150
\$\begingroup\$

Your task, if you wish to accept it, is to write a program that outputs a positive integer (higher than 0). The tricky part is that if I duplicate your source code, the output must be double the original integer.

Rules

  • You must build a full program. That is, your output has to be printed to STDOUT.

  • The initial source must be at least 1 byte long.

  • Both the integers must be in base 10 (outputting them in any other base or with scientific notation is forbidden).

  • Your program must not take input (or have an unused, empty input).

  • Outputting the integers with trailing / leading spaces is allowed.

  • You may not assume a newline between copies of your source.

  • This is , so the shortest (original) code in each language wins!

  • Default Loopholes apply.

Example

Let's say your source code is ABC and its corresponding output is 4. If I write ABCABC instead and run it, the output must be 8.

Leaderboard

This uses uses @manatwork's layout.

/* Configuration */

var QUESTION_ID = 132558; // Obtain this from the url
// It will be like https://XYZ.stackexchange.com/questions/QUESTION_ID/... on any question page
var ANSWER_FILTER = "!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe";
var COMMENT_FILTER = "!)Q2B_A2kjfAiU78X(md6BoYk";
var OVERRIDE_USER = 8349457; // This should be the user ID of the challenge author.

/* App */

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,]*[^\s,]),.*?(-?\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,
      });
    
  });
  
  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;
    if (! /<a/.test(lang)) lang = '<i>' + lang + '</i>';
    lang = jQuery(lang).text().toLowerCase();

    languages[lang] = languages[lang] || {lang: a.language, user: a.user, size: a.size, link: a.link, uniq: lang};
  });

  var langs = [];
  for (var lang in languages)
    if (languages.hasOwnProperty(lang))
      langs.push(languages[lang]);

  langs.sort(function (a, b) {
    if (a.uniq > b.uniq) return 1;
    if (a.uniq < b.uniq) 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: bold;
}

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/Sites/codegolf/all.css?v=617d0685f6f3">
<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>
<div id="language-list">
  <h2>Winners 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>
<table style="display: none">
  <tbody id="answer-template">
    <tr><td>{{PLACE}}</td><td>{{NAME}}</td><td>{{LANGUAGE}}</td><td><a href="{{LINK}}">{{SIZE}}</a></td></tr>
  </tbody>
</table>
<table style="display: none">
  <tbody id="language-template">
    <tr><td>{{LANGUAGE}}</td><td>{{NAME}}</td><td><a href="{{LINK}}">{{SIZE}}</a></td></tr>
  </tbody>
</table>

\$\endgroup\$
33
  • 2
    \$\begingroup\$ @Mr.Xcoder Then I'll just have to include one in my own source. \$\endgroup\$
    – steenbergh
    Commented Jul 15, 2017 at 17:55
  • 3
    \$\begingroup\$ @Mr.Xcoder I think that you should have prevented reading your own source code. \$\endgroup\$ Commented Jul 15, 2017 at 18:09
  • 2
    \$\begingroup\$ It only needs to work when doubled once? We don't need to support n many doublings? \$\endgroup\$ Commented Jul 16, 2017 at 8:36
  • 8
    \$\begingroup\$ @Daniel Let's say your source is `` (empty program) and it produces 5. If you double it, your source is `` (empty program) and that produces 5 as well, no matter what you do. That being said, an empty program duplicated is still the empty program, and always produces the same output, except for the case where the empty program means something else (a random number generator, for example), which could not be valid anyway. \$\endgroup\$
    – Mr. Xcoder
    Commented Jul 17, 2017 at 9:11
  • 1
    \$\begingroup\$ This shouldn't be hard for esolangs that automatically dump the top of stack upon program termination. \$\endgroup\$
    – MD XF
    Commented Jul 18, 2017 at 22:26

162 Answers 162

1
\$\begingroup\$

JavaScript, 41 chars

clearTimeout(1);x=setTimeout('alert(x)');

clearTimeout(1);x=setTimeout('alert(x)');clearTimeout(1);x=setTimeout('alert(x)');

\$\endgroup\$
2
  • \$\begingroup\$ Shows 3 and 4 for me using the Run code snippet button \$\endgroup\$
    – THC
    Commented Jul 18, 2017 at 20:07
  • \$\begingroup\$ @FliiFe, try in incognito window with all browser extensions disabled. \$\endgroup\$
    – Qwertiy
    Commented Jul 18, 2017 at 20:54
1
\$\begingroup\$

><>, 9 bytes

l";3"+10p

Try it online!

Explanation

l";3"+10p
l              | Push the length of the stack to the stack; STACK[0]
 ";3"          | Push 59 and 51 to the stack;               STACK[0, 59, 51]
     +         | Add the stack top 2 items;                 STACK[0, 110]
      10p      | At codebox[0,1] put the stack top item;    STACK[0]
                   NEW CODEBOX; ln;3"+10p
ln;            | Push length, print stack top and end;

When doubled we run through the code twice, it edits the print function in the same spot but doubles the amount of items we push to the stack.

><>, 18 bytes Doubled version

l";3"+10pl";3"+10p

Try it online!

\$\endgroup\$
1
\$\begingroup\$

Bash 35 bytes

trap 'echo $a' 0
a=$((2**${a:-1}))

single source output:

2

double source output:

4

This uses bash default value parameter expansion to set a to either 1 or the result of 2^a. An exit trap is defined that will print the current value of a, so each additional copy of the source will double the output.

\$\endgroup\$
1
\$\begingroup\$

Python 3, 62 60 bytes

This time, back to stdout and no cheating.

class X:
 def __del__(s):print(y)
try:y*=2
except:y=1;z=X()

Saved 2 bytes thanks to @OldBunny2800!

\$\endgroup\$
5
  • \$\begingroup\$ Can't you remove the newline/indent in front of class? \$\endgroup\$
    – AAM111
    Commented Jul 25, 2017 at 13:28
  • \$\begingroup\$ There shouldn't be one in front of class; do you mean the def? I think that's necessary. I don't know how to put method definition on the same line as the class definition. \$\endgroup\$
    – lungj
    Commented Jul 25, 2017 at 13:32
  • \$\begingroup\$ I mean, would class X():def __del__(s):print(y) work? \$\endgroup\$
    – AAM111
    Commented Jul 25, 2017 at 13:33
  • \$\begingroup\$ Also, don't think you need parentheses after a class declaration. \$\endgroup\$
    – AAM111
    Commented Jul 25, 2017 at 13:34
  • \$\begingroup\$ The one liner does not work on Python 3.6.0 on macOS installed via homebrew, but it's possible works on some other Python 3. \$\endgroup\$
    – lungj
    Commented Jul 25, 2017 at 13:35
1
\$\begingroup\$

Python 3, 48 bytes

Apparently, file I/O is acceptable as an output format.

try:y*=2
except:y=1
open('x','w').write(str(y))

The 48th character is a new-line character at the end of this program.

\$\endgroup\$
1
\$\begingroup\$

WinDBG, 40 bytes

r$t0=d;r$t0=@$t0*2;.printf"\r%d",@$t0;$$

Outputs 26, and that doubles each time the source is appended.

How it works:

r $t0 = d;                           Initialize psuedo-register t0 to 13
r $t0 = @$t0 * 2;                    Double t0
.printf "\r%d", @$t0;                Move caret to start of output line and print t0
$$                                   Comment until the next ; (comment r $t0 = d)

Sample output:

0:000> r$t0=d;r$t0=@$t0*2;.printf"\r%d",@$t0;$$
26

0:000> r$t0=d;r$t0=@$t0*2;.printf"\r%d",@$t0;$$r$t0=d;r$t0=@$t0*2;.printf"\r%d",@$t0;$$
52

0:000> r$t0=d;r$t0=@$t0*2;.printf"\r%d",@$t0;$$r$t0=d;r$t0=@$t0*2;.printf"\r%d",@$t0;$$r$t0=d;r$t0=@$t0*2;.printf"\r%d",@$t0;$$
104

0:000> r$t0=d;r$t0=@$t0*2;.printf"\r%d",@$t0;$$r$t0=d;r$t0=@$t0*2;.printf"\r%d",@$t0;$$r$t0=d;r$t0=@$t0*2;.printf"\r%d",@$t0;$$r$t0=d;r$t0=@$t0*2;.printf"\r%d",@$t0;$$
208
\$\endgroup\$
1
\$\begingroup\$

Scala (interpreted), 85 Bytes

if(System getProperty "v"eq "")print(2)else{System.setProperty("v","");print("1\r")};

Commented

if(System getProperty "v"eq "") // If the system property "v" is set to ""
  print(2)                      // Print 2
else{                           // Otherwise
  System.setProperty("v","");   // Set the system property "v" to ""
  print("1\r")                  // Print 1 with a carriage return
};


Text that ends with a carriage return and not a newline will be overwritten if anything else is written to the line.


Note: likely doesn't work on all consoles, tested on Windows 8.1 command prompt. For example the TIO console.

\$\endgroup\$
1
\$\begingroup\$

Lua, 63 bytes

a=1+(a or 0)t=setmetatable(t or{},{__gc=function()print(a)end})

This code does not have a newline at the end.
When program starts, variable a is nil, then it gets modified from nil to 1 and (if program code is doubled) from 1 to 2.
The finalizer of table t gets executed when program finishes (when memory of the table is released), it simply prints last value of variable a.
Lua 5.2+ is required.

\$\endgroup\$
7
  • 1
    \$\begingroup\$ Seems like it prints nothing. Checked it at lua.org/cgi-bin/demo. Isn't it too late to print when the finalizer is called? \$\endgroup\$
    – Qwertiy
    Commented Jul 19, 2017 at 9:29
  • \$\begingroup\$ @Qwertiy - Thanks for bugreport. It is a bug on Lua demo web page. \$\endgroup\$ Commented Jul 22, 2017 at 20:43
  • \$\begingroup\$ Yep, fine: ideone.com/5xuJhk & ideone.com/poh01N \$\endgroup\$
    – Qwertiy
    Commented Jul 22, 2017 at 21:26
  • \$\begingroup\$ You can golf this down to 64 bytes (from 76 bytes at current) by removing whitespace (see Here) \$\endgroup\$ Commented Aug 1, 2017 at 12:24
  • 2
    \$\begingroup\$ @TaylorScott - All the bytes are <s>belong to us</s> counted! They are 63. Newline was removed. \$\endgroup\$ Commented Aug 3, 2017 at 18:53
1
\$\begingroup\$

Ruby, 21 bytes

$.+=1
END{p$.
exit!};

1x: Try it online! 2x: Try it online!

\$\endgroup\$
1
\$\begingroup\$

Ruby, 31 bytes

x||=1;x*=2;END{p x;exit!};

Try it online!

\$\endgroup\$
3
  • 1
    \$\begingroup\$ You can use END instead of at_exit for -7 bytes. \$\endgroup\$
    – Jordan
    Commented Dec 17, 2017 at 23:02
  • \$\begingroup\$ @Jordan nice, thanks! I've edited to incorporate that, but your answer is still shorter so I hope that's okay with you :) \$\endgroup\$ Commented Dec 18, 2017 at 15:49
  • \$\begingroup\$ Totally fine. Nice answer! \$\endgroup\$
    – Jordan
    Commented Dec 18, 2017 at 15:50
1
\$\begingroup\$

Implicit on TIO, 2 bytes

:.

Hmm, not particularly positive how this works. I'll explain it to myself:

:.
    (implicit read integer input), duplicate stack (stack: 0, 0)
 .  increment                                      (stack: 0, 1)

Ok, now I see. : doesn't read input if the stack has a value on it.

:.:.
:     stack: 0, 0
 .    stack: 0, 1
  :   stack: 0, 1, 1
   .  stack: 0, 1, 2

Only works on TIO (:., :.:.) when the input box is empty.

\$\endgroup\$
5
  • \$\begingroup\$ Would this not also work if you redirect STDIN to an empty file, or if you press Ctrl + Shift + 2 when prompted for input? \$\endgroup\$
    – LyricLy
    Commented Sep 7, 2017 at 5:32
  • \$\begingroup\$ @LyricLy Yes, but those require extra bytes. \$\endgroup\$
    – MD XF
    Commented Sep 8, 2017 at 2:42
  • \$\begingroup\$ I don't understand. Doesn't it work that way already? Both methods would be sending empty input, just like TIO does. Or am I misunderstanding the reason that it only works on TIO? \$\endgroup\$
    – LyricLy
    Commented Sep 8, 2017 at 2:45
  • \$\begingroup\$ @LyricLy Yes, it works that way. But redirecting to STDIN would require, in a shell, ./interpreter sourcefile < emptyfile, and assuming you have an empty file is a standard loophole. Pressing Ctrl+Shift+2 would require user input that is not specified in the challenge, another standard loophole. \$\endgroup\$
    – MD XF
    Commented Sep 8, 2017 at 2:46
  • 1
    \$\begingroup\$ I wouldn't consider merely requiring EOF to be piped to the program a standard loophole, but I see where you're coming from. \$\endgroup\$
    – LyricLy
    Commented Sep 8, 2017 at 2:52
1
\$\begingroup\$

Pyt, 2 bytes

Explanation:

0       Push 0 onto the stack [0]
 Ł      Get the length of the stack (1)


Try it online!


Doubled:

0Ł0Ł

Explanation:

0           Push 0 onto the stack [0]
 Ł          Get the length of the stack (1)
  0         Push 0 onto the stack [1,0]
   Ł        Get the length of the stack (2)


Try it online!

\$\endgroup\$
1
\$\begingroup\$

Chip, 11+5 = 16 bytes

+5 bytes for -wc1

<b
*\ae*f
`

The active code:

*\ae*f

Activates a, e, and f which means 0x00110001.

Active code when doubled:

 b
*\ae*f
`<
*\a

Prevents activation of both a's, activates b, e, and f, which means 0x00110010.

Try it online!

\$\endgroup\$
1
\$\begingroup\$

tinylisp, 16 bytes

((q(G(i G 2 1)))

Try it online!

Try it doubled!

How?

(q (G (function-body))) creates an unnamed lambda function that takes a variable number of arguments and names the arglist G. In our case, the body of the function is (i G 2 1), which tests if G is empty or not; if it is, we return 1; if it isn't, we return 2. In other words, if we call this function with zero arguments, it returns 1; otherwise, 2.

In the single version, therefore, we call the function without arguments: (function (with an implicit closing parenthesis). In the double version, we call the function, and then call it again with that result as an argument: (function (function (with two implicit closing parentheses).

\$\endgroup\$
1
\$\begingroup\$

Newline 3 bytes

i\n

Note: running newline in TC mode

Output is the red text up top

Try it online

\$\endgroup\$
8
  • \$\begingroup\$ You should indicate a newline by leaving an empty line in the code block, not by using an escape sequence. \$\endgroup\$
    – Jakob
    Commented Aug 22, 2017 at 22:38
  • \$\begingroup\$ You're referring to the last line of code, correct? That is an actual newline; the string literal is 2 bytes (i and a newline), but its representation in JavaScript source is three (i\n). Your submission here should be encoded as text, not JavaScript source. \$\endgroup\$
    – Jakob
    Commented Aug 23, 2017 at 1:32
  • \$\begingroup\$ Could you provide a reference to the thread on Meta? You posted a 3-byte solution here, but the solution you link to is 2 bytes. This doesn't make sense to me. \$\endgroup\$
    – Jakob
    Commented Aug 23, 2017 at 21:24
  • \$\begingroup\$ continuing in chat \$\endgroup\$
    – Jakob
    Commented Aug 24, 2017 at 21:25
  • 1
    \$\begingroup\$ @Christopher This has nothing to do with rewriting the interpreter in another language. The source code of your program in memory is 2 bytes, not 3. Whether you create that string with a literal (which takes more than 2 characters) or read it from a file (which would contain 2 characters) doesn't matter. \$\endgroup\$ Commented Feb 13, 2018 at 14:15
1
\$\begingroup\$

Microscript, 2 bytes

1 

(Note the trailing space.)

\$\endgroup\$
1
\$\begingroup\$

SNOBOL4 (CSNOBOL4), 48 bytes

	INPUT('I',1,,'.code.tio');	OUTPUT =size(I);end;

Try it online!

Equivalent to this python answer.

Because the existence of an END label terminates the program, reading the source file is the only way to get this to work, as the program ABCABC is otherwise equivalent to ABC.

SNOBOL always reads input one line at a time so we have to use ; rather than newlines.

\$\endgroup\$
1
\$\begingroup\$

Stax, 2 bytes

|X

Run online, doubled

Added for completeness. |X in Stax means increment register x and push. Register x is implicitly initialized with 0. The top of stack is implicitly output.

\$\endgroup\$
1
\$\begingroup\$

SmileBASIC, 11 10 bytes

CLS?X+1X=1
\$\endgroup\$
1
\$\begingroup\$

Yabasic, 22 bytes

An Anonymous function that takes no input and outputs to STDOUT in graphics mode.

Clear Screen
n=n+1
?n

Note: Because this answer uses graphics mode, it does not function on TIO

\$\endgroup\$
1
\$\begingroup\$

Fission, 13 bytes

O\aL;
+
$
SV;

Returns 1

Try it online!

Doubled:

O\aL;
+
$
SV;O\aL;
+
$
SV;

Returns 2

Try it online!

Fun twist with this language, that doubling source will double atoms, and each atom is in general command pointer. So my idea was to build such code that second atom will be destroyed.

  L  Created atom, moving left (mass 1, energy 0)
 a   stores in it ascii code of 'a' (97)
\    mirrors
 V   fission reactor, spliting atom into two with halved masses (48)
  ;  destroy right atom
S    conditional mirror - mirrors right, if energy is zero (default energy level)
$    increase energy by 1 (mass 48, energy 1)
+    increase mass by one (mass 49)

if one copy on source:

O    print ascii by mass of atom, destroy atom (prints 1)

if code was copied

S    conditional mirror - energy is 1 now, so goes throw
$    increase energy by 1 (mass 49, energy 2)
+    increase mass by one (mass 50)
O    print ascii by mass of atom, destroy atom (prints 2)

Second atom is created with copied code \aL; on 4th row. Mirror \ sends it to the ; on the first row, where atom is destroyed

\$\endgroup\$
1
\$\begingroup\$

Reflections, 10 bytes

_~#  _#_v

Test it! Test it double!

Explanation:

  • _~: read own source and push size
  • # _: convert to string
  • #_: print the first digit
  • v: reflects the IP down

Then, the program does either end when hitting the other v or when leaving the grid.

\$\endgroup\$
0
1
\$\begingroup\$

QBasic, 12 bytes

A script that takes no input and outputs to the console. Outputs 1 when single, outputs 2 when doubled.

CLS
n=n+1
?n
\$\endgroup\$
1
\$\begingroup\$

Small Basic, 45 bytes

A script which takes no input and outputs to the TextWindow console.

n=n+1
TextWindow.Clear()
TextWindow.Write(n)

Try it at SmallBasic.com

n=n+1
TextWindow.Clear()
TextWindow.Write(n)
n=n+1
TextWindow.Clear()
TextWindow.Write(n)

Try it Doubled at SmallBasic.com

SmallBasic.com depends on Silverlight, and thus the links must be opened in IE to function.

\$\endgroup\$
1
\$\begingroup\$

Python 2, 25 24 bytes

x=1
print 4*'\b',x,;x=2#

Both values of x (1 and 2) get printed when the code is repeated. However the second time, the backspace character backspaces/"erases" the 1 and prints the 2 over the top of it.

The behaviour of printing the backspace escape character \b seems quite system dependent (and it doesn't seem to work on many web REPLs...).

Trailing comment idea inspired by W W's answer.

Edit: byte saved, see comment.

\$\endgroup\$
3
  • \$\begingroup\$ Welcome to the site! \$\endgroup\$
    – Wheat Wizard
    Commented Jul 17, 2018 at 17:52
  • \$\begingroup\$ Why don't you put x=2 after the print and before the comment instead of multiplying by two before the print? \$\endgroup\$
    – Wheat Wizard
    Commented Jul 17, 2018 at 17:54
  • \$\begingroup\$ Thanks: I think that will save a byte or two... \$\endgroup\$
    – Harry King
    Commented Jul 17, 2018 at 19:46
1
\$\begingroup\$

Z80Golf, 6 bytes

00000000: 3676 f630 3ce5                           6v.0<.

Try it online!

Doubled

00000000: 3676 f630 3ce5 3676 f630 3ce5            6v.0<.6v.0<.

Try it online!

Disassembly

start:
  ld (hl), $76
  or $30
  inc a
  push hl

The trick is to exclude any call or rst instructions, so that the execution exactly follows the order:

  • The program (one or two copies)
  • The nop slide, then putchar at address $8000 (exactly once), and then
  • halt which should be hit on return from putchar.

or $30; inc a sets up the ASCII '1' = $31 to print. push hl sets up the stack; ld (hl), $76 writes the halt instruction on the return address.

When doubled, the second inc a changes the value to ASCII '2' = $32. The other instructions are effective no-ops; the return address and the halt instruction on return don't change.

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1
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Ahead, 7 bytes

1~@O+K~

Prints 1 on its own, 2 when doubled, 3 when tripled etc.

Try it online!

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1
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Aheui, 23 bytes

분아떠망히
아뷴

outputs 4.

Doubled:

분아떠망히
아뷴
분아떠망히
아뷴

outputs 8.

Try it on jsaheui

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1
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W, 2 bytes

1+

Explanation

1+   % Adds an (implicit) 0 (on empty input)
  1+ % Add the constant 1 by 1
     % Implicit output
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1
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MathGolf, 1 byte

Added just for completeness.

)

Try it online!

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