127
\$\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\$
  • 2
    \$\begingroup\$ @Mr.Xcoder Then I'll just have to include one in my own source. \$\endgroup\$ – steenbergh Jul 15 '17 at 17:55
  • 3
    \$\begingroup\$ @Mr.Xcoder I think that you should have prevented reading your own source code. \$\endgroup\$ – caird coinheringaahing Jul 15 '17 at 18:09
  • 1
    \$\begingroup\$ It only needs to work when doubled once? We don't need to support n many doublings? \$\endgroup\$ – Cody Gray Jul 16 '17 at 8:36
  • 6
    \$\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 Jul 17 '17 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 Jul 18 '17 at 22:26

130 Answers 130

1
\$\begingroup\$

Ohm, 3 bytes

Original:

0Wl

Duplicated:

0Wl0Wl

Owles?

Explanation

0Wl    Main wire
0      Push 0
 W     Wrap in an array
  l    Push Length (1)
0Wl0Wl
0Wl    Push 1 ^
   0   Push another 0
    W  Wrap in an array ([0,1])
     l Push the length (2)

Try it online!
Try it online!Try it online!

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

PowerShell, 47 bytes

(gc $MyInvocation.MyCommand.Definition).Length#

Try it online! (Output: 47)

Output:

47

Explanation

(gc $MyInvocation.MyCommand.Definition).Length#
 ^  ^             ^         ^           ^     ^^
 |  |             |         |           |     ||
 |  |             |         |           |     |No newline and/or whitespace at the end
 |  |             |         |           |     |
 |  |             |         |           |     Hash - Marks all next text as comment
 |  |             |         |           |
 |  |             |         |           Length - Returns length of string
 |  |             |         |
 |  |             |         Definition - Returns full path to script file
 |  |             |
 |  |             MyCommand - Contains information about the script file
 |  |
 |  $MyInvocation - Auto variable containing information about the current script
 |
 Alias for Get-Content commandlet, returns content of file
\$\endgroup\$
1
\$\begingroup\$

JavaScript, 41 chars

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

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

\$\endgroup\$
  • \$\begingroup\$ Shows 3 and 4 for me using the Run code snippet button \$\endgroup\$ – FliiFe Jul 18 '17 at 20:07
  • \$\begingroup\$ @FliiFe, try in incognito window with all browser extensions disabled. \$\endgroup\$ – Qwertiy Jul 18 '17 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\$

Common Lisp REPL, 25 bytes

(if(boundp'z)6(setq z 3))

Try it online!

Try the double version 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\$
  • \$\begingroup\$ Can't you remove the newline/indent in front of class? \$\endgroup\$ – OldBunny2800 Jul 25 '17 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 Jul 25 '17 at 13:32
  • \$\begingroup\$ I mean, would class X():def __del__(s):print(y) work? \$\endgroup\$ – OldBunny2800 Jul 25 '17 at 13:33
  • \$\begingroup\$ Also, don't think you need parentheses after a class declaration. \$\endgroup\$ – OldBunny2800 Jul 25 '17 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 Jul 25 '17 at 13:35
1
\$\begingroup\$

Klein, 5 + 3 = 8 bytes

\+@
2

Single, Double

The single program puts a two on the stack and attempts to add it. There is nothing to add so it is a noop and outputs 2. In the doubled program the \+@ section is not encountered but we do hit an additional 2 meaning that when we add again we add two 2s instead of a 2 and a zero. This results in 4. 2 can be replaced with any single digit number and this will still work, + can also be replaced with a * as long as we keep the 2, and \ can be replaced with a /.

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

Python 3, 74 67 57 bytes

Quick and dirty... Output via text file is allowed per meta.

try:f+=[2]
except:f=[2]
open('o','w').write(str(sum(f)))

\$\endgroup\$
  • 1
    \$\begingroup\$ f.append(2) can be f+=[2] and try: f can be try:f and except: f=[] can be except:f=[] \$\endgroup\$ – Mr. Xcoder Jul 15 '17 at 15:36
  • \$\begingroup\$ @Dennis happier with this version? \$\endgroup\$ – wrymug Jul 16 '17 at 20:27
  • 1
    \$\begingroup\$ Much better. You need to count a trailing newline though, as open... and try... mustn't wind up on the same line. \$\endgroup\$ – Dennis Jul 17 '17 at 21:15
  • \$\begingroup\$ The rules allow assuming a trailing newline between copies. \$\endgroup\$ – zbw Jul 20 '17 at 15:52
  • \$\begingroup\$ @zbw the rules say explicitly the opposite of that \$\endgroup\$ – Taylor Scott Aug 2 '17 at 1:21
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\$
  • 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 Jul 19 '17 at 9:29
  • \$\begingroup\$ @Qwertiy - Thanks for bugreport. It is a bug on Lua demo web page. \$\endgroup\$ – Egor Skriptunoff Jul 22 '17 at 20:43
  • \$\begingroup\$ Yep, fine: ideone.com/5xuJhk & ideone.com/poh01N \$\endgroup\$ – Qwertiy Jul 22 '17 at 21:26
  • \$\begingroup\$ You can golf this down to 64 bytes (from 76 bytes at current) by removing whitespace (see Here) \$\endgroup\$ – Taylor Scott Aug 1 '17 at 12:24
  • 2
    \$\begingroup\$ @TaylorScott - All the bytes are <s>belong to us</s> counted! They are 63. Newline was removed. \$\endgroup\$ – Egor Skriptunoff Aug 3 '17 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\$
  • 1
    \$\begingroup\$ You can use END instead of at_exit for -7 bytes. \$\endgroup\$ – Jordan Dec 17 '17 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\$ – Simon George Dec 18 '17 at 15:49
  • \$\begingroup\$ Totally fine. Nice answer! \$\endgroup\$ – Jordan Dec 18 '17 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\$
  • \$\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 Sep 7 '17 at 5:32
  • \$\begingroup\$ @LyricLy Yes, but those require extra bytes. \$\endgroup\$ – MD XF Sep 8 '17 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 Sep 8 '17 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 Sep 8 '17 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 Sep 8 '17 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\$

Julia 0.6, 41 bytes

My first two ideas were already done in python answers (check file length, use a finalizer on an object). So here I check if π is an integer to gate the code that sets π=1 and registers the print to occur on exit. Then just double π each time.

π%1>0?(π=1;atexit(()->show(π))):π*=2;

Try it online!

Try twice it online!

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

Julia 0.6, 24 bytes

Riffing off of lungj's Python 3 answer. This works on my terminal on my mac, but not on TIO. The \r should reset back to the start of the line, so it always overwrites the previous output. Uses string interpolation with $ to execute π=2π÷1 which doubles π and uses integer division to round it to an integer. It does output "WARNING: imported binding for π overwritten in module Main" as well, but thats not an error, and is pretty close to a compiler warning.

print("\r$(π=2π÷1)");

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\$
  • \$\begingroup\$ You should indicate a newline by leaving an empty line in the code block, not by using an escape sequence. \$\endgroup\$ – Jakob Aug 22 '17 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 Aug 23 '17 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 Aug 23 '17 at 21:24
  • \$\begingroup\$ continuing in chat \$\endgroup\$ – Jakob Aug 24 '17 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\$ – Martin Ender Feb 13 '18 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\$

Rust, 81 bytes

mod x{pub const Y:i8=1;}use x::*;fn main(){print!("{}",Y)}/*
const Y:i8=2;// *///

Doubled:

mod x{pub const Y:i8=1;}use x::*;fn main(){print!("{}",Y)}/*
const Y:i8=2;// *///mod x{pub const Y:i8=1;}use x::*;fn main(){print!("{}",Y)}/*
const Y:i8=2;// *///

Rust pretty much doesn't allow duplicate items in a source code. For instance, following code causes an error due to an item defined multiple times.

const Y: i32 = 2;
const Y: i32 = 2;

There are three exceptions to this rule.

  • Macros - which are pretty much useless, as macros don't follow the usual visibility rules - code cannot refer to a macro later in the code.
  • Wildcard imports - if there is a non-wildcard import, it has precedence while resolving item references. This is to help avoid incompatibilities caused by other crates (including std itself) adding more public items.
  • Overriding a builtin item or item from prelude - see bonus below, this way turned out to be longer, but also has more potential for improvements.

I decided to go with duplicating an item by using a glob import. This necessiated making a module.

mod x { ... }

That had a public item in it. Not public items aren't accessible outside of module that defined them. i8 type was chosen because it's the shortest integer type -- a type needs to be declared for const items, this cannot be skipped.

A string literal wasn't used as &str is 2 bytes longer, and also quotes would be necessary, not saving bytes even with removal of "{}", from main function. Adding 6 bytes is not worth it for removing 5 bytes.

pub const Y: i8 = 1;

Later I glob import this constant. Note that the constant can be overridden by a different non-wildcard declaration.

use x::*;

And a function prints whatever value Y holds. Note that Y can be overridden by a non-wildcard declaration. This is important when doubling source code.

print! ends with an exclamation mark as it is a macro. As the first parameter is a formatting pattern, I cannot use an integer directly, instead I have to specify "{}", formatting pattern.

Missing semicolon at the end of block means that this block returns a value. This is fine, as main returns () (implicit, due to not specifying another return type), and print! macro returns () (as in, it doesn't have an useful return value).

fn main() {
    print!("{}", Y)
}

Later, I want there to be a second declaration of Y, but I don't want it come into play for the first pass, so I comment it out. There is a newline, so that a line comment that will be declared later won't skip over a constant declaration. There is a space after // as there is nested comments feature, otherwise Rust would see /* inside block comment and start a nested comment.

/*
const Y:i8=2;// */

At end, I put //, so that first line of code of code is skipped when the code is doubled. It involves items that cannot be defined multiple times.

//

The execution may conntinue from second line. If it does, value of Y used by main function will be different.

 const Y: i8 = 2;

A line comment is included so that closing block comment that was needed for first pass won't cause issues.

// *///

And that's it, a program that detect it being duplicated in Rust done in 81 characters. Thank you for reading this explanation.

Bonus (alternative way for possible improvements, 93 92 bytes)

Instead of using glob imports, it's possible to override one builtin items from either a builtin type or something imported from prelude.

fn main(){println!("{}",i8::max_value())}/*
enum i8{}impl i8{fn max_value()->u8{254}}// *///

However, this solution is currently longer than the solution above.

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

JavaScript (console), 30 bytes

[$^=1]+{valueOf:x=>alert(2-$)}

In console environments, we have the $ default global variable, which is two characters shorter than Map, which is the shortest in other environments.

The way this works:

  1. $ is coerced to a number, and has bitwise XOR applied to it, making it 1.
  2. The object at the end is part of an addition equation, so it is also coerced into a number, calling the valueOf function which alerts 2-$, which equals 1.

When the code is repeated, it looks like this:

[$^=1]+{valueOf:x=>alert(2-$)}[$^=1]+{valueOf:x=>alert(2-$)}

The first instance of the object now has a member operator attached, so instead of the object's valueOf being called, instead we have (object)[$^=1 /* 0 */ ], which is undefined. The first instance doesn't alert anything, but the second one does, and by now $ has been changed to 0 because $^=1 has run twice, so it alerts 2.

Or, for non-console environments, we can use Map instead of $ (34 bytes):

[Map^=1]+{valueOf:x=>alert(2-Map)}

Or, we could use...

JavaScript, 32 bytes

-0?alert(2)+'':[,a=alert(1)]=[1]

When not repeated:

When repeated:

-0?alert(2)+'':[,a=alert(1)]=[1]-0?alert(2)+'':[,a=alert(1)]=[1]

  • -0 is still false, so we go to [,a=alert(1)]=[1]-0?alert(2)+'':[,a=alert(1)]=[1]
  • [1]-0 returns 1 which is true, so we alert(2) and coerce the undefined result to a string by adding an empty string. This returns "undefined". a is still being set to the "undefined"[1] which is this case is the letter "n". Since "n" is a value, the default alert(1) isn't run.
\$\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\$

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