1, 2, Fizz, 4, Buzz

Question

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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Answer 1

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!

Answer 2

Kotlin, 119 115 bytes

fun main(a:Array<String>){for(i in 1..100)println(if(i%3<1)"Fizz" else ""+if(i%5<1)"Buzz" else if(i%3<1)"" else i)}

Try it Online!

Answer 3

Python 2, 91

1;exec"print'FizzBuzz'if _%3==_%5==0else'Fizz'if _%3==0else'Buzz'if _%5==0else _;_+=1;"*100

Answer 4

Python 91 Bytes

for n in range(100):
    c=""
    if n%3==0:c+="fizz"
    if n%5==0:c+="buzz"
    if c=="":c=n
    print c

Answer 5

Elixir, 182 bytes

import Stream
f=fn(n)->(zip(cycle(["","","fizz"]),cycle(["","","","","buzz"]))|>zip(iterate(1,&(&1 + 1)))|>map(fn{{"",""},n}->n
{{p,o},_}->p<>o end))|>take(n)|>Enum.each(&IO.puts/1)end

LiveDemo

Calling: f.(100)

Answer 6

Ruby, 72 bytes

(1..100).each{|n|puts "#{n%3==0?'Fizz':''}#{n%5==0?'Buzz':n%3!=0?n:''}"}

Explanation

For each number from 1 to 100, print 'Fizz' if the number mod 3 is 0, then if the number mod 5 is 0, print 'Buzz' else if the number mod 3 is 0, print the number.

Answer 7

Groovy, 61 bytes

100.times{a=++it%3?"":"fizz"
a+=it%5?"":"buzz"
println a?:it}

Answer 8

Mathematica, 61 Bytes

Heavily inspired by this post on the mathematica stack exchange

fizzbuzz[#,,fizz,,buzz][[#~GCD~15~Mod~15]]&~Array~100//Colum‌​n

Answer 9

Plain Javascript (no console.log() & no alert()), 64 bytes

for(f=b='';f++<100;b+=(f%5?z||f:z+'Buzz')+'\n')z=f%3?'':'Fizz';b

Just copy/paste into any javascript console and hit enter.

Answer 10

REXX, 94 bytes

f.=
b.=
f.0='Fizz'
b.0='Buzz'
do i=1 to 100
  f=i//3
  b=i//5
  r=f.f||b.b
  if r<i then r=i
  say r
  end

Alternative 94 byte solution:

f='FizzBuzz'
do i=1 to 100
  r=left(f,(i//3=0)*4)right(f,(i//5=0)*4)
  if r='' then r=i 
  say r
  end

Answer 11

Lua, 126 bytes

for i=1,100 do if i%15==0 then print"FizzBuzz"elseif i%3==0 then print"Fizz"elseif i%5==0 then print"Buzz"else print(i)end;end

Answer 12

K, 52 Bytes

-1@{,/$$[#i:&~.q.mod[x;3 5];`Fizz`Buzz i;x]}'1+!100;

Thanks

Answer 13

Turtlèd, 221 bytes

99;,:[*,l]u[*d{*u,dr}ul' l[ l]ru]u3;[ [ r]l[ ' l]ll"Fizz";]<97:>5[ (*[ r]l[ ' l]ll)(zr)"Buzz";]<104:>[ lll'0rrr[ r]l{*' l{*l}{ l}(9'0l( r"1!")(9"10!"))(8'9)(7'8)(6'7)(5'6)(4'5)(3'4)(2'3)(1'2)(0'1)(!'0):{ l}}[ l]drrrr{zd}]

due to a bug (having two close curly brackets next to each other causes the second one to match to the first ones open bracket), the Try it online version is different for now (I put a space between the two close curly brackets)

99;,:[*,l]u[*d{*u,dr}ul' l[ l]ru]u3;[ [ r]l[ ' l]ll"Fizz";]<97:>5[ (*[ r]l[ ' l]ll)(zr)"Buzz";]<104:>[ lll'0rrr[ r]l{*' l{*l}{ l}(9'0l( r"1!")(9"10!"))(8'9)(7'8)(6'7)(5'6)(4'5)(3'4)(2'3)(1'2)(0'1)(!'0):{ l} }[ l]drrrr{zd}]

Try it online!

I might write an explanation if I get time motivated

Answer 14

Ruby, 68 65 bytes

q='Buzz';(1..100).map &->e{p e%3<1?'Fizz'+(e%5<1?q:''):e%5<1?q:e}

Answer 15

braingasm, 40 bytes

The language is coming along nicely. A few recent features allows for a quite decent FizzBuzz, if I may say so myself:

100[>#3p["Fizz".+]#5p["Buzz".+]z[#:]10.]

Here's how it works:

100[               One hundred times:
    >                Go to the next cell.
    #3p[             If current cell number is divisble by 3:
        "Fizz".        Print "Fizz".
        +              Increment current cell
    ]
    #5p["Buzz".+]    Same thing for 5 and "Buzz".
    z[               If the current cell is 0 (hasn't been incremented):
      #:               Print current cell number
    ]
    10.              Print a newline
]

Answer 16

Python 2, 106 bytes

for n in range(1,101):
    a="Fizz"*int(n%3==0)+"Buzz"*int(n%5==0)
    if a=="": print n
    else: print a

Try it online!

Answer 17

Python 2.7, 59 Bytes

for i in range(100):print i%3//2*'Fizz'+i%5//4*'Buzz'or i+1

Explanation:

for i in range(100)

This will generate a list of numbers from 1 to 100 and assign it to the i variable.

print i%3//2*'Fizz'+i%5//4*'Buzz'or i+1

Note that this loop will repeat 100 times, and in each time, the loop will do what this line is commanding.

Answer 18

AsciiDots, 152 bytes

 #
/$""
\~\
 @\-\
[=]@*-\
/@----*-~&
|0/---*[>]
|@\<>$_"Buzz"@1)
|*-~/ /-/
||[%]-\>#100)
v\-*#5*/
|.//
| \<>$_"Fizz"@1)
|/-~/
@|[%]-\
0\-*#3/
*-[+]
\#1/

Try it online!

Answer 19

Recursiva, 46 bytes

EmB100"%a15:%a5:%a3:a!'Fizz'!'Buzz'!'FizzBuzz'

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Explanation:

EmB100"%a15:%a5:%a3:a!'Fizz'!'Buzz'!'FizzBuzz'
E                                              - Stringify and join with new-lines
 m                                             - map with
  B100                                         - Range [1,2..100]
      "                                        - mapping function start
       %a15:%a5:%a3:a!'Fizz'!'Buzz'!'FizzBuzz' - string as per divisibility

%a15:%a5:%a3:a!'Fizz'!'Buzz'!'FizzBuzz' roughly translates to this if-else block:

if(a%15):
    elif(a%5):
        elif(a%3):
            return a
        else:
            return 'Fizz'
    else:
        return 'Buzz'
else:
     return 'FizzBuzz'  

Answer 20

Extremely late answer and for a language that has already a bit too many answers, however, i've not seen a single function answer that actually RETURNS the expected string, instead of printing it to console, so here it is:

Javascript 73 bytes

(i=0,r="")=>i>99?r:f(++i,r+=(i%5?i%3?i:'Fizz':i%3?'Buzz':'FizzBuzz')+'
')

Answer 21

Funky, 59 bytes

for(i=1i<100i++)print((("Fizz"*1&!i%3)+("Buzz"*1&!i%5))ori)

Try it online!

Answer 22

nodejs repl, 60 bytes

for(i=0;++i<101;util.puts(i%5?s||i:s+'Buzz'))s=i%3?'':'Fizz'

With latest node, this prints a deprecation warning to the console. This presumably violates the rule about printing to STDERR, but using an older version of node will fix this.

Answer 23

APL Nars 93 chars

h;i;j
i←0
L:j←0⋄i+←1⋄→A×⍳0≠3∣i⋄⍞←'Fiz'⋄j←1
A:→B×⍳0≠5∣i⋄⍞←'Buz'⋄j←1
B:→C×⍳j=0⋄→D
C:⍞←i
D:⎕←''⋄→L×⍳i≤99

Indented:

 ∇ h;i;j
  i←0
L:j←0⋄i+←1⋄→A×⍳0≠3∣i⋄⍞←'Fiz'⋄j←1
A:         →B×⍳0≠5∣i⋄⍞←'Buz'⋄j←1
B:         →C×⍳j=0⋄→D
C:    ⍞←i
D:    ⎕←''⋄→L×⍳i≤99 ∇

Answer 24

Clean, 148 + 2 = 150 bytes

+2 for the -b compiler flag

module m
import StdEnv
f[0,0,_]="FizzBuzz"
f[a,b,n]|a==0="Fizz"|b==0="Buzz"=fromInt n
Start=flatlines[fromString(f[n rem 3,n rem 5,n])\\n<-[1..100]]

Try it online!

Answer 25

C, 152 bytes

#include<stdio.h>
#define p printf
int i,f;main(){while(i++<100){if(i%3==0&&(f=1))p("Fizz");if(i%5==0&&(f=1))p("Buzz");if(f==0||(f=0))p("%d",i);p("\n");}}

Ungolfed one:

#include<stdio.h>
#define p printf
int i,f;
main()
{
    while(i++<100)
    {
        if(i%3==0&&(f=1))p("Fizz");
        if(i%5==0&&(f=1))p("Buzz");
        if(f==0||(f=0))p("%d",i);
        p("\n");
    }
}

Answer 26

C (gcc), 101 bytes

#define P printf
main(i){for(i=0;i++<100;P("\n"))(i%3<1&&P("fizz"))+(i%5<1&&P("Buzz"))<1&&P("%u",i);}

It could have one indefinite behavior in the use + operation, the order of evaluation could be swapped. Try it online!

Answer 27

Kotlin, 115 bytes

fun main(a:Array<String>){(1..100).map{println(when(it%15){0->"FizzBuzz";3,6,9,12->"Fizz";5,10->"Buzz";else->it})}}

Try it online!

Answer 28

face, 152 bytes

(FizzBuzz%d@
)\F*,c'Fo>m+*1+m=*m?*m1+m3+m4+m5+11334455$BF"B4$%B"%4$N%"N3:~0?%=+3?=31?w=F4>:3%=+5?=51?w=B4>:5??+p="%c+w="=>:+w=N1>+++1*=55*==4+==1-=+=?=~

Annotated:

(FizzBuzz%d@
)
\F*,c'Fo>               ( obtain data pointer, a number, and stdout )
m+*1+                   ( counter variable, goes from 1 to 100 )
m=*m?*                  ( result and temp variables )
m1+m3+m4+m5+11334455    ( relevant constants )
$BF"B4$%B"%4$N%"N3      ( get pointers to Fizz, Buzz, %d, and \n )
:~
    0?                  ( has either Fizz or Buzz been printed? )
    %=+3?=3             ( skip to label 3 if not divisible by 3 )
    1?w=F4>:3           ( otherwise, print "Fizz" )
    %=+5?=51?w=B4>:5    ( same for "Buzz" )
    ??+p="%c+w="=>:+    ( if we didn't print anything, print the number )
    w=N1>               ( print a newline )
    +++1                ( increment the counter )
    *=55*==4+==1-=+=    ( set the result variable to 100-n )
?=~

Try it online! (The trailing newline is required on TIO due to a bug that's been fixed in a newer version of face.)

Answer 29

J, 73 Bytes

}.":`('Fizz'"_)`('Buzz'"_)`('FizzBuzz'"_)@.((0:=3&|)++:@(0:=5&|))"0 i.101

Definitely not the greatest, looking for ways to improve this.

Explanation:

                                                                 "0 i.101  | To each number from 0 to 100
                                           ((0:=3&|)++:@(0:=5&|))          | 1 if divisble by three, 2 if by 5, 3 if by both
                                         @.                                | Using the result of the previous verb, select a verb from the following gerund
  ":`('Fizz'"_)`('Buzz'"_)`('FizzBuzz'"_)                                  | Apply the appropriate Fizz-Buzz string
}.                                                                         | Remove the first entry (since we counted 0)

Answer 30

17, 229 bytes

17 was made after the challenge was made, but not designed to be any good at it. This code could probably be golfed a bit smaller, but it is a reasonably hard language to do anything with. Golfed version(removed unneeded space and duplicated numbers where useful):

0{2 #
1 +
:
2 @
5g ==
1 +
0 @}1{2 # 3 % ! 3 *
2 # 5 % ! 5 *
+ 10 + 0 @}13{42 $ 63 $ 73 : $ $ a $
0 0 @}15{3f $ 6f $ 73 : $ $ a $
0 0 @}18{42 $ 63 $ 73 : : : $ $ 3f $ 6f $ $ $ a $
0 0 @}10{2 # $$ a $
0 0 @}}777{0 2 @
0 1 @
0 0 @}

Ungolfed version:

0 {
2 #
1 +
:
2 @
5g ==
1 +
0 @
}

1 {
2 # 3 % ! 3 *
2 # 5 % ! 5 *
+ 10 + 0 @
}

13 {
42 $ 63 $ 73 $ 73 $ a $
0 0 @
}

15 {
3f $ 6f $ 73 $ 73 $ a $
0 0 @
}

18 {
42 $ 63 $ 73 $ 73 $ 3f $ 6f $ 73 $ 73 $ a $
0 0 @
}

10 {
2 # $$ a $
0 0 @
}


777 {
0 2 @
0 1 @
0 0 @
}

Explanation:

Block 0: Load value at 2, add 1, duplicate, store at 2, if value == 100 push 1, else 0, add 1, store at 0.

Block 1: Will only be ran after block zero if value != 100. Load from 2, mod 3, not, times 3. If value at 2 is a multiple of 3 it will be 3, else 0. Same again for 5, add them and 17(looks like 10, but remember it uses base 17) and stores value at 0.

Block 13(or 20): Will only be run after 1 if value is a multiple of 3, not 5. Push and print ascii values for Fizz\n

Block 15(or 22): Will only be run after 1 if value is a multiple of 5, not 3. Push and print ascii values for Buzz\n

Block 18(or 25): Will only be run after 1 if value is a multiple of 3 and 5. Push and print ascii values for FizzBuzz\n

Block 10(or 17): Will only run after 1 if value is not multiple of 3 or 5. Prints numeric form of value.

Block 777(or 2149): Like main for C++, first block to be run