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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body{text-align:left!important}#answer-list{padding:10px;width:290px;float:left}#language-list{padding:10px;width:290px;float:left}table thead{font-weight:700}table td{padding:5px}

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <link rel="stylesheet" type="text/css" href="//cdn.sstatic.net/codegolf/all.css?v=83c949450c8b"> <div id="language-list"> <h2>Shortest Solution by Language</h2> <table class="language-list"> <thead> <tr><td>Language</td><td>User</td><td>Score</td></tr> </thead> <tbody id="languages"> </tbody> </table> </div> <div id="answer-list"> <h2>Leaderboard</h2> <table class="answer-list"> <thead> <tr><td></td><td>Author</td><td>Language</td><td>Size</td></tr> </thead> <tbody id="answers"> </tbody> </table> </div> <table style="display: none"> <tbody id="answer-template"> <tr><td>{{PLACE}}</td><td>{{NAME}}</td><td>{{LANGUAGE}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr> </tbody> </table> <table style="display: none"> <tbody id="language-template"> <tr><td>{{LANGUAGE}}</td><td>{{NAME}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr> </tbody> </table>

Answer 1

Pyt, 91 bytes

11`ŕĐĐ5%⇹3%?ŕ:ŕ8⬡2+ƇĐ2ᴇ5+Ƈ7⬠Ƈ5Ș;?ŕ:ŕ8⬡2+ƇĐ9⬠Ƈ6⬡Ƈ4Ș;ǰĐĐąɬɔĐƩ?ŕ⇹ą⇹*ǰ⇹ąɬą\ǰƖ⇹ƥ:ŕŕŕƖĐƥ;⁺Đ2ᴇ⁺<łĉ

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Code	Action
1	Pushes 1
1`ŕ...ł	Do... while top of stack is truthy
ĐĐ5%⇹3%	Gets n mod 5 and n mod 3
?ŕ:ŕ8⬡2+ƇĐ2ᴇ5+Ƈ7⬠Ƈ5Ș;	If n mod 3 is 0, push "Fizz"
?ŕ:ŕ8⬡2+ƇĐ9⬠Ƈ6⬡Ƈ4Ș;	If n mod 5 is 0, push "Buzz"
ǰĐĐąɬɔĐƩ?ŕ⇹ą⇹*ǰ⇹ąɬą\ǰƖ⇹ƥ:ŕŕŕƖĐƥ;	If the stack contains any letters, print only the letters; otherwise, print the number
⁺Đ2ᴇ⁺<	Is the number less than 100?
ĉ	Clears the stack

Answer 2

Thunno H N, \$ 27 \log_{256}(96) \approx \$ 22.22 bytes

e1+D35d%0="FBiuzzzz"Zlz*Js~

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Thunno H d, \$32 \log_{256}(96) \approx\$ 26.34 bytes

R{1+zt3%!"Fizz"*s5%!"Buzz"*+s~ZK

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Explanations:

           # The H flag implicitly pushes 100 to the stack
e1+        # Map over range(100) and add one each time:
   D       #   Duplicate the number
    35d    #   Push the digits of 35 - [3, 5]
       %0= #   Push [divisible by 3, divisible by 5]
"FBiuzzzz" #   Push string "FBiuzzzz"
Zl         #   Uninterleave to get ["Fizz", "Buzz"]
  z*J      #   Multiply element-wise and join
     s~    #   Swap and logical or with the number
           # After the map, the N flag joins by newlines

           # The H flag implicitly pushes 100 to the stack
R{         # Loop through range(100):
  1+       #   Add one
    zt     #   Triplicate so it is on the stack three times
      3%!  #   Is it divisible by 3?
"Fizz"*    #   Multiply this by the string "Fizz"
           #   i.e. push "Fizz" if it's divisible by 3, "" otherwise
       s   #   Swap so the number is back on top
5%!        #   Is it divisble by 5?
   "Buzz"* #   Multiply this by the string "Buzz"
           #   i.e. push "Buzz" if it's divisible by 5, "" otherwise
+          #   Add them togther, so we have "Fizz", "Buzz", "FizzBuzz", or ""
 s~        #   Swap and perform `or`, so we get the number if it's ""
   ZK      #   Print with a newline
           # After the loop, the d flag stops the implicit output

Answer 3

Minecraft Function, 666 bytes

scoreboard objectives add c dummy
scoreboard players add c c 1
scoreboard players set 0 c 0
scoreboard players set 3 c 3
scoreboard players set 5 c 5
scoreboard players operation f c = c c
scoreboard players operation b c = c c
scoreboard players operation f c %= 3 c
scoreboard players operation b c %= 5 c
execute if score f c > 0 c if score b c > 0 c run tellraw @a {"score":{"name":"c","objective":"c"}}
execute if score f c = 0 c if score b c > 0 c run tellraw @a "Fizz"
execute if score f c > 0 c if score b c = 0 c run tellraw @a "Buzz"
execute if score f c = 0 c if score b c = 0 c run tellraw @a "FizzBuzz"
execute if score c c matches ..99 run function a:b

Must be run as a function named a in a data pack named b. I think this is about optimal for a scoreboard based solution.

Commented version:

# Create scoreboard to store variables
scoreboard objectives add c dummy
# Increment counter by 1 (starts at 0)
scoreboard players add c c 1
# Set constants 0, 3, and 5 for comparison and modulo
scoreboard players set 0 c 0
scoreboard players set 3 c 3
scoreboard players set 5 c 5

# Copy value of counter to f and b and modulo them by 3 and 5 respectively.
scoreboard players operation f c = c c
scoreboard players operation b c = c c
scoreboard players operation f c %= 3 c
scoreboard players operation b c %= 5 c

# If f and b are both > 0, print the counter
execute if score f c > 0 c if score b c > 0 c run tellraw @a {"score":{"name":"c","objective":"c"}}
# If f = 0 and b > 0, print Fizz
execute if score f c = 0 c if score b c > 0 c run tellraw @a "Fizz"
# If f > 0 and b > 0, print Buzz
execute if score f c > 0 c if score b c = 0 c run tellraw @a "Buzz"
# If both are 0 print FizzBuzz (Minecraft doesn't support string concatenation so this is necessary)
execute if score f c = 0 c if score b c = 0 c run tellraw @a "FizzBuzz"
# Run the function again if the counter is <= 99
execute if score c c matches ..99 run function a:b

Answer 4

Aubergine, 225 bytes ( 222 bytes with error )

:Ba=aA-A1=ba-b1:BA-Ai+Ai+a1=bi=oA+a1:bA+a1+iA=a1-ii?		\0B*Fizz\0\0FBuzz\0[-aa=aA-aA+ai-ai=bi-bi+iA=ab-a1=oA=oB=bi-bi+B1+b1=oB+b1-aa:aB+ii=a1=aA+A1+A1-aa=aA-a1-a1=bA-a1-bA=Ab-ib-a1-AA-a1-AA-aa-a1-A1-a1=bA-a1-Ab-a1+A1-a1-iAo01

* four \0 mean NUL(ASCII 0) character

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The equivalent pseudo code in C is like below;

#include <stdio.h>
#include <stdint.h>

#define I(x) ((intptr_t)&&x)
#define P(x) *(void*)(x)

static void nop(void) {}

int main(void) {
  intptr_t
    *cp,*cpt,t,p,
    pvars0[] =
      { I(Tj3)-I(Tj1), I(Tj2)-I(Tj1), I(Tj2)-I(Tj1), 0, I(Tc1)-I(Tj3),
        I(Lb), 3, 'F', 'i', 'z', 'z', 0, I(Lb)-I(Tx2), },
    pvars1[] =
      { I(Lj), 5, 'B', 'u', 'z', 'z', 0, I(Lc)-I(Tx2), },
    dvars[] = { I(Tc2)-I(Tj3), '0', '1', '\n', 10, },
    *pbase[] = { &pvars0[6], &pvars1[1] },
    *dbase3=&dvars[2], *dbase1=&dvars[4];
  int i=0;

Lx: // Fizz/Buzz print ( i=0 for Fizz, i=1 for Buzz )
  // decrement counter
  cp = pbase[i];
  *cp-=1;
  if ( *cp!=0 ) goto P(*(cp-1));  // skip printing and jump to Lb/Jj
  *cp+=3; // restore the counter to 3 ( add more 2 for Buzz in Lc )
  cp+=1;
  p=I(Tx1);
Tx1: // putchar loop
  putchar(*cp);
  cp+=1;
  if ( *cp!=0 ) goto P(p); // jump to Tx1
  goto P(I(Tx2)+*(cp+1)); // jump to Lb/Lc
Tx2:

Lb: // Buzz print
  i=1;
  goto Lx;

Lj: // number print and post process
  cp=pbase[0];
  // select &pvars[0] thru &pvars[2] as jump offset
  cp-=*cp;
  cp-=3;
  cpt=dbase3;
  goto P( I(Tj1)+*cp ); // jump to Tj3(when Fizz) or Tj1/Tj2
Tj1: // print the upper digit
  cp=cpt-1;
  putchar(*cp);
Tj2: // print the lower digit
  putchar(*cpt);
Tj3: // print a new line and increment the lower digit
  cp=dbase3;
  *cp+=1;
  cp+=1;
  putchar(*cp);
  cp+=1;
  i=0;
  if ( *cp!=0 ) goto Lx; // next iteration
  return 0; // exit

Lc: // after Buzz
  // add more 2 to the Buzz counter
  cp=pbase[1];
  *cp+=2;
  // switch pvars0[3] ( jump offset ) between 0 and pvars0[4]
  cp=pbase[0]-2;
  t=*cp;
  cp-=1;
  t-=*cp;
  *cp=t;
  goto P( I(Tc1)-t ); // jump to Tc1/Tj3
Tc1: // carry up
  // reset jump offsets in pvars[1],pvars[2]
  // so that the 1st "goto" in Lj jump to Tj1 instead of Tj2
  cp-=1;
  *cp=0;
  cp-=1;
  *cp=0;
  // decrement global loop counter
  cp=dbase1;
  *cp-=1;
  // decrease the lower digit by 10
  cp-=1;
  t=*cp;
  cp-=1;
  *cp-=t;
  // increment the upper digit
  cp-=1;
  *cp+=1;
  cp-=1;
  goto P(I(Tc2)-*cp); // jump to Tj3
Tc2:
  nop(); // not reached
}

Each code block handles;

• Label Lx
  Print Fizz ( when i=0 ) or Buzz ( when i=1 ) if n is the multiple of 3 or 5.
• Label Lb
  Just jump to Lx after setting i=1.
• Label Lj
  Print n itself ( when Fizz or Buzz are not printed ) and do post-process.
• Label Lc
  Carry-up operation.

In Aubergine code, each address range corresponds to;

• 0 thru 2
  The 1st instruction :Ba has no effect, but stores pbase[].
  pbase[0] is ord(':')=58, pbase[1] is ord('B')=66.
• 3 thru 50
  The code block with label Lx.
  Lx: =aA-A1=ba-b1:BA-Ai+Ai+a1=bi=oA+a1:bA+a1+iA, Tx1: =a1-ii
  * set i=0 ( -ii ) to implement goto Lx
• 51
  A blank area.
• 52 thru 72
  Stores pvars0[] and pvars1[]
• 73 thru 135
  The code block with Label Lj.
  Lj: -aa=aA-aA+ai-ai=bi-bi+iA, Tj1: =ab-a1=oA, Tj2: =oB, TJ3: =bi-bi+B1+b1=oB+b1-aa:aB+ii
  The last instruction +ii jumps to an address out of range to stop the program.
• 136 thru 219
  The code block with Label Lc.
  Lc: =a1=aA+A1+A1-aa=aA-a1-a1=bA-a1-bA=Ab-ib, Tc1: -a1-AA-a1-AA-aa-a1-A1-a1=bA-a1-Ab-a1+A1-a1-iA
• 220 thru 224
  Stores dvars[]
  This area is accessed via minus addresses such as dbase1=-1 or dbase3=-3.

The 222 bytes version, which is shown below, outputs some error.

:Ba=aA-A1=ba-b1:BA-Ai+Ai+a1=bi=oA+a1:bA+a1+iA=a1-ii?		\0Q*Fizz\0\0\x9aBuzz\0=a1=aA+A1+A1-aa=aA-a1-a1=bA-a1-bA=Ab+ib-a1-AA-a1-AA-aa-a1-A1-a1=bA-a1-Ab-a1+A1-a1+iA-aa=aA-aA+ai-ai=bi-bi+iA=ab-a1=oA=oB=bi-bi+B1+b1=oB+b1-aa:aB$01

* \0 is NUL, and \x9a is a byte 232

This code cannot be run on TIO because TIO requires codes to be UTF-8 clean. ( a single \x9a is not valid as UTF-8 byte sequence )

The differences between 225B version and this 222B version are;

• Exchange the position in code of the Lj code block and the Lc code block.
• Replace relative backward jump instructions -ib, -iA in Lc code block to forward jumps +ib, +iA.
• Omit the instruction +ii, which works to stop the code, from Lj code block.
  * The code stops in error when the instruction pointer points the dvars area just after the Lj code block.

Answer 5

sclin, 63 bytes

100I-a \; tap
"Fizz""Buzz", over3 5, % ! ** c>< dup \pop |# n>o

Try it on scline!

My collection grows once again

Answer 6

gnuplot, 72 68 bytes

• -4 bytes thanks to noodle person

se pr'-'
do for[i=1:100]{pr i%15?i%5?i%3?i:"Fizz":"Buzz":"FizzBuzz"}

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Not sure whether the statement se pr'-' (set output to the console) would be allowed to be moved to the header, so, let's keep it in the body.

Answer 7

J, 47 bytes

echo(0{[:\:~":,:'FizzBuzz'#~4#0=3 5&|)"0>:i.100

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or, if you assume implicit echo like in Jconsole, as the other leading J solution here does, then you can remove the echo from the start for 43 bytes.

There seem to be shorter solutions on code.golf but solutions are private there so I don't know what they are… For now this can be the shortest we have on CGCC.

   NB. this runs for each 1 to 100, consider the 80th line.

   (0=3 5&|) 80
0 1
   (4#0=3 5&|) 80
0 0 0 0 1 1 1 1
   ('FizzBuzz'#~4#0=3 5&|) 80
Buzz
   (":,:'FizzBuzz'#~4#0=3 5&|) 80
80
Buzz
   ([:\:~":,:'FizzBuzz'#~4#0=3 5&|) 80
   NB. sort reverse-lexicographically; n only wins if s is empty
Buzz
80
   (0{[:\:~":,:'FizzBuzz'#~4#0=3 5&|) 80
Buzz

Answer 8

tinylisp 2, 80 bytes

(quiet(m(\(N)(w(?(% N 3)(?(% N 5)N"Buzz")(?(% N 5)"Fizz""FizzBuzz"))nl))(1to 100

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Ungolfed/explanation

We map a lambda function over the numbers from 1 to 100. For each number N:

• Is it nonzero mod 3?
  • Is it nonzero mod 5? Then return N
  • Else, return "Buzz"
• Else:
  • Is it nonzero mod 5? Then return "Fizz"
  • Else, return "FizzBuzz"

We then write this value, followed by a newline. Since write returns nil, the result of the map is a list of 100 copies of nil. We prevent the interpreter from outputting this list by passing it to quiet.

(quiet
 (map
  (lambda (N)
   (write
    (if (% N 3)
     (if (% N 5)
      N
      "Buzz")
     (if (% N 5)
      "Fizz"
      "FizzBuzz"))
    nl))
  (1to 100)))

Answer 9

Boo, 114 bytes

for i in range(1,101):
 if i%15==0:print'FizzBuzz'
 elif i%5==0:print'Buzz'
 elif i%3==0:print'Fizz'
 else:print i

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Answer 10

Iogii, 29 bytes

"Fizz","Buzz"@r3,5%0q*u100}@O

Little peculiar that the last value next to the register sets all the other values. You can try it online by putting it in the program box here: https://golfscript.com/iogii/online_interpreter.html

Explained

I haven't quite figured out how to structure Iogii explanations yet, so some lines are somewhat oversimplified. If it helps, this is my standard approach of "multiply ["Fizz", "Buzz"] by [n % 3 == 0, n % 5 == 0], join into string, and logical or.

"Fizz","Buzz"@r3,5%0q*u100}@O­⁡​‎‎⁡⁠⁢⁢⁤‏⁠‎⁡⁠⁢⁣⁡‏⁠‎⁡⁠⁢⁣⁢‏⁠‎⁡⁠⁢⁣⁣‏⁠‎⁡⁠⁢⁣⁤‏‏​⁡⁠⁡‌⁢​‎‎⁡⁠⁤⁢‏⁠‎⁡⁠⁤⁣‏⁠‎⁡⁠⁤⁤‏⁠‎⁡⁠⁢⁡⁡‏⁠‎⁡⁠⁢⁡⁢‏⁠‎⁡⁠⁢⁡⁣‏‏​⁡⁠⁡‌⁣​‎‎⁡⁠⁢⁡⁤‏⁠‎⁡⁠⁢⁢⁡‏‏​⁡⁠⁡‌⁤​‎‎⁡⁠⁡‏⁠‎⁡⁠⁢‏⁠‎⁡⁠⁣‏⁠‎⁡⁠⁤‏⁠‎⁡⁠⁢⁡‏⁠‎⁡⁠⁢⁢‏⁠‎⁡⁠⁢⁣‏⁠‎⁡⁠⁢⁤‏⁠‎⁡⁠⁣⁡‏⁠‎⁡⁠⁣⁢‏⁠‎⁡⁠⁣⁣‏⁠‎⁡⁠⁣⁤‏⁠‎⁡⁠⁤⁡‏⁠‎⁡⁠⁢⁢⁢‏‏​⁡⁠⁡‌⁢⁡​‎‎⁡⁠⁢⁢⁣‏‏​⁡⁠⁡‌⁢⁢​‎‎⁡⁠⁢⁣⁤‏⁠‎⁡⁠⁢⁤⁡‏‏​⁡⁠⁡‌­
                       100}@   # ‎⁡For each number in the range [1, 100]:
             @r3,5%            # ‎⁢  Push [number % 3, number % 5]
                   0q          # ‎⁣  Do those equal 0?
"Fizz","Buzz"        *         # ‎⁤  Correspond those to the list ["Fizz", "Buzz"] via replication.
                      u        # ‎⁢⁡  Join into a single string (either "Fizz", "Buzz", "FizzBuzz", or "")
                           @O  # ‎⁢⁢Logical or each with the original number from the range
💎

Created with the help of Luminespire.

Answer 11

BrainChild, 97 bytes

include*;var p=print;while(int i++<100){if!(i%3)p("Fizz")if!(i%5)p("Buzz")if i%3&&i%5p(i)p("\n")}

Readable

include io.bc;
int i = 0;
while(i<100){
    i++;
    if (!(i%3))
        print("Fizz")
    if (!(i%5))
        print("Buzz")
    if (i%3 && i%5)
        print(i)
    print("\n")
}

Try It Online!

Answer 12

ibe, 197 bytes

adadadadakqafQmwafWmQafEmQsjWmrsjEmyafRmWafRmEshRmqmwsliqslwqyslweeslweesluuslwwislweeslweeslwqaltakwshWmqmesliqslwqyslweeslweeslwqaltakeshEmqmrsluuslwwislweeslweeslwqaltakrasQaktskWskEskRssQmwqqmq

code explained: (since ibe doesn't support comments, this code is not valid)

adadadad        | initialize 4 items in tape: Q, W, E, and R. they all start out as 0
                | Q is the initial counter 
                | W is Q % 3
                | E is Q % 5
                | R is W + E
akq             | start of loop (point 0 aka 'q')
afQmw           |   increase Q by 1
afWmQ           |   add Q to W
afEmQ           |   add Q to E
sjWmr           |   modulo W by 3
sjEmy           |   modulo E by 5
afRmW           |   add W to R
afRmE           |   add E to R
shRmqmw         |   if R is not equal to 0, goto point 1 (aka 'w'). else...
sliq            |       write F
slwqy           |           ..i
slwee           |           ..z
slwee           |           ..z
sluu            |           ..B
slwwi           |           ..u
slwee           |           ..z
slwee           |           ..z
slwq            |           ..newline
alt             |       goto point 4 (aka 't')
akw             |   point 1 (aka 'w')
shWmqme         |   if W is not equal to 0, goto point 2 (aka 'e'). else...
sliq            |       write F
slwqy           |           ..i
slwee           |           ..z
slwee           |           ..z
slwq            |           ..newline
alt             |       goto point 4 (aka 't')
ake             |   point 2 (aka 'e')
shEmqmr         |   if E is not equal to 0, goto point 3 (aka 'r'). else...
sluu            |       write B
slwwi           |           ..u
slwee           |           ..z
slwee           |           ..z
slwq            |           ..newline
alt             |       goto point 4 (aka 't')
akr             |   point 3 (aka 'r')
asQ             |   print Q, since nothing isn't true if it can reach this
akt             |   point 4 (aka 't')
skW             |   set W to 0
skE             |   set E to 0
skR             |   set R to 0
ssQmwqqmq       |   if Q < 100, goto point 0 (aka 'q')

Answer 13

Python 2, 72 bytes

for x in range(100):print('Fizz'*(x%3>1)+'Buzz'*(x%5>3)or str(x+1))+'\n'

Not as clever as feersum's solution, but it avoids casting exec magic.

EDIT: with just two more parentheses, it works in Python 3 AND Python 2:

for x in range(100):print(('Fizz'*(x%3>1)+'Buzz'*(x%5>3)or str(x+1))+'\n')

Answer 14

APL, 56 50 bytes

⊣{1+⍵⊣⎕←∊2↑((0=3 5|⍵)/'Fizz' 'Buzz'),''(⍕⍵)}⍣100⊢1

Note: The very first ⊣ should suppress output in GNU APL. Replace it with → to get correct results in Dyalog, etc, or with portable assignment to some variable X← adding one byte.

A+, 51 bytes

(x←100)do↓⊃2↑((0=3 5|1+x)/4⊂'FizzBuzz'),⌽2↑<1↓⍕1+x;

Answer 15

Javascript ES6, 77 chars

"\n".repeat(100).replace(/(?=\n)/g,(m,i)=>(++i%3?m:"Fizz")+(i%5?m:"Buzz")||i)

Test:

"\n".repeat(100).replace(/(?=\n)/g,(m,i)=>(++i%3?m:"Fizz")+(i%5?m:"Buzz")||i) == document.querySelector("pre").textContent

Answer 16

///, 229 bytes

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

Answer 17

JavaScript (using external library) (90 bytes)

x=>_.Range(1,100).Select(x=>x%3==0?(x%5==0?"FizzBuzz":"Fizz"):x%5==0?"Buzz":x).WriteLine()

Link to library: https://github.com/mvegh1/Enumerable/

Code explanation: Creates a range starting at 1 and extending 100 elements. Select maps the item according to the predicate. First check if divisible by 3, if so check if divisible by 5. If so, select "FizzBuzz", else select "Fizz". Otherwise,check if divisible by 5. If so, select "Buzz", else just select the number. WriteLine is built into the library, and it writes each element on its own line

Answer 18

DUP, 76 bytes

[$$3/%$[]['F,'i,'z,'z,]?\5/%$[]['B,'u,'z,'z,]?*[$.][]?10,]c:0[$100<][1+c;!]#

Explanation of program flow, code blocks:

DUP has two default truth values: -1 (true, all bits set) and 0 (false, no bits set). For conditional evaluations every nonzero value is accepted as true, comparisons themselves lead to either -1 or 0.

DUP has two conditional expressions, if-then-else and a while loop:

[execute if not false][execute if false]?. E.g. 1['t,]['f,]? prints t to STDOUT, because 1 is not false.

[condition][execute while not false]#. E.g. 5[$0>][$.1-]# prints the numbers 5,4,3,2,1 to STDOUT. The condition block tests the top stack value if it’s greater than 0, the execute block prints the number and decrements it.

The condition that’s checked is always the topmost stack element, which gets popped off the stack during the check.

[
    $$3/%$                              {dup,dup,push 3, moddiv, pop, dup}
    []                                  {if condition true do nothing}
      ['F,'i,'z,'z,]?                   {if condition false print Fizz}
    \5/%$                               {swap, push 5, moddiv, pop, dup}
    []                                  {if condition true do nothing}
      ['B,'u,'z,'z,]?                   {if condition false print Buzz}
    *                                   {mul}
    [$.]                                {if condition true dup, print to STDOUT}
        []?                             {if condition false do nothing}
    10,                                 {push 10, print char to STDOUT (newline)}
]c:                                     {define function c}

0                                       {push 0}

[$100<]                                 {while dup < push 100 true}
       [1+c;!]#                         {do push 1, add, execute function c}

Step by step:

instruction             stack
0                       0
[
$                       0,0
100                     0,0,100
<                       0,-1
]                       0
[
1                       0,1
+                       1               increment counter
c;                      1,0             0=start address of function c
!                       execute c
[
$$3                     1,1,1,3
/                       1,1,1,0         mod,div operator (n mod 3, n div 3)
%                       1,1,1           /% → mod
$                       1,1,1,1         result of n%3 used as flag for later check
[]                      1,1,1           condition true→do nothing
['F,'i,'z,'z,]?                         skipped
\                       1,1,1           swap top and 2nd stack elements
5                       1,1,1,5
/                       1,1,1,0         mod,div operator
%                       1,1,1           /% → mod
$                       1,1,1,1         n%5 flag
[]                      1,1,1           condition true → do nothing
['B,'u,'z,'z,]?                         skipped
*                       1,1             multiply %3 and %5 flags, result is only >0
                                        if neither %3 nor %5 are 0.
[                       1               if condition (flag)>0
 $                      1,1             dup
  .]                    1               print number to STDOUT: '1'
[]?                                     skipped

back to while loop

[
$                       1,1
100                     1,1,100
<                       1,-1
]                       1
[
1                       1,1
+                       2               increment counter
c;                      2,0             0=start address of function c
!                       execute c

...

A full description of all operators and program flow can be found at my GitHub repository, together wtih my Julia interpreter of DUP.

An online Javascript implementation showing useful debug information can be found here.

A rather terse introduction can be found on the esolangs.org DUP page.

Answer 19

S.I.L.O.S, 141 bytes

lbla
a+1
c=1-(a%3)*(a%5)
if c b
printIntNoLine a
lblb
c=(a%3)
if c d
print Fizz
lbld
c=(a%5)
if c c
print Buzz
lblc
printLine 
b=100-a
if b a

Try it online!

Unfortunately SILOS is incredibly verbose, and an interpreter bug cost me a few extra bytes, but I think SILOS is "superior" only to scratch in this challenge.

Answer 20

Fourier, 81 bytes

|`Fizz`|F|`Buzz`|B1~i100(i~X%15{0}{1~XFB}X%5{0}{1~XB}X%3{0}{1~XF}X{i}{Xo}10ai^~i)

Try it on FourIDE

Who knew such a simple task would yield such a mammoth of a program!

Pseudocode of the program:

Function F {
    Print "Fizz"
}

Function B {
    Print "Buzz"
}

i = 1
While i != 100
    X = i
    If X % 15 == 0
        X = 1
        Function F
        Function B
    End if
    If X % 5 == 0
        X = 1
        Function B
    End If
    If X % 3 == 0
        X = 1
        Function F
    End If
    If X == i
        Print X
    End if
    Print "\n"
    i += 1
End while

Answer 21

///, 209 208 bytes

/I/
EFizz
//H/FizzE//G/
Fizz
//E/Buzz
/1
2G4I7
8GE11G13
14
H16
17G19I22
23GE26G28
29
H31
32G34I37
38GE41G43
44
H46
47G49I52
53GE56G58
59
H61
62G64I67
68GE71G73
74
H76
77G79I82
83GE86G88
89
H91
92G94I97
98GBuA

Answer 22

Micro, 72 bytes

:i{i1+:i"":d
i3/i3%=if("Fizz":d,)
i5/i5%=if(d"Buzz"+:d,)
d:\i100=if(,a)}

Answer 23

Jq 1.5, 90 bytes

def f:if.%15<1then"FizzBuzz"elif.%5<1then"Buzz"elif.%3<1then"Fizz"else. end;range(100)+1|f

Try it online!

Answer 24

Modula-2, 222 bytes

MODULE F;IMPORT InOut;VAR i:INTEGER;BEGIN
FOR i:=1 TO 100 DO
CASE i*i*i*i MOD 15 OF
0:InOut.WriteString("FizzBuzz")|1:InOut.WriteInt(i,0)|6:InOut.WriteString("Fizz")|10:InOut.WriteString("Buzz")END;InOut.WriteLn
END END F.

Tested with Amsterdam Compiler Kit. The program needs 4-byte INTEGER. If using 2-byte INTEGER, i*i*i*i overflows and the program fails with a case error.

With indentation:

MODULE F;
IMPORT InOut;
VAR i: INTEGER;
BEGIN
  FOR i := 1 TO 100 DO
    CASE i * i * i * i MOD 15 OF
      0: InOut.WriteString("FizzBuzz")
    | 1: InOut.WriteInt(i, 0)
    | 6: InOut.WriteString("Fizz")
    | 10: InOut.WriteString("Buzz")
    END;
    InOut.WriteLn
  END
END F.

ACK compiles the PIM3 dialect of Modula-2 and comes with an InOut module. Other compilers for Modula-2 or Oberon might have shorter syntax or come with shorter modules.

• I don't like having 3 calls to InOut.WriteString, but adding a PROCEDURE p(s:ARRAY OF CHAR);BEGIN InOut.WriteString(s)END p; would cost 61 bytes and save only 48 bytes in the calls.

• I would write FOR i:=1TO 100DO but ACK rejects it as a syntax error.

• I had used i*i*i*i MOD 15 in my AppleScript answer.

Answer 25

MY-BASIC, 116 bytes

A basic response.

For i=1 To 100
S=""
If i Mod 3=0 Then S="Fizz"
If i Mod 5=0 Then S=S+"Buzz"
If S="" Then Print i; Else Print S;
Next

Try it online!

Answer 26

05AB1E, 25 bytes

тLv”FizzÒÖ”#y3 5‚ÖÏJy‚éθ,

Try it online!

I figured I'd add a fourth 05AB1E answer to this challenge. This uses my favourite "vectorise is-divisible and then repeat the fizbuzz list accordingly" approach.

Explained

тLv”FizzÒÖ”#y3 5‚ÖÏJy‚éθ,
тL                        # The range 1 .. 100
  v                       # For each N in that range:
   ”FizzÒÖ”#              #     Push the list ["Fizz", "Buzz"]
            y3 5‚Ö        #     Push N % [3, 5] == 0 (vectorised)
                  Ï       #     "Multiply" the two lists as you would in Vyxal
                   J      #     And concatenate that into a single string
                    y‚    #     And pair that with N
                      é   #     Sorting that by length, so that the value we want is always last
                       θ, #     And print it with a newline

Answer 27

D, 115 bytes

import std;void main(){string x;for(int i;i++<100;x=((i%3?"":"Fizz")~(i%5?"":"Buzz")),writeln(x?x:i.to!string)){};}

Try it online!

Adapted from this code:

import std;void main(){auto fizzbuzz(in uint i){string r;if(i%3==0)r~="Fizz";if(i%5==0)r~="Buzz";if(r.length==0)r~=i.to!string;return r;}enum r=1.iota(101).map!fizzbuzz;r.each!writeln;}

Answer 28

Fortran (GFortran), 187 bytes

DO I=1,100
IF(MOD(I,3)==0.AND.MOD(I,5)==0)THEN;PRINT'(A)','FizzBuzz'
ELSEIF(MOD(I,3)==0)THEN;PRINT'(A)','Fizz'
ELSEIF(MOD(I,5)==0)THEN;PRINT'(A)','Buzz'
ELSE;PRINT'(I0)',I
ENDIF;ENDDO;END

Try it online!

Answer 29

Pascal (FPC), 180 bytes

program f(o);var i:integer;begin for i:=1to 100do if i mod 15=0then writeln('FizzBuzz')else if i mod 3=0then writeln('Fizz')else if i mod 5=0then writeln('Buzz')else writeln(i)end.

Try it online!

Answer 30

Morse Code, 1777 bytes

.---- / ..--- / ..-. .. --.. --.. / ....- / -... ..- --.. --.. / ..-. .. --.. --.. / --... / ---.. / ..-. .. --.. --.. / -... ..- --.. --.. / .---- .---- / ..-. .. --.. --.. / .---- ...-- / .---- ....- / ..-. .. --.. --.. -... ..- --.. --.. / .---- -.... / .---- --... / ..-. .. --.. --.. / .---- ----. / -... ..- --.. --.. / ..-. .. --.. --.. / ..--- ..--- / ..--- ...-- / ..-. .. --.. --.. / -... ..- --.. --.. / ..--- -.... / ..-. .. --.. --.. / ..--- ---.. / ..--- ----. / ..-. .. --.. --.. -... ..- --.. --.. / ...-- .---- / ...-- ..--- / ..-. .. --.. --.. / ...-- ....- / -... ..- --.. --.. / ..-. .. --.. --.. / ...-- --... / ...-- ---.. / ..-. .. --.. --.. / -... ..- --.. --.. / ....- .---- / ..-. .. --.. --.. / ....- ...-- / ....- ....- / ..-. .. --.. --.. -... ..- --.. --.. / ....- -.... / ....- --... / ..-. .. --.. --.. / ....- ----. / -... ..- --.. --.. / ..-. .. --.. --.. / ..... ..--- / ..... ...-- / ..-. .. --.. --.. / -... ..- --.. --.. / ..... -.... / ..-. .. --.. --.. / ..... ---.. / ..... ----. / ..-. .. --.. --.. -... ..- --.. --.. / -.... .---- / -.... ..--- / ..-. .. --.. --.. / -.... ....- / -... ..- --.. --.. / ..-. .. --.. --.. / -.... --... / -.... ---.. / ..-. .. --.. --.. / -... ..- --.. --.. / --... .---- / ..-. .. --.. --.. / --... ...-- / --... ....- / ..-. .. --.. --.. -... ..- --.. --.. / --... -.... / --... --... / ..-. .. --.. --.. / --... ----. / -... ..- --.. --.. / ..-. .. --.. --.. / ---.. ..--- / ---.. ...-- / ..-. .. --.. --.. / -... ..- --.. --.. / ---.. -.... / ..-. .. --.. --.. / ---.. ---.. / ---.. ----. / ..-. .. --.. --.. -... ..- --.. --.. / ----. .---- / ----. ..--- / ..-. .. --.. --.. / ----. ....- / -... ..- --.. --.. / ..-. .. --.. --.. / ----. --... / ----. ---.. / ..-. .. --.. --.. / -... ..- --.. --..