Thread: help me understanding this code

  1. #1
    Registered User
    Join Date
    Jan 2004

    help me understanding this code

    a friend of mine gave me this code to explain it to him
    but i can't read it too even with knowing the output
    it would be nice if someone could explain me this code

    for(char *c="89?9;?8>8::<89@8?<8;:9<<89@8?>8>8::<9<>8?<8;:9 <<8A?9;?8>89<<",i=0;i<=57;i+=3,(i%(*c&1<<3| ~~1 <<2))?*c<<2&i>>1|*c*2: putchar(10)) for( int n=7;n>=0;n--)((*(c+i)-56)*100+((*(c+i+1))-56)*10+(*(c+i+2)-56))&(1<<n)?putchar('@'):putchar(' ');
    big thx in advance

  2. #2
    and the hat of int overfl Salem's Avatar
    Join Date
    Aug 2001
    The edge of the known universe
    Nothing useful is ever gained from writing code like that, it's just there to tease
    If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
    If at first you don't succeed, try writing your phone number on the exam paper.

  3. #3
    ATH0 quzah's Avatar
    Join Date
    Oct 2001
    Basicly it just goes through a string, does a bit of math on each character, and prints out the resulting character. Probably just to print some message like "I am 1337.". I haven't bothered compiling it, but that's the gist of it.

    All of the "*(c+X)" stuff is dereferencing the string 'c' plus X number of bytes down the string. Such as:

    char *c = "abcde";

    *(c + 3) gives you the letter 'd' from the string. So all they really do is jump to specific spots in the string, pull out that number. Subtract a bit to get a letter or number, or do a bit of other math to get whatever value they want from it, and display that character. The '10' they're putting out is the newline character. It's a way to obfuscate the fact that they're putting a new line to without blatently saying: "putchar( '\n' );

    Curses, foiled again. And as Salem said, it never has a practical purpose. Unless you're trying to win an obfuscated code contest.

    Hope is the first step on the road to disappointment.

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