '15 at 14:53 # rs, 92 91 bytes (_)^^(100) +^(_+)(_)/\1 \1\2 \b((___)+)\b/Fi;\1 \b(_{5})+\b/Bu; ;_*/zz \b(_+)\b/(^^\1) /\n  Saved 1 byte thanks to @MartinBüttner! Live demo. (It may take a bit to run!) • Do you not have \0 in rs? You can probably save some bytes either way by using the trick from my Retina answer: \b((___)+)\b/Fi;\1 ... \b(_{5})+\b/Bu; ... ;_+/zz – Martin Ender Sep 28 '15 at 14:38 • @MartinBüttner Thanks! I updated the post. What exactly do you mean by \0? – kirbyfan64sos Oct 1 '15 at 19:44 • Something to reference the entire match, not just a capturing group. – Martin Ender Oct 1 '15 at 19:59 • @MartinBüttner I guess not. :( – kirbyfan64sos Oct 1 '15 at 20:05 • @MartinBüttner I'm just using Python's built-in substitution thing. I can easily override it, though... – kirbyfan64sos Oct 1 '15 at 20:06 # OCaml, 106 for i=1to 100do let(!)n=i mod n<1and p=Printf.printf in!3&p"Fizz"=();!5&p"Buzz"=()or!3||p"%d"i=();p" "done  Apparently this isn't a very good attempt as the shortest one on anarchy golf is only 97. # Scala, 90 bytes for(i<-1 to 100)println{var s="";if(i%3==0)s="Fizz";if(i%5==0)s+="Buzz";if(s=="")i else s}  # Dart, 88 main({i:0}){while(i<100)print(["FizzBuzz","Fizz","Buzz",++i][(i%3).sign*2+(i%5).sign]);}  Dart is somewhat hampered in the golfing by not having conversion between bool and int, but the sign getter on integers helps a little. # Groovy, 83 80 bytes (1..100).each{def f=it%3,b=it%5;println!f&&!b?'FizzBuzz':!f?'Fizz':!b?'Buzz':it}  # scg, 51 bytes 1á01°r{d[[d"Buzz"]["Fizz"d"Buzz"+]]\3%!@\5%!@" "}m  So, does this mean that scg is a real language now? Explanation: 1 .- adds 1 to the stack á01 .- adds 101 to the stack °r .- range, adds array with 1-100 on the stack { .- start function for use in map d .- duplicates number [ .- array [ d .- duplicate number again, ends up in array "Buzz" .- wonder what this does ] .- end array [ "Fizz" d .- duplicate fizz "Buzz"+ .- ends up with "FizzBuzz" ] ] .- end array. Ends up with a 2D array \ .- gets number to calculate to the top 3% .- mod 3 ! .- not, so any above 0 int turns to 0 and 0 turns to1 @ .- get array value. Now you have two choices for output \5%!@ .- same as above but for 5. .- now we have the correct fizzbuzz value "\n" .- pushes newline. I do not have variables yet so no shortcuts }m .- end function, map. output is implicit  # Rust, 145137 131 bytes ## Golfed fn main(){for i in 1..101{let s=i.to_string();println!("{}",match i%15{0=>"FizzBuzz",3|6|9|12=>"Fizz",5|10=>"Buzz",_=>&(s)[..]});}}  ## Ungolfed fn main() { for i in 1..101 { let s = i.to_string(); println!("{}", match i % 15 { 0 => "FizzBuzz", 3|6|9|12 => "Fizz", 5|10 => "Buzz", _ => &(s)[..] }); } }  Uses the current stable version of Rust (1.5.0). • You can remove a semicolon after println! call, as println! returns (). Also, it's possible to replace &(s)[..] with &s. – Konrad Borowski Jul 6 '18 at 7:53 # 𝔼𝕊𝕄𝕚𝕟, 32 chars / 47 bytes ⩥ṤⓜᵖFizzBuzzė⧺_%3⅋4,_%5?4:8)⋎_  Try it here (Firefox only). # C#, 174 bytes void A(){for(int x=1;x<101;x++){if(x%15<1)Console.Write("FizzBuzz\n");else if(x%3<1)Console.Write("Fizz\n");else if(x%5<1)Console.Write("Buzz\n");else Console.WriteLine(x);}}  Ungolfed: void A(){ for (int x = 1; x < 101; x++) { if (x % 15 < 1) Console.Write("FizzBuzz\n"); else if (x % 3 < 1) Console.Write("Fizz\n"); else if (x % 5 < 1) Console.Write("Buzz\n"); else Console.WriteLine(x); } }  # Racket, 125 122 bytes (for([x(range 1 101)])(define(m n)(=(modulo x n)0))(displayln(cond[(and(m 3)(m 5))'FizzBuzz][(m 3)'Fizz][(m 5)'Buzz][x])))  Simplest approach, took some work to get it lower than 130 bytes. Inspired by the Java example. ## Pretty-printed code (for ([x (range 1 101)]) (define (m n) (= (modulo x n) 0)) (displayln (cond [(and (m 3) (m 5)) 'FizzBuzz] [(m 3) 'Fizz] [(m 5) 'Buzz] [x])))  # TCL, 208 bytes Golfed: set i 1;while {$i<101} {set p "";if {[expr $i % 3]==0} {set p [concat$p {fizz}]};if {[expr $i % 5]==0} {set p [concat$p {buzz}]};if {[expr $i % 3]>0&&[expr$i % 5]>0} {set p [concat $p$i]};puts $p;incr i;}  Ungolfed: set i 1 while {$i<101} {
set p ""
if {[expr $i % 3]==0} {set p [concat$p {fizz}]}
if {[expr $i % 5]==0} {set p [concat$p {buzz}]}
if {[expr $i % 3]>0 && [expr$i % 5]>0} {set p [concat $p$i]}
puts \$p
incr i
}


## Kotlin, 145 bytes

fun main(a:Array<String>){IntRange(1,100).forEach{val f=it%3==0;val b=it%5==0;var x="";if(f)x+="Fizz";if(b)x+="Buzz";if(!f&&!b)x+=it;println(x)}}


Try it online!