86
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Write a program that outputs the lyrics to 99 Bottles of Beer, in as few bytes as possible.

Lyrics:

99 bottles of beer on the wall, 99 bottles of beer.
Take one down and pass it around, 98 bottles of beer on the wall.

98 bottles of beer on the wall, 98 bottles of beer.
Take one down and pass it around, 97 bottles of beer on the wall.

97 bottles of beer on the wall, 97 bottles of beer.
Take one down and pass it around, 96 bottles of beer on the wall.

96 bottles of beer on the wall, 96 bottles of beer.
Take one down and pass it around, 95 bottles of beer on the wall.

95 bottles of beer on the wall, 95 bottles of beer.
Take one down and pass it around, 94 bottles of beer on the wall.

....

3 bottles of beer on the wall, 3 bottles of beer.
Take one down and pass it around, 2 bottles of beer on the wall.

2 bottles of beer on the wall, 2 bottles of beer.
Take one down and pass it around, 1 bottle of beer on the wall.

1 bottle of beer on the wall, 1 bottle of beer.
Go to the store and buy some more, 99 bottles of beer on the wall.

Rules:

  • Your program must log to STDOUT or an acceptable alternative, or be returned from a function (with or without a trailing newline).
  • Your program must be a full, runnable program or function.
  • Languages specifically written to submit a 0-byte answer to this challenge are allowed, just not particularly interesting.

    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.

  • This is different from the output by HQ9+ or 99. Any answers written in these languages will be deleted.

As this is a catalog challenge, this is not about finding the language with the shortest solution for this (there are some where the empty program does the trick) - this is about finding the shortest solution in every language. Therefore, no answer will be marked as accepted.

Catalogue

The Stack Snippet at the bottom of this post generates the catalogue from the answers a) as a list of shortest solution per language and b) as an overall leaderboard.

To make sure that your answer shows up, please start your answer with a headline, using the following Markdown template:

## Language Name, N bytes

where N is the size of your submission. If you improve your score, you can keep old scores in the headline, by striking them through. For instance:

## Ruby, <s>104</s> <s>101</s> 96 bytes

If there you want to include multiple numbers in your header (e.g. because your score is the sum of two files or you want to list interpreter flag penalties separately), make sure that the actual score is the last number in the header:

## Perl, 43 + 2 (-p flag) = 45 bytes

You can also make the language name a link which will then show up in the snippet:

## [><>](http://esolangs.org/wiki/Fish), 121 bytes

<style>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: bold; } table td { padding: 5px; }</style><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><script>var QUESTION_ID = 64198; var ANSWER_FILTER = "!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe"; var COMMENT_FILTER = "!)Q2B_A2kjfAiU78X(md6BoYk"; var OVERRIDE_USER = 36670; var answers = [], answers_hash, answer_ids, answer_page = 1, more_answers = true, comment_page; function answersUrl(index) { return "//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 "//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(42), 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); } }</script>

\$\endgroup\$
25
  • 14
    \$\begingroup\$ Whoever closevoted this, the other one is a popularity contest... \$\endgroup\$ – TheDoctor Nov 18 '15 at 15:19
  • 22
    \$\begingroup\$ This is a subtask of the Create an HQ9+ Interpreter golf. \$\endgroup\$ – Geobits Nov 18 '15 at 15:25
  • 4
    \$\begingroup\$ @CᴏɴᴏʀO'Bʀɪᴇɴ I haven't VTCd it yet, still deciding how I feel about that. Either way, I think some answers (or snippets from them) from the other can be directly copy/pasted while remaining competitive, which is a usual test for dupe-ness. \$\endgroup\$ – Geobits Nov 18 '15 at 15:54
  • 13
    \$\begingroup\$ Note that quartata had already sandboxed a 99BB catalog several weeks ago and was planning to post it. \$\endgroup\$ – Alex A. Nov 18 '15 at 17:07
  • 14
    \$\begingroup\$ @GamrCorps Please do not unnecessarily change the rules after the challenge has started. None of the characters in the output should be optional. \$\endgroup\$ – feersum Nov 18 '15 at 18:42

161 Answers 161

1 2 3 4 5
6
0
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Javascript ES6, 247 237 bytes

m=" bottles of beer",j=m.replace("s",""),k=" on the wall",a="";for(i=99;i>1;)a+=i+m+k+`, ${i+m}.
Take one down and pass it around, ${(--i)+(i-1?m:j)+k}.

`;a+=`1${j+k}, 1${j}. Go to the store and buy some more, 99${m+k}.`;console.log(a);

-10 bytes thanks to Jo King.

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0
0
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Actually, 1 byte

N

Try it online!

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1
  • \$\begingroup\$ There's already a Seriously answer here, and I'd say that it borders on breaking a standard loophole anyway, since the Bottles of Beer built-in was added to the language specifically for this challenge, which is why they are incorrect in the same ways (no zero bottle verse) \$\endgroup\$ – Jo King Sep 4 '19 at 3:55
0
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Hexadecimal Stacking Pseudo-Assembly Language, 2208 bytes

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

Try it online!

That was quite some work, because the language has a few difficulties (or maybe just interpreter bugs)

  • You can only jump to labels you have been before, so basically you have to run through the code, skipping every jump and print command, and then you can run it

  • Subtracting is "backwards". For example, 100-1 is push 1, push 100, sub, what makes loops a bit difficult.

  • It can't handle negative values, so using -1 + 100 instead, isn't possible. If you try to calculate 0-1, you will get 0 as the result.

  • The functions NOT and XOR aren't working (I didn't try if AND and OR are working)

  • The only conditional is "skip next command, if popped value is not zero", which is exactly opposite to all other esoteric languages I know so far.

  • You have to push return-labels by yourself (and you can't return to a label where you haven't been before)

So what the code does is, it sets up a list of 1's on the initializing-stack and uses them to skip every jump and print command, then it deletes the list, fills it with 0's and runs the main loop.

Stack 0: bottle count, Stack 1: return labels, Stack 2: Strings, Stack 3: Initialize flag, Stack 4: Initialize Queue

200001  1
400300  push initialize = 1

000006  initializer:
201000  0x1000 (init_counter)
400000  push
000016  initialize_loop:
420300  peek initialize
400400  push init_queue

410000  pop init_counter
400100  push on other stack
200001  1
400000  push
410100  pop init_counter
400000  push to counter_stack
220000  sub
420000  peek init_counter
400000  push
400000  push 0
300000  equals
030000  if init_counter > 0
010016  go to initialize loop

200063  99
400000  push
030400  if not initializing
010000  go to main

000001  Label wall:
200011  0x11 (return address)   
400100  push retstack
030400  if not initializing
010002  jmp beer
000011  Label Return_beer_wall
20006C  'l'
400200  push
20006C  'l'
400200  push
200061  'a'
400200  push
200077  'w'
400200  push
200020  ' '
400200  push
200065  'e'
400200  push
200068  'h'
400200  push
200074  't'
400200  push
200020  ' '
400200  push
20006E  'n'
400200  push
20006F  'o'
400200  push
200020  ' '
400200  push
030400  if not initializing
140200  print string
030400  if not initializing
020100  return

000002  Label beer:
200012  0x12 (return address)   
400100  push retstack
030400  if not initializing
010003  jmp bottles
000012  Label Return_bottles_beer
200072  'r'
400200  push
200065  'e'
400200  push
200065  'e'
400200  push
200062  'b'
400200  push
200020  ' '
400200  push
200066  'f'
400200  push
20006F  'o'
400200  push
200020  ' '
400200  push
030400  if not initializing
140200  print string
030400  if not initializing
020100  return

000003  Label bottles:
420000  peek bottle count
400000  push
030400  if not initializing
120000  print number
200065  'e'
400200  push
20006C  'l'
400200  push
200074  't'
400200  push
200074  't'
400200  push
20006F  'o'
400200  push
200062  'b'
400200  push
200020  ' '
400200  push
030400  if not initializing
140200  print string
420000  get bottle count
400000  push
420000  get bottle count
400000  push
200001  1
400000  push
300000  equals
200000  0
400000  push
300000  equals
030000  skip if true
020100  return
200073  's'
400200  push
030400  if not initializing
140200  print string
030400  if not initializing
020100  return

000004  Label Take:
420000  get bottle count
400000  push
200000  0
400000  push
300000  equals (bottle count = 0)
200000  0
400000  push
300000  equals (bottle count != 0)
030000  if bottle count = 0
010005  jmp Store
200064  d
400200  push
20006E  n
400200  push
200075  u
400200  push
20006F  o
400200  push
200072  r
400200  push
200061  a
400200  push
200020
400200  push
200074  t
400200  push
200069  i
400200  push
200020
400200  push
200073  s
400200  push
200073  s
400200  push
200061  a
400200  push
200070  p
400200  push
200020
400200  push
200064  d
400200  push
20006E  n
400200  push
200061  a
400200  push
200020
400200  push
20006E  n
400200  push
200077  w
400200  push
20006F  o
400200  push
200064  d
400200  push
200020
400200  push
200065  e
400200  push
20006E  n
400200  push
20006F  o
400200  push
200020
400200  push
200065  e
400200  push
20006B  k
400200  push
200061  a
400200  push
200054  T
400200  push
030400  if not initializing
140200  print string
030400  if not initializing
020100  return

000005  Label Store:
200020  ' '
400200  push
20002C  ','
400200  push
200065  e
400200  push
200072  r
400200  push
20006F  o
400200  push
20006D  m
400200  push
200020
400200  push
200065  e
400200  push
20006D  m
400200  push
20006F  o
400200  push
200073  s
400200  push
200020
400200  push
200079  y
400200  push
200075  u
400200  push
200062  b
400200  push
200020
400200  push
200064  d
400200  push
20006E  n
400200  push
200061  a
400200  push
200020
400200  push
200065  e
400200  push
200072  r
400200  push
20006F  o
400200  push
200074  t
400200  push
200073  s
400200  push
200020
400200  push
200065  e
400200  push
200068  h
400200  push
200074  t
400200  push
200020
400200  push
20006F  o
400200  push
200074  t
400200  push
200020
400200  push
20006F  o
400200  push
200047  G
400200  push
030400  if not initializing
140200  print string
200063  99
400000  push bottle counter
200015  0x15 (return address)   
400100  push retstack
030400  if not initializing
010001  jmp wall
000015  Label Return_wall_Store
20002E  '.'
400200  push
030400  if not initializing
140200  print string
030400  if not initializing
010027  go to end_program

000000  Label Main:
200010  0x10 (return address)   
400100  push retstack
030400  if not initializing
010001  jmp wall
000010  Label Return_wall_main
200020  ' '
400200  push
20002C  ','
400200  push
030400  if not initializing
140200  print string
200020  0x20 (return address)   
400100  push retstack
030400  if not initializing
010002  jmp beer
000020  Label Return_beer_main
20000A  <nl>
400200  push
20002E  '.'
400200  push
030400  if not initializing
140200  print string
410000  pop bottle count
400100  push on other stack
200001  1
400000  push
410100  pop bottle count
400000  push to original stack
220000  sub
200030  0x30 (return address)   
400100  push retstack
030400  if not initializing
010004  jmp Take
000030  Label Return_Take_main
200020  ' '
400200  push
20002C  ','
400200  push
030400  if not initializing
140200  print string
200040  0x40 (return address)   
400100  push retstack
030400  if not initializing
010001  jmp wall
000040  Label Return_wall_main_2
20000A  <nl>
400200  push
20000A  <nl>
400200  push
20002E  '.'
400200  push
030400  if not initializing
140200  print string
030400  if not initializing
010000  jmp main

000007  flush_strings:
410200  pop character
430200  get strings-stack size
400000  push
400000  push 0
300000  equals
030000  if characters left
010007  go to flush_strings
000017  flush_init_queue:
410400  pop entry
430400  get init_queue_stack size
400000  push
400000  push 0
300000  equals
030000  if entries left
010017  go to flush_init_queue

000027  end_program:
400300  push 0
300303  equals
420300  peek initialize
400300  push initialize
030300  if not initialize
010006  go to initializer
\$\endgroup\$
0
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Whitespace, 1977 1342 1013 877 bytes

(very much whitespace)

Try it online!

Finally, under 1000 bytes.

-635 bytes by manually correcting the automatic whitelips labels

-329 bytes by offsetting the letters and using 0 for spaces

-136 bytes by pushing whole sentences in one big integer and using divmod to calculate the letters.

visible code:

sssttsssttl push 99

lssssl      label main:
lsttl       call wall
lststl      call comma_space
lsttsl      call beer
lstsl       call stop_nl
ssstl       push 1
tsst        sub
lstttl      call take
lststl      call comma_space
lsttl       call wall
lstsl       call stop_nl
ssststsl    push newline (10)
tlss        printc
lslssl      jump main

lssttl      label take:
sls         dup
ltssssl     jz store
ssstststssl "T"
tlss        printc
ssstsstsstttssttsssttsststtstststststssststssttsttttttstsssstttstsssttstttssstssssttsssssssssttsstttststttsssttsssttststtstsssttttstssssttsstssssttl
            "ake one down and pass it around"
lsll        jmp print

lsssssl     label store:
sll         drop
sssttsssttl push 99
ssstssstttl "G"
tlss        printc
ssstsstststttttsttssssststtstttttssttsstsststttssssttsssttsssssttstssttstststtsttsttstssssstsssttstsstttttttssssttsttsttstststsststtssttsstssttttsttstttl
            "o to the store and buy some more"
lstl        call print
lststl      call comma_space
lsttl       call wall
lstsl       call stop_nl
lll         end

lsssl       label stop_nl:
ssststttsl  "."
tlss        printc
ssststsl    <lf>
tlss        printc
ltl         return

lssstl      label comma_space:
ssststtssl  ","
tlss        printc
sssl        0
lsll        jmp print

lsstl       label wall:
lsttsl      call beer
ssststssststssssttttttsstttsstststttstssttsstsssssstttttttsl
            " on the wall"
lsll        jmp print

lsstsl      label beer:
lststtl     call bottles
ssstssstssssstsssttsttsttstttssstttsttttsl
            " of beer"
lsll        jmp print

lsssttl     label bottles:
sls         dup bottlecount
tlst        printi
sssttsststsstssssstttsttssttttssssl
            " bottle"
lstl        call print
sls         dup bottlecount
ssstl       push 1
tsst        sub
ltststl     jz return
ssstssttl   's'

lssl        label print:
sssttstsl   26
stsstl      copy 1
stsstl      copy 1
tstt        mod
sls         dup
ltstssl     jz space
ssstssssssl push 64
tsss        add
lsstssl     label space:
ssstsssssl  push 32
tsss        add
tlss        printc
tsts        div
sls         dup
ltsstsl     jz end_print
lsll        jump print
lssstsl     label end_print:
sll         drop
lsststl     label return:
ltl         return
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0
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C# (Visual C# Interactive Compiler), 243 bytes

for(int i=99;i>0;){string s="s",a=" bottle",b=" of beer",w=" on the wall",c=$"{i+a+(i>1?s:"")+b}";Write(@$"{c+w}, {c}.
{(--i>0?$"Take one down and pass it around, {i+a+(i>1?s:"")+b+w}":$"Go to the store and buy some more, 99{a+s+b+w}")}.

");}

Try it online!

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0
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C (gcc), 229 bytes

#define B" bottle%c of beer"
#define W" on the wall"
f(x){for(x=99;x;)printf("%d"B W", %1$d"B".\n%s, %d"B W".\n\n",x+1,'s'*!!x,'s'*!!x,x?"Take one down and pass it around":"Go to the store and buy some more",x?x:99,--x-1?'s':0);}

Needs more golf...

Try it online!

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1
0
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JavaScript ES6, 179 bytes

for(a=99,b=" bottles of beer",c=" on the wall";0<a;)alert(a+b+c+", "+a+b+".\nTake one down, pass it around, "+(a-1)+b+c),a--;alert("Go to the store and buy some more, 99"+b+c+".")
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  • \$\begingroup\$ Doesn't seem to have the correct behavior: 1. supposed to be "1 bottle" and not "1 bottles". 2. At the end, there should be no "0 bottles of beer on the wall". 3. You are missing a few stops. 3. It's "Take one down and pass it around"; not "Take one down, pass it around". Though this seems to be the shortest JavaScript answer, it has some mistakes. \$\endgroup\$ – Recursive Co. May 18 at 15:31
0
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Pxem, Filename: 131 bytes + Content: 61 bytes = 192 bytes.

I escaped spaces too (\040).

  • Filename (escaped): c.z.e,.o\040.oq.e.s..o\n.o\001.-.c\001.y.sGo\040to\040the\040store\040and\040buy\040some\040more,\040.pc.\n.e.d.a.tTake\040one\040down\040and\040pass\040it\040around,\040.p.m.e..o\n.o\n.o.a
  • Content (escaped): .cd.y.t\040on\040the\040wall.mXXq.a.s.c.n.t\040of\040beer.m\001.zXXs.a\040bottle.p

Actual source

Filename

c.z.e,.o .oq.e.s..o
.o.-.c.y.sGo to the store and buy some more, .pc.
.e.d.a.tTake one down and pass it around, .p.m.e..o
.o
.o.a

Content

.cd.y.t on the wall.mXXq.a.s.c.n.t of beer.m.zXXs.a bottle.p

With comments

Inspired by Dorian's answer in Whitespace at first, then golfed well. Though both Whitespace and Pxem are stack-oriented programming language, their features are different, as in these:

  • Whitespace has label and (conditional) go-to commands; Pxem has while-done loop.
  • Both of them have subroutine, but Pxem can have only one subroutine. Also that of Pxem works as a function rather than a subroutine.
  • Whitespace has commands that accesses to non-top items of its stack; Pxem can reverse items on stack.
  • And more, I think.

Overall of the program

  • Iteration from 99 to 1 on filename.
  • Content stands for a subroutine to output strings that are frequently recycled.   - It depends on top of stack. If the item is more than 99, it outputs " on the wall" in addition to "[0-9]+ bottles? of beer".

Filename

XX.z
# push 99;
# while size<2||[ $(pop) -ne $(pop) ]; do
.ac.zXX.z
  # for i in $(reverse $(call $stack)); do
  #   push $i; done
  ## NOTE let it be "call" next time
  .a.eXX.z
  # push 44; putc $(pop); push 32; putc $(pop)
  ## NOTE DOTo is actually:
  ## if !empty; then : do above; fi
  .a,.o .oXX.z
  # push 113; call; pop
  ## NOTE 113>=100
  .aq.e.sXX
  # push 46; putchar $(pop);
  # push 10; putchar $(pop)
  .a..o\n.oXX.z
  # push 1;
  # push $(( (x=$(pop)-$(pop))<0?-x:x ))
  ## NOTE arithmetic operators actually:
  ## check if size>=2 before;
  ## NOP if size<2
  .a\001.-XX.z
  # dup; push 1;
  # while size<2||[ $(pop) -gt $(pop) ]; do
  .a.c\001.yXX.z
    # pop;
    ## NOTE literal is pushed from backwards
    # for i in 32 44 --- 71; do push $i; done
    .a.sGo to the store and buy some more, XX.z
    # while !empty; do putchar $(pop); done
    .a.pXX.z
    # push 10; push 46; push 99;
    # call
    .ac.\n.eXX.z
    # exit
    .a.dXX.z
  # done
  .a.aXX.z
  # if !empty; do heap=$(pop); fi
  # for i in 32 44 --- 84; do push $i; done
  .a.tTake one down and pass it around, XX.z
  # while !empty; do putc $(pop); done
  .a.pXX.z
  # case ${heap:+y} in (y)
  #   push $heap;; esac;
  # call; push 46; putc $(pop);
  # push 10; putc $(pop)
  # push 10; putc $(pop)
  .a.m.e..o\n.o\n.oXX.z
# done
.a.a

Content

XX.z
## NOTE subroutine stack is initialized
## to be same as that of its caller
## but its heap is initialized to be empty,
## which is not shared to its caller
## So, implicitly: stack="$*"
## NOTE when returning to subroutine,
## heap is discarded but items on its
## stack is pushed to its caller, from
## bottom to top
# dup; push 100;
# while size<2||[ $(pop) -gt $(pop) ]; do
.a.cd.yXX.z
  # heap=$(pop)
  .a.tXX.z
  # for i in 108 108 --- 32; do push $i; done;
  # push $heap;
  # push 113; push 88; push 88
  .a on the wall.mXXqXX.z
# done
.a.aXX.z
# pop; dup; printf %d $(pop)
.a.s.c.nXX.z
# heap=$(pop)
.a.tXX.z
# for i in 114 101 --- 32; do push $i; done;
# push $heap; push 1;
# while size<2||[ $(pop) -ne $(pop) ]; do
.a of beer.m\001.zXX.z
  # push 115; push 88; push 88
  .aXXXXs.z
# done
.a.aXX.z
# for i in 101 --- 32; do push $i; done;
# while !empty; do putc $(pop); done
.a bottle.pXX.z
# (* implicitly *) return $stack (* to its caller *)
.a

Try it online!

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0
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Duocentehexaquinquagesimal, 279 bytes

`Y7íY-þÌÅȵ3₃1œ₆"Σè‹×ê8ëbc¾½}tiQιδcâāιΖ´K2X©∍U7ΘËð»¦Õe˜ãΓh=_²₆G܃‚ÄÏ4ÒþO@æc†5ˆ‹₄ý¡)Esƶ’ĆË₁^rgÝyúÃćÆa^}λв?‰ZJ~‡çN©ƶƶнîõgoΈy«N§Lÿ£&Ćƶ¯…sHkHÆ#нåY~gûîf6!0§`zнÕlʒ3ǝú×…ù[9»¦\ÌååÌ}ι|~qsŒnNdΔd’â*›nQáîÃî~j™∍ît‹ó0uº5-˜ÐØîĆe&U!z©ÅñÒ(„†{k4Zéм›}%1₅cʒ€bιG6c‚¤[øÚYR:ñŒäηíMÝVηí\δÇñ5æƶÖ|›5Wì̦z

Try it online!

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0
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Pinecone, 216 bytes

l:" on the wall";k:" bottles of beer";i:99|i>0|i:i-1@(j:(i.String);print:j+k+l+", "+j+k+".";i=1?print:"Go to the store and buy some more, 99"+k+l+"."|print:"Take one down and pass it around, "+((i-1).String)+k+l+".")
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-1
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R, 269 bytes

R sometimes can process text without much pain. In this case can't, but the iteration system rocks:

cat(sprintf('\n%s bottles of beer on the wall, %s bottles of beer.Take one down and pass it around, %s bottles of beer on the wall.',99:2,99:2,98:1),'\n1 bottle of beer on the wall, 1 bottle of beer.\nGo to the store and buy some more, 99 bottles of beer on the wall.')
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  • 1
    \$\begingroup\$ Output does not match challenge example and when there's 2 left: "2 bottles of beer on the wall, 2 bottles of beer.Take one down and pass it around, 1 bottles of beer on the wall. " there's an extraneouss at "1 bottles". \$\endgroup\$ – Tensibai Feb 25 '16 at 9:20
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