# 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

var QUESTION_ID=58615;var ANSWER_FILTER="!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe";var COMMENT_FILTER="!)Q2B_A2kjfAiU78X(md6BoYk";var OVERRIDE_USER=30525;var answers=[],answers_hash,answer_ids,answer_page=1,more_answers=true,comment_page;function answersUrl(index){return"https://api.stackexchange.com/2.2/questions/"+QUESTION_ID+"/answers?page="+index+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+ANSWER_FILTER}function commentUrl(index,answers){return"https://api.stackexchange.com/2.2/answers/"+answers.join(';')+"/comments?page="+index+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+COMMENT_FILTER}function getAnswers(){jQuery.ajax({url:answersUrl(answer_page++),method:"get",dataType:"jsonp",crossDomain:true,success:function(data){answers.push.apply(answers,data.items);answers_hash=[];answer_ids=[];data.items.forEach(function(a){a.comments=[];var id=+a.share_link.match(/\d+/);answer_ids.push(id);answers_hash[id]=a});if(!data.has_more)more_answers=false;comment_page=1;getComments()}})}function getComments(){jQuery.ajax({url:commentUrl(comment_page++,answer_ids),method:"get",dataType:"jsonp",crossDomain:true,success:function(data){data.items.forEach(function(c){if(c.owner.user_id===OVERRIDE_USER)answers_hash[c.post_id].comments.push(c)});if(data.has_more)getComments();else if(more_answers)getAnswers();else process()}})}getAnswers();var SCORE_REG=/<h\d>\s*([^\n,<]*(?:<(?:[^\n>]*>[^\n<]*<\/[^\n>]*>)[^\n,<]*)*),.*?(\d+)(?=[^\n\d<>]*(?:<(?:s>[^\n<>]*<\/s>|[^\n<>]+>)[^\n\d<>]*)*<\/h\d>)/;var OVERRIDE_REG=/^Override\s*header:\s*/i;function getAuthorName(a){return a.owner.display_name}function process(){var valid=[];answers.forEach(function(a){var body=a.body;a.comments.forEach(function(c){if(OVERRIDE_REG.test(c.body))body='<h1>'+c.body.replace(OVERRIDE_REG,'')+'</h1>'});var match=body.match(SCORE_REG);if(match)valid.push({user:getAuthorName(a),size:+match[2],language:match[1],link:a.share_link,});else console.log(body)});valid.sort(function(a,b){var aB=a.size,bB=b.size;return aB-bB});var languages={};var place=1;var lastSize=null;var lastPlace=1;valid.forEach(function(a){if(a.size!=lastSize)lastPlace=place;lastSize=a.size;++place;var answer=jQuery("#answer-template").html();answer=answer.replace("{{PLACE}}",lastPlace+".").replace("{{NAME}}",a.user).replace("{{LANGUAGE}}",a.language).replace("{{SIZE}}",a.size).replace("{{LINK}}",a.link);answer=jQuery(answer);jQuery("#answers").append(answer);var lang=a.language;lang=jQuery('<a>'+lang+'</a>').text();languages[lang]=languages[lang]||{lang:a.language,lang_raw:lang.toLowerCase(),user:a.user,size:a.size,link:a.link}});var langs=[];for(var lang in languages)if(languages.hasOwnProperty(lang))langs.push(languages[lang]);langs.sort(function(a,b){if(a.lang_raw>b.lang_raw)return 1;if(a.lang_raw<b.lang_raw)return-1;return 0});for(var i=0;i<langs.length;++i){var language=jQuery("#language-template").html();var lang=langs[i];language=language.replace("{{LANGUAGE}}",lang.lang).replace("{{NAME}}",lang.user).replace("{{SIZE}}",lang.size).replace("{{LINK}}",lang.link);language=jQuery(language);jQuery("#languages").append(language)}}
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>

• 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

# SmileBASIC, 70 bytes

Printing "Fizz" and "Buzz" is easy, the slightly more difficult part is to only print the number when required. There are basically 2 ways to do this (and they end up being the same length)

1: Print the number when I isn't divisible by 3 or 5

FOR I=0TO 100A=I MOD 3B=I MOD 5?"Fizz"*!A;"Buzz"*!B;STR$(I)*(A&&B)NEXT  2: Print the number if the cursor is at column 0: FOR I=1TO 100?"Fizz"*!(I MOD 3);"Buzz"*!(I MOD 3); ?STR$(I)*!CSRX
NEXT


In a previous version of SB, % was used for MOD, making the program shorter:

# Petit Computer BASIC, 62 bytes

FOR I=0TO 100?"Fizz"*!(I%3);"Buzz"*!(I%5);
?STR$(I)*!CSRX NEXT  • The language name should probably be "SmileBASIC 2.x" or something. – snail_ Feb 6 '17 at 13:39 # KoopaScript, 250 characters def i 1 if \%va is \%vu set a 1;setath b \%va %% 3;setath c \%va %% 5;setath e \%va %% 15;if \%vb is 0 if \%vc not 0 print Fizz;if \%vb not 0 if \%vc is 0 print Buzz;if \%ve is 0 print FizzBuzz;if \%vb not 0 if \%vc not 0 print \%va;setath a \%va + 1  Guide to reading: Don't. Either learn KoopaScript more or less entirely by looking at the code (my documentation isn't that great) or just take my word for it that it works. By the way, KoopaScript is an interpreted language I made (not specifically for this) that runs inside ActionScript 2, and doesn't actually have else statements, or... most of the stuff that makes other examples short. All functions are one line, so this was pretty easy. Here's the GitHub repo. • This looks like quite the painful programming language to use haha. Welcome to PPCG! :) – James Mar 8 '18 at 19:55 • Thanks! Yeah, it is a bit painful, but I can't really be bothered to add stuff like arrays. I made a pi approximator and a prime number generator (both very slow), both were quite painful because of the lack of useful functions. – Jhynjhiruu Rekrap Mar 8 '18 at 20:45 # Husk, 35 bytes m§Ysλṁ!⁰₇ḣ3)ḣ100 ¶Σfm%⁰NC2¨¶¶¶⌈iΩZu  Try it online! I feel like I'm missing some opportunities for cleverness here. Like, I'm not really pleased with that lambda sitting there on the first line and having to be manually closed. But anyway, here we are. # Explanation I'll explain the second line first. It's a function TNum -> [[TChar]] that maps 1 -> [] 2 -> ["","","Fizz"] 3 -> ["","","","","Buzz"]  It works like this: ¶Σfm%⁰NC2¨¶¶¶⌈iΩZu ¨¶¶¶⌈iΩZu --The compressed string "\n\n\n\nFiBuzz" C2¨¶¶¶⌈iΩZu --cut; the list ["\n\n", "\n\n", "Fi", "Bu", "zz] m%⁰N --nats mod input, i.e, with input 3, [1,2,0,1,2,...] fm%⁰NC2¨¶¶¶⌈iΩZu --filter;so grab (1) nothing, (2) pairs 135, (3) pairs 1245 Σfm%⁰NC2¨¶¶¶⌈iΩZu --concat;now at (1) [] (2) ["\n\nFizz"] (3) ["\n\n\n\nBuzz"] ¶Σfm%⁰NC2¨¶¶¶⌈iΩZu --split by newlines to return claimed result  Next, the lambda in the first line. It has type TNum -> [TChar]; given an integer, it returns "", "Fizz", "Buzz", or "FizzBuzz" as appropriate. In detail: λṁ!⁰₇ḣ3) ḣ3 --heads; the list [1,2,3] ₇ḣ3 --call 2nd line with map overflow: [[],["","","Fizz"],["","","","","Buzz"]] ṁ!⁰₇ḣ3 --index the lambda argument to each list (modularly) and concat the results  And pulling it together: m§Ysλṁ!⁰₇ḣ3)ḣ100 λṁ!⁰₇ḣ3) --"","Fizz","Buzz",or"FizzBuzz" s --string representation of the number §Ysλṁ!⁰₇ḣ3) --apply both and take max; "" < numeric strings < alphabet m§Ysλṁ!⁰₇ḣ3)ḣ100 --map across [1,..,100]  Then Husk's default output style for an object of type [[TChar]] turns out to be just what we want. • This could have been -3 bytes if we didn't need to hardcode it for 100 lines! Ah well. – Sophia Lechner Mar 23 '18 at 23:48 # sed, 275272270260254249 245 bytes s/.*/t0u123456789/ :1 s/(tu?(.).*)/\1\n\2/ s/u(.)(t?)(.*)/\1\2u\3\1/ /9u/{s/t(.)/\1t/;s/u//;s/^/u/} /99/!b1 s/$/\n10/
s/[0-9][05]/&Buzz/g
s/[0369]{2}/&Fizz/g
s/[147][258]/&Fizz/g
s/[258][147]/&Fizz/g
s/[0-9]+([FB])/\1/g
s/\n0/\n/g
s/[^\n]+\n//
q


Try it Online

This is a pure sed script which discards all input, if any, and then prints all the necessary output lines and quits.

# Explanation

The script is divided into a sequence of three main parts: 1. Numeral generation; 2. "Fizz/Buzz/FizzBuzz" insertion; and 3. Formatting.

1. Numeral generation (lines 1 through 6)

The purpose of this part is to generate 99 lines containing the base 10 numerals corresponding to numbers 1 through 99. For that, we first set the entire pattern space to the following string:

t0u123456789


We will call the above string our state string. Next, we enter a loop in which each iteration goes like this:

1. A newline is appended to the pattern space;
2. A copy of the first numeral character after the "t" in the state string is appended to the pattern space;
3. A copy of the first numeral character after the "u" in the state string is appended to the pattern space;
4. The "u" in the state string is moved from its current position to the immediate right of the first numeral character after it or, if a "t" is already in said position, the "u" is instead moved to the immediate right of that "t";
5. If the "u" in the state string is immediately at the right of the "9" in the state string, the "t" is moved from its current position to the immediate right of the first character after it, and the "u" is moved from its current position to the immediate left of the "0" in the state string;
6. If there isn't a "99" anywhere in the pattern space (i.e., the loop has not finished its job of generating all the 99 lines), control goes back to line 2 (label :1) and so this enumeration of procedures is repeated from step 1; otherwise, control flow continues into the next line.

2. "Fizz/Buzz/FizzBuzz" insertion (lines 7 through 11)

The purpose of this part is to append "Fizz" immediately after each two-digit numeral in the pattern space which corresponds to a number which is a multiple of 3 but not of 5; to append "Buzz" immediately after each two-digit numeral in the pattern space which corresponds to a number which is a multiple of 5 but not of 3; and to append "FizzBuzz" after each two-digit numeral in the pattern space which corresponds to a number which is a multiple of both 3 and 5. This is how the computations go:

1. We append a newline followed by "10" to the pattern space.
2. Next, we search for all substrings of the pattern space formed by a digit between 0-9 on the left and either a 0 or a 5 on the right. We insert "Buzz" into the pattern space immediately after each such substring;
3. Finally, we insert "Fizz" into the pattern space immediately after each substring which matches either of the following criteria:
• Substrings formed by two digits which may be 0, 3, 6, or 9;
• Substrings formed by a digit which is either 1, 4 or 7 on the left and a digit which is either 2, 5 or 8 on the right;
• Substrings formed by a digit which is either 2, 5 or 8 on the left and a digit which is either 1, 4 or 7 on the right.

3. Formatting (lines 12 through 15)

This part is straight forward. Here we remove all of the following substrings from the pattern space:

1. Every contiguous sequence of numeral characters on the immediate left of a "F" or a "B";
2. All 0's on the immediate right of a newline character;
3. All characters from the beginning of the pattern space up to and including the first newline character (remember that the state string is still there and there's a newline immediately before the first output numeral).

And then sed just prints the final contents of the pattern space and calls it quits.

# ///, 198 bytes

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


with Ada.Text_IO;use Ada.Text_IO;procedure T is begin for I in Integer range 1..100 loop if I mod 15 = 0 then Put_Line("FizzBuzz");elsif I mod 3 = 0 then Put_Line ("Fizz");elsif I mod 5 = 0 then Put_Line ("Buzz");else Put_Line (Integer'Image (I)(2 ..Integer'Image(I)'Last));end if; end loop; end T;


Try it online!

Ungolfed:

with Ada.Text_IO;use Ada.Text_IO;

procedure Test is begin
for I in Integer range 1 .. 100 loop
if I mod 15 = 0 then
Put_Line ("FizzBuzz");
elsif I mod 3 = 0 then
Put_Line ("Fizz");
elsif I mod 5 = 0 then
Put_Line ("Buzz");
else
Put_Line (Integer'Image (I)(2 .. Integer'Image(I)'Last));
end if;
end loop;
end Test;


Pretty vanilla, but I didn't see an Ada solution yet. Probably because Ada might just be the worst real-world language to golf with!

• Ada isn't bad, it's more that hardly anyone even knows about it – ASCII-only Apr 18 '18 at 22:06
• Oh, I love Ada. It is definitely my favourite language. It just sucks for code golf. And yeah, I wish more people knew about it. :) – LambdaBeta Apr 18 '18 at 22:11
• Well, it can't possibly be as bad as Java :P – ASCII-only Apr 18 '18 at 22:12
• also on your profile: "am familiar with a majority of commonly used programming languages". which ones exactly? :P – ASCII-only Apr 18 '18 at 23:05
• 207 – ASCII-only Apr 24 '18 at 9:56

# QB64, 102 94 bytes

FOR i=1TO 100
o$=MID$("Fizz",i*5MOD 15)+MID$("Buzz",i*5MOD 25) IF""<o$THEN?o$ELSE WRITE i NEXT  Doesn't work on actual QBasic; see below for why. This program has one problem: QBasic/QB64 outputs to an 80x24 window, not a terminal, so the results can't be scrolled back. If you run the above code as-is, all you'll see is the lines from 78 onward. To prove that the code does 1 to 100 correctly, you can add the line SLEEP 1 right before NEXT for a 1-second delay on each iteration. ### Ungolfed code and explanation FOR i = 1 TO 100 index = 5 * (i MOD 3) o$ = MID$("Fizz", index) index = 5 * (i MOD 5) o$ = o$+ MID$("Buzz", index)
IF "" < o$THEN PRINT o$
ELSE
WRITE i
END IF
NEXT


On each iteration, we put the appropriate fizzes and buzzes into the string o$, check if it's empty, and output o$ or the number accordingly. The main question is how to get "Fizz" when i is divisible by 3 and "" otherwise. Here are the approaches I tried:

IF i MOD 3THEN o$=""ELSE o$="Fizz"
o$="":IF i MOD 3=0THEN o$="Fizz"
o$=MID$("Fizz",5*(i MOD 3))
o$=MID$("Fizz",i*5MOD 15)


The approach with MID$ is much shorter. This function takes 3 arguments--string, start index, and number of characters--and returns the appropriate substring. When the third argument is omitted, it takes everything from the start index to the end of the string. Here, when i is exactly divisible, the start index is 0 and we get the whole string; otherwise, it's something larger that's past the end of the string, so MID$ gives "".1

The other tricky part is printing numbers according to the spec. QBasic's PRINT command outputs positive numbers with leading spaces, which is occasionally useful but usually just annoying. The WRITE command, however, does not add a leading space--perfect for our purposes here.

1 Strings are 1-indexed in QBasic--i.e., in the string "abcd", a is at index 1 and d is at index 4. This is why I'm multiplying the mod result by 5: MID$("Fizz",4) gives "z". In actual QBasic, 0 isn't a legal index and gives Illegal function call; but in QB64, MID$("Fizz",0) happily returns the whole string instead of complaining.

# Whitespace, 307 bytes

[S S S N
_Push_0][N
S S N
_Create_Label_LOOP][S S S T N
_Push_1][T  S S S _Add][S N
S _Duplicate][S N
S _Duplicate][S S S T   T   N
_Push_3][T  S T T   _Modulo][N
T   S T N
_Jump_to_Label_FIZZ_if_0][N
S S S N
_Create_Label_RETURN_FIZZ][S N
S _Duplicate][S N
S _Duplicate][S S S T   S T N
_Push_5][T  S T T   _Modulo][N
T   S T T   N
_Jump_to_Label_BUZZ_if_0][N
S S S S N
_Create_Label_RETURN_BUZZ][S S S T  T   S S T   S S N
_Push_100][T    S S T   _Subtract][N
T   S T T   T   N
_Jump_to_Label_EXIT_WITH_ERROR_if_0][N
S T T   T   T   T   N
_Call_Label_PRINT_INT][S S S T  S T S N
_Push_10][T N
S S _Print_as_character][N
S N
N
_Jump_to_Label_LOOP][N
S S T
_Create_Label_FIZZ][S S S T S S S T T   S N
_Push_70][T N
S S _Print_as_character][S S S T    T   S T S S T   N
_Push_105][T    N
S S _Print_as_character][S S S T    T   T   T   S T S N
_Push_122][S N
S _Duplicate][T N
S S _Print_as_character][T  N
S S _Print_as_character][N
S N
S
_Return_to_Label_FIZZ_RETURN][N
S S T   T   N
_Create_Label_BUZZ][S S S T S S S S T   S N
_Push_66][T N
S S _Print_as_character][S S S T    T   T   S T S T N
_Push_117][T    N
S S _Print_as_character][S S S T    T   T   T   S T S N
_Push_122][S N
S _Duplicate][T N
S S _Print_as_character][T  N
S S _Print_as_character][N
S N
S S N
_Return_to_Label_BUZZ_RETURN][N
S S T   T   T   T   N
_Create_Label_PRINT_INT][S N
S _Duplicate][S S S T   T   N
_Push_3][T  S T T   _Modulo][N
T   S S T   N
_Jump_to_Label_LOOP_if_0][S N
S _Duplicate][S S S T   S T N
_Push_5][T  S T T   _Modulo][N
T   S S T   N
_Jump_to_Label_LOOP_if_0][T N
S T _Print_as_integer][N
T   N
_Return][N
S S S T N
_Create_Label_RETURN][N
T   N
_Return]


Letters S (space), T (tab), and N (new-line) added as highlighting only.
[..._some_action] added as explanation only.

Try it online (with raw spaces, tabs and new-lines only).

Can definitely be golfed. If-checks are rather annoying in Whitespace, and I still have to get used to them some more.

EDIT: Fixed TIO version. Will try to golf it some more later.

General explanation in Pseudo-code:

Integer i = 0
Start LOOP
Increase i by 1
If i modulo 3 is 0: Call function FIZZ()
If i modulo 5 is 0: Call function BUZZ()
If i is 100: Stop program
Call function PRINT_INT(i)
Print new-line
Go to next iteration of the LOOP

function FIZZ():
Print "Fizz"
Return

function BUZZ():
Print "Buzz"
Return

function PRINT_INT(integer i):
If i modulo 3 is 0: Return
If i modulo 5 is 0: Return
Print i
Return


Example run:

Command        Explanation                  Stack                 STDOUT    STDERR

SSSN           Push 0                       [0]
NSSN           Create Label_LOOP            [0]

SSSTN          Push 1                       [0,1]
SNS            Duplicate (1)                [1,1]
SNS            Duplicate (1)                [1,1,1]
SSSTTN         Push 3                       [1,1,1,3]
TSTT           Modulo (1%3)                 [1,1,1]
NSSSSN         Create Label_RETURN_FIZZ     [1,1]
SNS            Duplicate (1)                [1,1,1]
SNS            Duplicate (1)                [1,1,1,1]
SSSTSTn        Push 5                       [1,1,1,1,5]
TSTT           Modulo (1%5)                 [1,1,1,1]
NSSSSN         Create Label_RETURN_BUZZ     [1,1,1]
SSSTTSSTSSN    Push 100                     [1,1,1,100]
TSST           Subtract (1-100)             [1,1,-99]
NSTTTTTN       Call Label_PRINT_INT         [1,1]
NSSTTTTN      Create Label_PRINT_INT       [1,1]
SNS           Duplicate (1)                [1,1,1]
SSSTTN        Push 3                       [1,1,1,3]
TSTT          Modulo (1%3)                 [1,1,1]
SNS           Duplicate (1)                [1,1,1]
SSSTSTN       Push 5                       [1,1,1,5]
TSTT          Modulo (1%5)                 [1,1,1]
TNST          Print as integer             [1]                      1
NTN           Return                       [1]
SSSTSTSN       Push 10                      [1,10]
TNSS           Print as character           [1]                      \n

SSSTN          Push 1                       [1,1]
SNS            Duplicate (2)                [2,2]
SNS            Duplicate (2)                [2,2,2]
SSSTTN         Push 3                       [2,2,2,3]
TSTT           Modulo (2%3)                 [2,2,2]
NSSSSN         Create Label_RETURN_FIZZ     [2,2]
SNS            Duplicate (2)                [2,2,2]
SNS            Duplicate (2)                [2,2,2,2]
SSSTSTn        Push 5                       [2,2,2,2,5]
TSTT           Modulo (2%5)                 [2,2,2,2]
NSSSSN         Create Label_RETURN_BUZZ     [2,2,2]
SSSTTSSTSSN    Push 100                     [2,2,2,100]
TSST           Subtract (2-100)             [2,2,-98]
NSTTTTTN       Call Label_PRINT_INT         [2,2]
NSSTTTTN      Create Label_PRINT_INT       [2,2]
SNS           Duplicate (2)                [2,2,2]
SSSTTN        Push 3                       [2,2,2,3]
TSTT          Modulo (2%3)                 [2,2,2]
SNS           Duplicate (2)                [2,2,2]
SSSTSTN       Push 5                       [2,2,2,5]
TSTT          Modulo (2%5)                 [2,2,2]
TNST          Print as integer             [2]                      2
NTN           Return                       [2]
SSSTSTSN       Push 10                      [2,10]
TNSS           Print as character           [2]                      \n

SSSTN          Push 1                       [2,1]
SNS            Duplicate (3)                [3,3]
SNS            Duplicate (3)                [3,3,3]
SSSTTN         Push 3                       [3,3,3,3]
TSTT           Modulo (3%3)                 [3,3,0]
NSST           Create Label_FIZZ            [3,3]
SNS            Duplicate (3)                [3,3,3]
SNS            Duplicate (3)                [3,3,3,3]
SSSTSSSTTSN   Push 70                      [3,3,3,3,70]
TNSS          Print as character           [3,3,3,3]                F
SSSTTSTSSTN   Push 105                     [3,3,3,3,122]
TNSS          Print as character           [3,3,3,3]                i
SSSTTTTSTSN   Push 122                     [3,3,3,3,122]
SNS           Duplicate (122)              [3,3,3,3,122,122]
TNSS          Print as character           [3,3,3,3,122]            z
TNSS          Print as character           [3,3,3,3]                z
NSSSSN         Create Label_RETURN_FIZZ     [3,3,3,3]
SSSTSTN        Push 5                       [3,3,3,3,5]
TSTT           Modulo (3%5)                 [3,3,3,3]
NSSSSN         Create Label_RETURN_BUZZ     [3,3,3]
SSSTTSSTSSN    Push 100                     [3,3,3,100]
TSST           Subtract (3-100)             [3,3,-97]
NSTTTTTN       Call Label_PRINT_INT         [3,3]
NSSTTTTN      Create Label_PRINT_INT       [3,3]
SNS           Duplicate (3)                [3,3,3]
SSSTTN        Push 3                       [3,3,3,3]
TSTT          Modulo (3%3)                 [3,3,0]

Stack contains additional leading [3, but we'll ignore it in this explanation
SSSTN          Push 1                       [3,1]
SNS            Duplicate (4)                [4,4]
SNS            Duplicate (4)                [4,4,4]
SSSTTN         Push 3                       [4,4,4,3]
TSTT           Modulo (4%3)                 [4,4,1]
NSSSSN         Create Label_RETURN_FIZZ     [4,4]
SNS            Duplicate (4)                [4,4,4]
SNS            Duplicate (4)                [4,4,4,4]
SSSTSTN        Push 5                       [4,4,4,4,5]
TSTT           Modulo (4%5)                 [4,4,4,4]
NSSSSN         Create Label_RETURN_BUZZ     [4,4,4]
SSSTTSSTSSN    Push 100                     [4,4,4,100]
TSST           Subtract (4-100)             [4,4,-96]
NSTTTTTN       Call Label_PRINT_INT         [4,4]
NSSTTTTN      Create Label_PRINT_INT       [4,4]
SNS           Duplicate (4)                [4,4,4]
SSSTTN        Push 3                       [4,4,4,3]
TSTT          Modulo (4%3)                 [4,4,1]
SNS           Duplicate (4)                [4,4,4]
SSSTSTN       Push 5                       [4,4,4,5]
TSTT          Modulo (4%5)                 [4,4,4]
TNST          Print as integer             [4]                      4
NTN           Return                       [4]
SSSTSTSN       Push 10                      [4,10]
TNSS           Print as character           [4]                      \n

SSSTN          Push 1                       [4,1]
SNS            Duplicate (5)                [5,5]
SNS            Duplicate (5)                [5,5,5]
SSSTTN         Push 3                       [5,5,5,3]
TSTT           Modulo (5%3)                 [5,5,2]
NSSSSN         Create Label_RETURN_FIZZ     [5,5]
SNS            Duplicate (5)                [5,5,5]
SNS            Duplicate (5)                [5,5,5,5]
SSSTSTN        Push 5                       [5,5,5,5,5]
TSTT           Modulo (5%5)                 [5,5,5,0]
NSSTTN        Create Label_BUZZ            [5,5,5]
SSSTSSSSTSN   Push 66                      [5,5,5,66]
TNSS          Print as character           [5,5,5]                  B
SSSTTTSTSTN   Push 117                     [5,5,5,117]
TNSS          Print as character           [5,5,5]                  u
SSSTTTTSTSN   Push 122                     [5,5,5,122]
SNS           Duplicate (122)              [5,5,5,122,122]
TNSS          Print as character           [5,5,5,122]              z
TNSS          Print as character           [5,5,5]                  z
NSSSSN         Create Label_RETURN_BUZZ     [5,5,5]
SSSTTSSTSSN    Push 100                     [5,5,5,100]
TSST           Subtract (5-100)             [5,5,-95]
NSTTTTTN       Call Label_PRINT_INT         [5,5]
NSSTTTTN      Create Label_PRINT_INT       [5,5]
SNS           Duplicate (5)                [5,5,5]
SSSTTN        Push 3                       [5,5,5,3]
TSTT          Modulo (5%3)                 [5,5,2]
SNS           Duplicate (5)                [5,5,5]
SSSTSTN       Push 5                       [5,5,5,5]
TSTT          Modulo (4%5)                 [5,5,0]

... etc. etc.

SSSTN          Push 1                       [99,1]
SNS            Duplicate (100)              [100,100]
SNS            Duplicate (100)              [100,100,100]
SSSTTN         Push 3                       [100,100,100,3]
TSTT           Modulo (100%3)               [100,100,1]
NSSSSN         Create Label_RETURN_FIZZ     [100,100]
SNS            Duplicate (100)              [100,100,100]
SNS            Duplicate (100)              [100,100,100,100]
SSSTSTN        Push 5                       [100,100,100,100,5]
TSTT           Modulo (100%5)               [100,100,100,0]
NSSTTN        Create Label_BUZZ            [100,100,100]
SSSTSSSSTSN   Push 66                      [100,100,100,66]
TNSS          Print as character           [100,100,100]            B
SSSTTTSTSTN   Push 117                     [100,100,100,117]
TNSS          Print as character           [100,100,100]            u
SSSTTTTSTSN   Push 122                     [100,100,100,122]
SNS           Duplicate (122)              [100,100,100,122,122]
TNSS          Print as character           [100,100,100,122]        z
TNSS          Print as character           [100,100,100]            z
NSSSSN         Create Label_RETURN_BUZZ     [100,100,100]
SSSTTSSTSSN    Push 100                     [100,100,100,100]
TSST           Subtract (100-100)           [100,100,0]


# Groovy, 63 bytes

Stand-alone program, prints result to STDOUT.

100.times{n->s=(++n%3?'':'Fizz')+(n%5?'':'Buzz')
println s?s:n}


I'd have preferred to use (1..100).each{ instead of 100.times{, so that I didn't have to ++n at the start of each iteration, but this is golf and that saves me two bytes.

Other than that, a pretty standard truthy-based submission. 'Fizz' and 'Buzz' are added when remainder is 0 because 0 is falsy, and '' is a falsy string, so I can print n when the first line hasn't had either of 'Fizz' or 'Buzz' added.

Try it online!

# Brachylog, 48 bytes

100⟦₁{f{∋15∧"FizzBuzz"|∋3∧"Fizz"|∋5∧"Buzz"|t}ẉ}ᵐ


Try it online!

Probably not golfed too well but it works.

## Scheme, 118 bytes

(for-each(lambda(i)(printf"~a~a~%"(if(=(mod i 3)0)'Fizz"")(if(=(mod i 5)0)'Buzz(if(=(mod i 3)0)""i))))(cdr(iota 101)))


Try it online!

Ungolfed:

(for-each
(lambda (i)
(printf
"~a~a~%"
(if (= (mod i 3) 0) 'Fizz "")
(if (= (mod i 5) 0) 'Buzz
;; else
(if (= (mod i 3) 0) "" i))))
(cdr (iota 101)))


# C, 98 94 bytes

i;main(){for(;++i<101;)i%3*i%5?printf("%d\n",i):printf("%s%s\n",i%3?"":"Fizz",i%5?"":"Buzz");}


Pretty simple stuff...

Thanks to Yoris Fresh for helping me save a couple bytes. Thanks to Jerry Jeremiah for pointing out a mistake in the ungolfed version.

Ungolfed:

i;
main() {
for (; ++i < 101;)
i % 3 && i % 5 ?
printf("%d\n",i) :
printf("%s%s\n", i % 3 ? "" : "Fizz", i % 5 ? "" : "Buzz");
}

• the ungolfed version increments i twice each loop. – Jerry Jeremiah Jun 4 '19 at 3:33

# 05AB1E, 24 bytes

тÝ3Å€"Fizz"}5Å€á”ÒÖ”J}¦»


Try it online!

Explanation:

тÝ                  # range 0..100
3Å€      }        # for every 3rd element...
"Fizz"         # replace it with Fizz
5Å€      }        # for every 5th element...
á              # keep only letters...
”ÒÖ”J         # and append "Buzz"
¦       # drop the first element
»      # join with newlines
# implicit output


Or alternatively:

тÝ35vyÅ€á”FizzÒÖ”#NèJ]¦»


Try it online!

35vy iterates over the digits of 35, which avoids repeating Å€}. ”FizzÒÖ” is the string Fizz Buzz, and then #Nè selects the appropriate element.

Legacy 05AB1E doesn’t have Å€, so neither of those work for it. However, legacy Ã (set intersection) implicitly splits numbers, which lets us get a 26 that only works on legacy:

тLεDÑ35Ãvá”Fizz ÒÖ”#yèJ},


Try it online!

тLε                         # for y in 1..100
DÑ                       # divisors of y
35Ã                    # keep only those in [3, 5]
v               }   # for each...
á                  # keep only letters
”Fizz ÒÖ”#yèJ     # append either Fizz or Buzz
,  # print with newline


# APL (Dyalog Unicode), 37 bytesSBCS

↑{∨/d←4/0=3 5|⍵:d/'FizzBuzz'⋄⍕⍵}¨⍳100


Try it online!

⍳100ɩndices 1…100

{}¨ apply the following anonymous lambda to each of those:

⍵ the argument; e.g. 20

3 5| the division remainder when that is divided by 3 and 5; e.g. [2,0]

0= Boolean mask where that is equal to 0; e.g. [0,1]

4/ replicate those numbers for 4 copies of each; e.g. [0,0,0,0,1,1,1,1]

d← assign that to d

∨/: if any of those are true (OR-reduction); e.g. true:

d/'FizzBuzz' use d to mask the characters of the string; e.g. "Buzz"

⋄ else:

⍕⍵ stringify the argument; e.g. "20"

↑ mix the list of strings into a matrix, so it prints right

# Pip, 43 41 bytes

Fa1,101{i:0a%3?i:1O"FIZZ"a%5?i?PaPxP"BUZZ"}

Lh{i:0o%3?i:1O"FIZZ"o%5?i?PoPxP"BUZZ"++o}


Try it online!

# Ink, 777057 48 bytes

-(n){n%3:{n%5:{n}}|Fizz}{n%5: |Buzz}
{n<100:->n}


Try it online!

# Julia, 87 bytes

z(i)=(f=i .%[3,5] .==0;sum(f)>0 ? foldl(*,["Fizz","Buzz"][f]) : i)
println.(z.(1:100))


Or we could kind of cheat to drop 3 bytes and use show instead of println. Julia 83 bytes

z(i)=(f=i .%[3,5] .==0;sum(f)>0 ? foldl(*,["Fizz","Buzz"][f]) : i)
show(z.(1:100))


or a more legible FP style, poor performing, 91 bytes

F(x)=foldl(*,["Fizz","Buzz"][x .%[3,5] .==0])
N(y)=F(y)=="" ? y : F(y)
println.(N.(1:100))


• prod is better than foldr(* – H.PWiz Nov 23 '19 at 23:45
• Nice! I didn't think that'd work. Either way my solution is 10s of bytes away from the best Julia one :). – caseyk Nov 23 '19 at 23:51
• Sure. My score is here – H.PWiz Nov 24 '19 at 12:25
• Woah! How'd you do it? Or is that a secret? :) We have a thread that divulged into codegolf on julia discourse! – caseyk Nov 24 '19 at 13:10
• Sort of a secret. Although I have discussed julia golfing in a chat room on this site. – H.PWiz Nov 24 '19 at 21:56

## MC6000 ASM (Shenzhen I/O), 133 Bytes

add 1
mov acc x3
mov 51 x2
mov acc x2
mov 3 x2
teq x2 0
+mov 1 x1
mov 51 x2
mov acc x2
mov 5 x2
teq x2 0
+mov 2 x1
slp 1
mov -999 x1


Uses an MC4010 co-processor on x2, numeric display on x3, and a custom fizzbuzz display on x1.

"Hardware":

Custom LCD:

Technically doesn't follow the rules as MCxxxx don't have a STDOUT, but I did the best I could.

# APL, 52 chars/bytes

generate integers from 1 to n

⍳


b is a 2 by n boolean matrix highlighting multiples of 3 and 5

b←0=3 5∘.|⍵


c is a boolean vector highlighting multiples of either 3 or 5

c←∨⌿b


format each integer into a character vector

⍕¨⍵


where a number is a multiple of either 3 or 5...

@{c}


...put either Fizz or Buzz (i.e. discard the last 4, the first 4 or no character at all from the string "FizzBuzz" according to b)

(↓∘'FizzBuzz'¨¯4 4+.×c/b)


transform the nested vector into a matrix

↑


# sed 4.2.2, 129 bytes

A 400-rep bounty for those who outgolf this solution https://codegolf.meta.stackexchange.com/a/18428/

s/^/00,/;h
:
y/0123456789';,/1234567890;,'/
/0.$/!{x;G;s/..\n.//} h s/[05].$/&Buzz/
s/.*,/Fizz/
s/.*\WB/B/
s/[0';]*//gp
g;/00/d
t


Try it online!

s/^/00,/;h


Each number is stored as three characters, the first two store the two-digit padded decimal number, and the last character stores its modulo 3.

: ... t


In a loop,

y/0123456789';,/1234567890;,'/


the modulo is cycled, and the number is incremented using transliteration. Each digit is increased modulo 10, then

/0.$/!{x;G;s/..\n.//}  a simple conditional corrects the first digit on the number if necessary, using the fact each number is stored with exactly 3 characters. h  This incremented form is stored in the hold space. s/[05].$/&Buzz/


Buzzs are added by looking at the last base-10 digit of the number.

s/.*,/Fizz/


Fizzs are added by looking at the modulo-3.

s/.*\WB/B/


There is some cleanup of the number, remove the base-10 digits if needed and

s/[0';]*//gp


remove leading 0s and the modulo-3 so that it is print-ready, and print it.

g;/00/d


Finally retrieve the number from the hold space and exit if 00 is present, i.e. 100 has been reached

# Rust, 114 bytes

It's not the shortest one available (on code-golf.io someone managed to somehow solve it in 99 bytes, but I have no idea how).

fn main(){(1..101).for_each(|i|println!("{}",["FizzBuzz","Fizz","Buzz",&i.to_string()][1.min(i%5)+2.min(i%3*2)]))}


First off, the obvious part

fn main() {/*..*/}


Then we iterate from 1 to 100 (inclusive) and println! an element of the array

(1..101).for_each(|i| println!("{}", ["FizzBuzz", "Fizz", "Buzz", &i.to_string()][/*..*/]))


To choose the right index, we use the following, where i is the current number

1.min(i % 5) + 2.min((i % 3) * 2)


The first part is 0 if i is a multiple of 5. Otherwise it's 1.

The second part is 0 if i is a multiple of 3. Otherwise it's 2.

Examples:

1.min(1 % 5) + 2.min((1 % 3) * 2) == 1 + 2 == 3 => "1"
1.min(3 % 5) + 2.min((3 % 3) * 2) == 1 + 0 == 1 => "Fizz"
1.min(4 % 5) + 2.min((4 % 3) * 2) == 1 + 2 == 3 => "4"
1.min(5 % 5) + 2.min((5 % 3) * 2) == 0 + 2 == 2 => "Buzz"
1.min(15 % 5) + 2.min((15 % 3) * 2) == 0 + 0 == 0 => "FizzBuzz"


This way, if both parts produce 0, we get the string FizzBuzz. If the sum is 1 we get Fizz, if it's 2 Buzz, if it's 3 we get &i.to_string() (the stringified number).

• You can replace println! with print! and add a literal (as in press ENTER/RETURN) newline after the {} in the format string to save a byte. Also, if you're allowed to use a closure, it'd save quite a few bytes over using fn main(). Finally, you save some bytes using a for loop. 99 bytes: TIO – TehPers Aug 3 '20 at 22:15

## Javascript 78 (but kinda 58) bytes

Produces the required string, then prints it. The function that produces the string is 58 bytes, another 20 are used to call console.log.

z=n=>n?z(n-1)+((n%3?'':'Fizz')+(n%5?'':'Buzz')||n)+'\n':'';console.log(z(100))


## Javascript 72 71 bytes (thanks, Jo King!)

Prints the required lines as it goes. Interweaving console.log with the logic creates a shorter, though less elegant, program.

z=n=>n&&(z(n-1),console.log((n%3?'':'Fizz')+(n%5?'':'Buzz')||n));z(100)


It can also be written like the following, to the same effect and byte count.

z=n=>n&&console.log((z(n-1),(n%3?'':'fizz')+(n%5?'':'buzz')||n));z(100)


## De-golfed version:

z = n =>           // take n as argument
n && (           // if n is not falsey (e.g. if it is not 0)
z(n-1),        // use the comma operator to call the previous fizzbuzz
console.log(   // then print a line
(n%3 ? '' : 'Fizz')   // 'Fizz' if n is a multiple of 3, otherwise empty
+ (n%5 ? '' : 'Buzz') // Concatenated with 'Buzz' if multiple of 5
|| n         // Or just the number n if the previous string is empty, which implies that n is not divisible by 3 or 5
)
);
z(100)             // Run fizzbuzz from 100

• You can save 5 bytes (on every solution) by using alert instead of console.log – VFDan May 19 '20 at 0:46

# International Phonetic Esoteric Language, 61 60 bytes

{2T}1ɑeb3ⱱɐbʌɔ|a|"Fizz"u|a|q5ⱱɐbʌɔ|b|"Buzz"u|b|ɞʌue1sø"\n"uɒ


Explanation:

{2T}1                                                        (push loop bounds 1 to 2T == 101 base36)
ɑ                                                       (start loop)
eb                                                     (push and dup loop index)
3ⱱɐbʌɔ|a|"Fizz"u|a|                                  (print if fizzy)
q                                 (OVER)
5ⱱɐbʌɔ|b|"Buzz"u|b|              (print if buzzy)
ɞʌu           (else print the number)
e1sø       (increment loop index)
"\n"u  (print a newline)
ɒ (end loop)


-1 byte for using base 36 for 101.

# MAWP, 120 bytes

[!!3P3WA<75W2W;73W5W;65W2W1M2W;65W2W1M2W;~0~>%!!5P5WA<92M6W;94M9W;65W2W1M2W;65W2W1M2W;~0~>%~{~!:~}~!554WWA?.%1M25W;~(%)]


Try it!

Created this while testing my new online interpreter!

## Explanation:

[                                              Start loop
!!3P3WA                                        Is divisible by 3? (Homemade modulo
function)
<                                              If top of stack is not 0 (number is
75W2W;73W5W;65W2W1M2W;65W2W1M2W;               Print 'Fizz'
~0~                                            Add 0 to the bottom of stack by
double reversing
>                                              End conditional
%                                              Remove modulo function result
!!5P5WA<92M6W;94M9W;65W2W1M2W;65W2W1M2W;~0~>%  Do the same for 'Buzz'
~                                              Reverse stack
{                                              If there are no 0 on the top of
stack (not 'Fizz' nor 'Buzz' was
printed),
~!:~                                           then print the current number
}~                                             End conditional and reverse stack
!554WWA?.%                                     If the number is 100, then terminate
program
1M25W;                                         Print newline
~(%)                                           Remove all 0 from bottom of stack
]                                              End of loop


# Integral, 416 Bytes

⌡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⌡[j


Try it

Integral doesn't have very many ways of doing this challenge yet, so this'll have to do for now.

I'll update it once there's a better way.

• ah, beautiful solution. what an algorithm. – Razetime Sep 21 '20 at 12:57

# LiveScript, 57 bytes

I noticed nobody had done LiveScript. This may not be the shortest possible answer, but it's my best attempt.

for i to 99
console.log \Fizz *!(++i%3)+\Buzz *!(i%5)||i


# MAWP 2.0, 60 bytes

[!3%<"Fizz":1=M>!5%<"Buzz":1=M>M1-{/!:\}0=M10;!100-?.1+]


Explanation:

[                start of loop
!3%              is top of stack divisible by 3?
<"Fizz":1=M>    if yes, print "Fizz" and set variable M to 1
!5%              is top of stack divisible by 5?
<"Buzz":1=M>    if yes, print "Buzz" and set variable M to 1
M1-              push value of M and subtract 1
{/!:\}          if not 0 then print the current number
0=M              set variable M to 0
10;              print a newline
!100-?.         if current number is 100 then terminate program
1+               add 1 to current number
]                end of loop


# Charcoal, 37 29 bytes

Ｅ…¹¦¹⁰¹∨⁺×Fizz¬﹪ι³×Buzz¬﹪ι⁵Ｉι


Try it online!

How there isn't a charcoal answer yet amazes me. -8 bytes thanks to @Neil

Verbose

• There are other minor variants such as Print(Map(100, Or(Add(Ternary(Modulo(Incremented(i), 3), "", "Fizz"), Ternary(Modulo(Incremented(i), 5), "", "Buzz")), Cast(Incremented(i))))); also for 29 bytes, but I do have a 28 byte solution... – Neil Sep 5 '20 at 23:52
• I guess it's about time I reveal my 28 byte solution: Ｅ¹⁰⁰∨⭆⪪FiBu²⎇﹪⊕ι⁺³⊗μω⁺λzzＩ⊕ι. – Neil Oct 18 '20 at 18:04

# Add++-i, 46 bytes

L,100Rdd3€Ω%€!"Fizz"€*$5€Ω%€!"Buzz"€*z£+z£obUn  Try it online! The flag simply invokes the function for us and outputs its return value, so we don't have to include the invocation in the program. ## How it works L, ; Anonymous function 100R ; Push [1 2 3 4 ... 99 100] STACK = [[1 2 ... 99 100]] dd ; Triplicate STACK = [[1 2 ... 99 100] [1 2 ... 99 100] [1 2 ... 99 100]] € € ; Over the first, check ... 3 Ω% ! ; divisibility by 3 STACK = [[1 2 ... 99 100] [1 2 ... 99 100] [0 0 ... 1 0]] "Fizz"€* ; Replace 1 with "Fizz" and 0 with "" STACK = [[1 2 ... 99 100] [1 2 ... 99 100] ["" "" ... "Fizz" ""]$ €  €		; Over the second, check ...
5 Ω% !		;     divisibility by 5			STACK = [[1 2 ... 99 100] ["" "" ... "Fizz" ""] [0 0 ... 0 1]]
"Buzz"€*	; Replace 1 with "Buzz" and 0 with ""	STACK = [[1 2 ... 99 100] ["" "" ... "Fizz" ""] ["" "" ... "" "Buzz"]]
z£+		; Zip and concatenate			STACK = [[1 2 ... 99 100] ["" "" ... "Fizz" "Buzz"]]
z£o		; Replace "" with the right number	STACK = [[1 2 ... "Fizz" "Buzz"]]
bUn		; Join with newlines			STACK = ["1\n2\nFizz\n...\n98\nFizz\nBuzz"]
; Return the top of the stack		"1\n2\nFizz\n...\n98\nFizz\nBuzz"


l:100
Fl,f,i%3,f=0,"Fizz"*f,b,i%5,b=0,"Buzz"*b,p,f+b,p="",Ip,,Oi,c,f+b,P,p=0,IP,,Oc,
`

Try it online!

"But there's a shorter Add++ answer" I hear you say. Well, this is flagless and doesn't have a Lambda, so I'd thought I'd post it for comparison.