I double the source, you double the output!

Question

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 code-golf, 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.

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Answer 1

Cubix, 4 5 bytes

Of course as soon as I look at it again this presents itself. Not as nice as the previous, but shorter.

))O@

Try it online!

  )
) O @ .
  .

Increment, output and exit.

))O@))O@

Try it online!

    ) )
    O @
) ) O @ . . . .
. . . . . . . .
    . .
    . .

Increment twice, output and exit

Answer 2

TI-BASIC, 2 bytes

2!

There are also 5 more solutions, of equal or greater length.

(2

abs(2

int(2

iPart(2

round(2

These all work on the same principle. Something evaluates to 2, and twice it is merely 2*2.

Answer 3

Bash + coreutils, 43 bytes

x=$(history|cut -c 8-|tail -1);echo ${#x};#

Explanation

Outputs the length of the last command entered. Uses the same concept as the Python and PHP solutions to prevent execution of any further copies (finishes with a comment marker, so anything after it is ignored). Length of the command will be doubled when it is written twice, hence the output.

Answer 4

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!

Answer 5

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!

Answer 6

Wumpus, 6 bytes

$@1$O+

Try it online!

Doubled:

$@1$O+$@1$O+

Try it online!

The first program pushes 1, adds the top two items on the stack (the 1 and an implicit 0), reflects off the end of the line, outputs and exits.

The second program does the same, except it adds two 1s to the stack, meaning it prints 2 and exits. This works for any number from 1-9.

A shorter solution may exist that takes advantage of two lines making the pointer bounce differently.

Answer 7

///, 20 bytes

Prints 1:

/\\1\//\/1\/2\/\//\1

Try it online!

Duplicated prints 2:

/\\1\//\/1\/2\/\//\1/\\1\//\/1\/2\/\//\1

Try it online!

How it works

• The initial substitution /\\1\//\/1\/2\/\// (in both versions) searches for the string \1/ in the remainder and replaces it by /1/2//.
• In the single program there is no such string, and nothing is replaced. The program is now reduced to the final \1, which prints a 1.
• In the duplicated program \1/ crosses the boundary between the copies.
  • After substituting, the remaining program becomes /1/2//\\1\//\/1\/2\/\//\1, which is the substitution /1/2/ followed by the single program.
  • This substitution then replaces every 1 in the single program by 2, giving /\\2\//\/2\/2\/\//\2.
  • This then runs pretty much like the single program does, except for printing 2 instead of 1.
• The normally redundant \ before the final 1, and the corresponding \\ in the initial substitution, are needed because without them, the substitution would be applied again to the /1/2/ result, causing an infinite loop.

Answer 8

Momema, 12 bytes

00 1+1*1-8*1

Try it online! This outputs 1.

Try it doubled! This outputs 02.

Explanation

The ungolfed form of the singular program 00 1+1*1-8*1 is

0   0     # set the 0th cell to 0 (this has no effect).
1   +1*1  # set the 1st cell (initialized to 0) to itself plus one (i.e. 1).
-8  *1    # output the value of the first cell as a decimal (1).

The ungolfed version of the doubled program 00 1+1*1-8*100 1+1*1-8*1 is

0   0     # set the 0th cell to 0.
1   +1*1  # increment the 1st cell.
-8  *100  # output the value of the 100th cell (0).
1   +1*1  # increment the 1st cell.
-8  *1    # output the value of the 1st cell (2).

This submission hinges on Momema's syntax: in particular, it allows leading 0s in a numeric literal to be parsed as separate numbers. This allows the leading 00 in the program to be parsed as a pointless assignment statement.

When the program is doubled, however, the 0s are no longer leading a numeric literal—they are a continuation of the literal 1 at the end of the program, forming 100.

Answer 9

Labyrinth, 8 5 bytes

-3 bytes thanks to @MartinEnder

Single version

^)!@

Try it online!

Double version

^)!@
^)!@

Try it online!

Answer 10

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;// *///

• Try it online (single)
• Try it online (double)

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.

Answer 11

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:

• -0 is false, so we go to [,a=alert(1)]=[1], destructuring with a default value. a is to be assigned to [1][1], which isn't defined, so a defaults to alert(1).

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.

Answer 12

Ruby, 10 bytes

+1;a||=p 1

Try it online!

Try it online!Try it online!

Explanation:

• Single version (+1;a||=p 1):
  • +1 is ignored
  • a is assigned to 1 (printing 1 in the process) because a was previously undefined
• Double version (+1;a||=p 1+1;a||=p 1)
  • +1 is ignored
  • a is set to 1+1 (2) (printing 2 in the process) because a was undefined
  • The next part is not executed because a is already defined

Answer 13

CJam, 4 bytes

1]:+

Try it online!

Explanation:

1]:+

1    push 1 to stack
 ]:+ sum stack

CJam, 1 byte

)

Try it online!

I'd say this uses 2 bytes since there is a 0 in the header. I don't know if this is a valid loophole or forbidden.

Answer 14

Perl 5, 10 bytes

print-s$0#

Try it online!

Doubled, 20 bytes

print-s$0#print-s$0#

Try it online!

Inspired by Sriotchilism O'Zaic recursive comment approach (Nice work by the way).

Answer 15

Octave, 25 13 11 3 bytes

3;ans=fix(ans);ans=ans*2.

Try it online! Try it doubled!

Edit 1

Got rid of the fix

1;ans=3*ans-0

Try it online! Try it doubled!

Edit 2

Got rid of the *

3;ans=ans+0

Try it online! Try it doubled!

Edit 3

Got rid of ans ... 😅

3+0

Try it online! Try it doubled!

Answer 16

Google Sheets, 12

This answer is much shorter. But here's one for a "normal" = formula.

=LEN("&00001

Google Sheets auto-closes quotes and parens. This outputs 6 (the length of &000001). Here is the doubled version:

=LEN("&00001=LEN("&00001

This outputs 12 because &00001=LEN( is 11 characters, concatenated with 1 gives 12 characters.

Answer 17

BRASCA, 2 bytes

Original:

C}

Doubled:

C}C}

Try it!

Try it!Try it!

Explanation

Original:

C     - Sets implicit output to number mode
 }    - Increment top of stack (stack is infinite zeroes by default)

Doubled:

C     - Sets implicit output to number mode
 }    - Increment top of stack (stack is infinite zeroes by default)
  C   - Implicit output is already set to number mode, so it does nothing
   }  - Increment top of stack again

Answer 18

Zsh, 9 bytes

wc -l<<<x

Note there is no trailing newline.

Attempt This Online! (outputs 1)

Attempt This Online!Attempt This Online! (outputs 2)

Explanation:

• wc: output word-count information:
  • -l: the number of lines in the input
• <<<x: and input the single-line string x

ExplanationExplanation:

• wc: output word-count information:
  • -l: the number of lines in the input
• <<<xwc: and input the single-line string xwc
• -l: redundant duplication of the previous -l option
• <<<x: and input a second single-line string x

---

Zsh, 6 bytes

If output requirements followed standard rules, we could output via exit code for 6 bytes:

grep x

Again, no trailing newline.

• grep x searches for x in the (empty) input, which has no results and outputs with status code 1
• grep xgrep x searches for xgrep in the (non-existent) file x, which is an error and outputs with status code 2

Attempt This Online! (returns 1)

Attempt This Online!Attempt This Online! (returns 2)

There might be an even shorter option using exit codes.

Answer 19

Rust, 69 bytes

use std::u8::*;fn main(){print!("{}",MIN+1)}#[test]
const MIN:u8=1;//

Try it online

Try it online (doubled)

This follows the suggestion in Konrad's bonus solution to override a standard library item and uses #[test] instead to "disable" a declaration on the next line. It's not as short as this solution, but it doesn't require reading the source.

Answer 20

Backhand, 3 bytes

]h 

Try it online! Try it doubled!

Prints 1 as-is, and prints 2 when doubled. There is a trailing space.

There is no two-byte solution because:

• The first command should produce some nonzero number, and the second should be h (print the number and halt)
• but both ab and abab will execute a then b, so any such program will print the same number (not twice) when doubled.

For a three-byte program, abc runs a then b while abcabc runs aab in the order, so "a = increment = ], b = print and halt = h" works.

Answer 21

Lambda calculus (Nat), 5 bytes

(SBI)

Try it online!

Instructions: Copy-paste the code into the code box, select Nat, press Compile and then Run to see the Output.

Outputs in Church numeral.

SBI
\xy. SBIxy
\xy. Bx(Ix)y
\xy. Bxxy
\xy. x(xy)

Applying a Church numeral n to another m, i.e. n m, evaluates to m^n. Therefore, (SBI)(SBI) = 2 2 = 2^2 = 4.

Turns out that I already did the same thing in Flurry...

Answer 22

MathGolf, 1 byte

There are multiple answers possible:

• ): try it online or try it doubled - outputs \$1\$ and \$2\$;
• ⌠: try it online or try it doubled - outputs \$2\$ and \$4\$;
• ó: try it online or try it doubled - outputs \$1\$ and \$2\$.

Explanation:

If the stack is empty when a numeric operator is used, it'll use an implicit \$0\$. The three programs will respectively do the following:

• ): Pop the top of the stack \$t\$, and push \$t+1\$
• ⌠: Pop the top of the stack \$t\$, and push \$t+2\$
• ó: Pop the top of the stack \$t\$, and push \$2^t\$

After which the entire stack joined together is output implicitly as result.

Answer 23

Pip, 2 bytes

+1

Try it online!

Try the double version!

Answer 24

Ohm, 2 bytes

1Σ

Try it online!

Try the double version!

aaaaaaaaaaaaaaa i can't find a language where this is one byte aaaaaaaaaaaaaaa

Answer 25

QBIC, 12 bytes

p=p+q┘Z=!p$┘

Explanation

p=p+q        Adds 1 to p (q = 1 in QBIC, and p starts out as 0)
┘            (Syntactic linebreak)
Z=!p$        Set Z$ to a string representation of p.
┘            (Syntactic linebreak)       

At the end of a QBIC program, if Z$<> "" it gets printed. So when running this code once, p gets increased by 1, the result is saved in Z$, the program ends and we print 1. By running it twice, Z$ will get overridden on the second iteration by p, which is now 2.

Answer 26

Vim, 5 bytes

i0<esc><C-a>:

Since V is backwards compatible, you can Try it online! or Try it doubled!

Answer 27

LOGO, 16 bytes

 ct pr bf gensym

output 1.

When doubled, it becomes

 ct pr bf gensym ct pr bf gensym

output 2.

Explanation:

• ClearText clears all the output.
• Print just print whatever it is given.
• ButFirst return the input with the first item removed.
• GenSym (perhaps GenerateSymbol) return g1 the first time it is invoked, g2 the second time, etc.

Because the output of GenSym depends on previous outputs, the interpreter should be reset between runs.

Answer 28

JavaScript (Node.js), 60 bytes

console.log(require('fs').readFileSync(__filename).length)//

Inspired by this Python 2 answer, this reads the length of the current file and prints it, preventing any duplicates of this code from being executed with the trailing comment.

Try it online!

Try it online doubled!

Answer 29

BlitzMax, 13 bytes

Print..
+1..'

No trailing newline.

Newlines act as statement terminators in BlitzMax unless prevented by a ... The ' symbol introduces a line comment. So if run as-is, the program outputs the result of the expression +1, which is 1. If doubled, the second Print is commented out and the program outputs the result of +1+1, which is 2. If you double the program a second time, you get 4, etc.

Answer 30

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!