# I double the source, you double the output!

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.

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 COMMENT_FILTER = "!)Q2B_A2kjfAiU78X(md6BoYk";
var OVERRIDE_USER = 8349457; // This should be the user ID of the challenge author.

/* App */

return "https://api.stackexchange.com/2.2/questions/" +  QUESTION_ID + "/answers?page=" + index + "&pagesize=100&order=desc&sort=creation&site=codegolf&filter=" + ANSWER_FILTER;
}

}

jQuery.ajax({
method: "get",
dataType: "jsonp",
crossDomain: true,
success: function (data) {
data.items.forEach(function(a) {
});
comment_page = 1;
}
});
}

jQuery.ajax({
method: "get",
dataType: "jsonp",
crossDomain: true,
success: function (data) {
data.items.forEach(function(c) {
if (c.owner.user_id === OVERRIDE_USER)
});
else process();
}
});
}

var SCORE_REG = /<h\d>\s*([^\n,]*[^\s,]),.*?(-?\d+)(?=[^\n\d<>]*(?:<(?:s>[^\n<>]*<\/s>|[^\n<>]+>)[^\n\d<>]*)*<\/h\d>)/;

function getAuthorName(a) {
return a.owner.display_name;
}

function process() {
var valid = [];

var body = a.body;
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],
});

});

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;

.replace("{{NAME}}", a.user)
.replace("{{LANGUAGE}}", a.language)
.replace("{{SIZE}}", a.size)

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)
language = jQuery(language);
jQuery("#languages").append(language);
}

}
body { text-align: left !important}

width: 290px;
float: left;
}

#language-list {
width: 290px;
float: left;
}

font-weight: bold;
}

table td {
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<tr><td></td><td>Author</td><td>Language</td><td>Size</td></tr>

</tbody>
</table>
</div>
<div id="language-list">
<h2>Winners by Language</h2>
<table class="language-list">
<tr><td>Language</td><td>User</td><td>Score</td></tr>
<tbody id="languages">

</tbody>
</table>
</div>
<table style="display: none">
</tbody>
</table>
<table style="display: none">
<tbody id="language-template">
</tbody>
</table>

• @Mr.Xcoder Then I'll just have to include one in my own source. – steenbergh Jul 15 '17 at 17:55
• @Mr.Xcoder I think that you should have prevented reading your own source code. – caird coinheringaahing Jul 15 '17 at 18:09
• It only needs to work when doubled once? We don't need to support n many doublings? – Cody Gray Jul 16 '17 at 8:36
• @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. – Mr. Xcoder Jul 17 '17 at 9:11
• This shouldn't be hard for esolangs that automatically dump the top of stack upon program termination. – MD XF Jul 18 '17 at 22:26

# Python 2, 33 bytes

print len(open(__file__).read())#


Try it online!

Try it doubled

# Python 3, 28 bytes

print(len(*open(__file__)))#


Try it online!

Try it doubled

## Explanation

This opens up the source code using open(__file__) and gets its length using len the # prevents any additional code from being read. When the source is doubled so is the length.

• Wow, I'm stunned... That's so brilliant! – Mr. Xcoder Jul 15 '17 at 15:24
• 32 bytes. Works by using append mode, setting the current position to the end of the file. tell() returns the current position in the file – Halvard Hummel Aug 29 '17 at 5:34
• @HalvardHummel Nice. However I have no intention of updating this answer. If you would like to post it on your own it is substantially different in my opinion. – Wheat Wizard Aug 29 '17 at 5:44
• @WheatWizard That is understandable, I made a separate answer – Halvard Hummel Aug 29 '17 at 5:48

# Jelly, 1 byte

‘


Try it online!

I have no idea how this works, but apparently it does.

• That moment when you have no idea what you have written... – Mr. Xcoder Jul 15 '17 at 15:31
• Darn it, I just thought of this 8 minutes too late. – HyperNeutrino Jul 15 '17 at 15:40
• All links need an argument. If the first element of the chain is a nilad, its result becomes the argument and the link is executed monadically. If there is no leading nilad, 0 is used instead. – Dennis Jul 15 '17 at 20:03

Anonymous worksheet formula that takes no input and outputs into the cell which holds the formula

=4/(2


As a single formula this evaluates to a call stack that looks a little something like

=4/(2
=4/(2)
=4/2
=2
2


However when this worksheet formula is doubled this call stack evaluates down to

=4/(2=4/(2
=4/(2=4/(2)
=4/(2=4/(2))
=4/(2=2)
=4/(True)
=4/True
=4/1
=4
4


Of course, an implication of using this method is that once this is repeated more than once, at the third and all following iterations of the problem, the call stack will reach =4/(2=4) and thus evaluate down to =4/0 and throw a #DIV/0! error

• Never expected Google Sheets would have use in code golf. Clever solution – hucancode Jul 17 '17 at 3:11
• @hucancode the really interesting bit about this is that because Excel throws an error if you exclude the trailing )'s this answer is the only Google Sheets answer that I've Seen that does not translate into an Excel answer – Taylor Scott Jul 19 '17 at 13:18

# 05AB1E, 2 bytes

Original

XO


Try it online!

Doubled

XOXO


Try it online!

Explanation

X pushes 1 to the stack.
O sums the stack.

• XOXO, nice solution. – Mr. Xcoder Jul 15 '17 at 15:36
• You did that on purpose, while you knew you could've also used 1O! – Erik the Outgolfer Jul 15 '17 at 19:00
• Where do you guys find these ridiculous languages? – DaveTheMinion Jul 17 '17 at 4:01
• @DavidB Usually, they write them. – Federico Poloni Jul 20 '17 at 6:56
• @DavidB Yes, people do invent languages for codegolf, and yes, they can get impressively low scores, but doing silly things like inventing a language after the challenge to solve it in 1 byte are disallowed, and programming in these languages is usually far from easy. – Esolanging Fruit Jan 10 '18 at 4:03

## C (gcc), 37 bytes

i;main(){putchar(i+49);}/*
i=1;//*///


The file does not contain a trailing newline.

Doubled version, for syntax highlighting:

i;main(){putchar(i+49);}/*
i=1;//*///i;main(){putchar(i+49);}/*
i=1;//*///


• Can you please explain how this works? why would the comment ever be uncommented? – phil294 Jul 15 '17 at 20:29
• When you double the source code, the /* is commented out by the //, which means the following i=1 is uncommented. This is easier to see if you put the doubled version of the code in a syntax highlighter – musicman523 Jul 15 '17 at 20:50
• Whoa, a tentative definition trick. Nice. – aschepler Jul 18 '17 at 11:18

# Python 2, 21 bytes

+1
if id:id=0;print 1


Try it online!

Doubled:

+1
if id:id=0;print 1+1
if id:id=0;print 1


Try it online!

• Crazily creative! Congrats! – Mr. Xcoder Jul 16 '17 at 7:50
• What if it is doubled with a linefeed inbetween? – user19214 Mar 17 '18 at 2:45

# Hexagony, 7 bytes

/)!@.).


Prints 1 regularly then 2 doubled.

### Expanded versions:

Regular:

 / )
! @ .
) .


Doubled:

  / ) !
@ . ) .
/ ) ! @ .
) . . .
. . .


The regular program follows the path: /)!.@ which increments a memory edge (all are initialised to zero) then prints its numeric value. The doubled program follows: /.)/)!@ which increments the edge twice before printing, instead.

• Wow nice work. I assume you found that by hand? Since 6 bytes is in brute force range, I thought I'd give it a go, and there's actually a 4-byte solution: [@!) (and some 570 5-byte solutions). Since you actually went to the trouble of finding a solution by hand, I'm perfectly happy for you to post the 4-byte solution. – Martin Ender Jul 16 '17 at 8:16
• If you're interested, here is the full list including the number that is printed: pastebin.com/TtRujjA4 – Martin Ender Jul 16 '17 at 8:24

# Braingolf, 1 byte

+


Try it online!

Now we're talkin'!

Outputs 20, or 40 when source is doubled.

## Explanation

+ is of course the "sum", "add" or "plus" operator, in Braingolf, however it has dyadic, monadic and niladic functions.

When there are at least 2 items on the stack, it's dyadic, and will sum the top 2 items of the stack.

When there is only 1 item on the stack, it's monadic, and will double the item.

When there are no items on the stack, it's niladic, and pushes 20!

Why does it push 20? Well because an empty Braingolf program simply prints a newline, and the ASCII value of a newline is 10, so I figured I'd make niladic + push 20 so it's like it's actually being monadic on the implicit newline (even though it isn't at all)

Therefore:

+   No input
Implicit output


And when doubled up:

++  No input
+  Monadic sum, Double top of stack
Implicit output


main=print$0 +1--  Try it online! Doubled: main=print$0

### Doubled

clearTimeout((t=this).x),t.x=setTimeout(alert(${t.n=-~t.n}));clearTimeout((t=this).x),t.x=setTimeout(alert(${t.n=-~t.n}));

• I guess this would work also: window.i=++window.i||1; in the browser console. It ouputs 1. Browser refresh, window.i=++window.i||1;window.i=++window.i||1; ouputs 2. – Christiaan Westerbeek Jul 17 '17 at 13:13
• @ChristiaanWesterbeek True. But then it's a REPL answer (and you can actually just do this.i=++this.i||1;). – Arnauld Jul 17 '17 at 13:23
• I don't know what that means, REPL answer – Christiaan Westerbeek Jul 17 '17 at 13:26
• @ChristiaanWesterbeek REPL stands for Read-Eval-Print Loop. In that case, the final result is not explicitly printed by the code but by the shell it's running in (like the browser console). – Arnauld Jul 17 '17 at 13:31
• @ChristiaanWesterbeek (Actually, just +1 would work in REPL -- like this Python REPL answer does.) – Arnauld Jul 17 '17 at 13:33

## Batch, 13 bytes

@echo %~z0
:


Explanation: %~z0 expands to the length of the source file, so doubling the file simply doubles the length. The second line defines an empty label, which does nothing. When the file is doubled, it becomes a label named @echo %~z0 instead, while the third line is another empty label.

# QBasic,  44  28 bytes

There is no newline at the end. Outputs 4 when single, 8 when doubled.

4
?x+y
END
DATA 4,0


### Explanation

For the single version:

• 4 is a line number.
• READ x,y takes the first two values from the DATA statement and stores them in x and y. Thus, x gets 4 and y gets 0.
• ?x+y adds the two numbers and prints them.
• END exits the program.

In the doubled version, the DATA statement becomes DATA 4,04, which assigns 4 to both x and y, thus making x+y equal 8 instead.

# Befunge-98, 5 bytes

90g.@


Try it online!

g gets the character value at coordinate (9, 0) in Funge-Space; . prints it as an integer, and @ halts the program. In the un-doubled version, (9, 0) is out of bounds of the program, and Funge-Space outside the program is initialized to the default value of a space, so we print 32. In the doubled version, (9, 0) is the @ character, so we print 64.

• This was not what i expected the best Befunge-98 answer to look like... really creative answer! – MildlyMilquetoast Sep 7 '17 at 4:21
• – MildlyMilquetoast Sep 7 '17 at 4:36