-4
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In case the title was confusing, I mean something like the following C code:

main(){}znva(){}

When rot13 encoded, the code becomes

znva(){}main(){}

and still runs.

But that isn't a valid answer, because of these rules:

  • Your code must produce output. Code that doesn't produce output is disqualified.
  • The program can't be empty. That's disqualified too.
  • The program can't be the same or similar to the output. This means that you can't use Golfscript or PHP.

In a valid answer, the program produces some output without being rot13'd, such as:

Hello, World!

And when the program is rot13'd, the output changes to the following:

Uryyb, Jbeyq!

which is the rot13 of Hello, World!.

Have fun!

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  • 1
    \$\begingroup\$ Is this code that produces an output when it is rot13'd? Or is it code that produces an output of itself rot13'd (that produces the original program)? \$\endgroup\$ – Justin Nov 18 '13 at 22:38
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    \$\begingroup\$ Does the output have to contain letters? As it stands, . is a valid response in brainfuck. \$\endgroup\$ – Alex Gittemeier Nov 19 '13 at 1:55
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    \$\begingroup\$ BTW: what is the objective winning criteria? \$\endgroup\$ – Johannes Kuhn Nov 19 '13 at 7:57
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    \$\begingroup\$ @tbodt In the future, please refrain from changing a challenge's rules and/or judging criteria once it's been posted. If you would like to "test out" a challenge before posting it, in order to per-determine what rule modifications might be necessary, please use the current Sandbox thread in Meta. (Currently at Mk V.) \$\endgroup\$ – Iszi Nov 20 '13 at 1:01
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    \$\begingroup\$ -1 Congratulations. You now have successfuly created a chamelion quesion. The statement "assuming they [the rules] don't change again" only shows that you can't answer this without your answer beeing invalidated. \$\endgroup\$ – Johannes Kuhn Nov 20 '13 at 7:54
4
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Ruby

Cheaty answer:

$><<'Hello, world!'

Slightly less cheaty answer:

s='Hello, world!'
$-I?puts(s):chgf(s)

Golf (15 chars)

$-I?p(:a):c(:a)
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  • \$\begingroup\$ I changed the rules. Your second solution is the only one valid. \$\endgroup\$ – tbodt Nov 19 '13 at 19:20
  • \$\begingroup\$ This is now a code golf contest. Shorten it up. \$\endgroup\$ – tbodt Nov 19 '13 at 23:27
2
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Perl, 25 14 12 characters

I believe this adheres to all of the requirements:

$_=a;fnl;say

Outputs a. After rot13 the program looks like this:

$_=n;say;fnl

and outputs n.

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  • \$\begingroup\$ Does perl generate an error on encountering fnl? \$\endgroup\$ – tbodt Nov 19 '13 at 19:48
  • \$\begingroup\$ @tbodt Without strict mode (and warnings) turned on, Perl will parse fnl as "fnl" (i.e. as a string). The statement then becomes a no-op expression. \$\endgroup\$ – breadbox Nov 19 '13 at 23:18
  • \$\begingroup\$ Your solution is not disqualified. And you get an up vote. \$\endgroup\$ – tbodt Nov 19 '13 at 23:21
  • \$\begingroup\$ This is now a code golf contest. Shorten it up. \$\endgroup\$ – tbodt Nov 19 '13 at 23:22
1
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Brainf**k

Since you never said the output had to actually be affected by rot13, this is totally a valid answer:

-.

Not sure why it was downvoted, although it is kind of a cheap answer.

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  • \$\begingroup\$ I changed the rules. histocrat's second solution is the only one valid. \$\endgroup\$ – tbodt Nov 19 '13 at 19:19
  • \$\begingroup\$ add an letter at the end, and it will be valid. \$\endgroup\$ – Johannes Kuhn Nov 19 '13 at 21:04
  • \$\begingroup\$ @JohannesKuhn Re-changed the rules. No comments. I think that I won't have to change them again. \$\endgroup\$ – tbodt Nov 19 '13 at 23:13
  • \$\begingroup\$ You change the rules just to invalidate existing answers? \$\endgroup\$ – Johannes Kuhn Nov 20 '13 at 7:52
0
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PHP 6

H<?//a

The program and its output differ by 83%, so not similar.
a is not transformed into output, should satisfy all rules now.

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  • \$\begingroup\$ I changed the rules. histocrat's second solution is the only one valid. \$\endgroup\$ – tbodt Nov 19 '13 at 19:19
  • \$\begingroup\$ Because chameleon question. \$\endgroup\$ – Johannes Kuhn Nov 19 '13 at 19:33
  • \$\begingroup\$ There I fixed it. \$\endgroup\$ – Johannes Kuhn Nov 19 '13 at 19:36
  • \$\begingroup\$ What does that // do? And don't you need a ?> ending delimiter? \$\endgroup\$ – tbodt Nov 19 '13 at 19:52
  • \$\begingroup\$ // is a comment. and no, I don't need an ending ?> \$\endgroup\$ – Johannes Kuhn Nov 19 '13 at 21:03
0
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bash, 64 chars

_() { [[ "${1:0:1}" > U ]]&&rpub $@||echo $@; };_ Hello, Jbeyq.
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