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Calculating Pi (from International Obfuscated C Code Contest)
Can someone explain to me how the following program calculates Pi from its own area? It's a winner entry of the International Obfuscated C Contest.
link: http://www.cise.ufl.edu/~manuel/obfuscate/pi.c
#define _ F-->00 || F-OO--;
long F=00,OO=00;
main(){F_OO();printf("%1.3f\n", 4.*-F/OO/OO);}F_OO()
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(The preceding white spaces in each line are deleted by the forum posting system, otherwise the program body looks like a perfect circle. Please do check the provided web link to look at the program.)
I find it amazing but don't understand. I'm pretty sure that F counts the number of "_" characters, and OO counts the number of lines, i.e. the size of the square. Therefore 4*F/OO/OO is an approximation for Pi. But I have no idea how F and OO are actually counted in the program.
(The "void" keyword is omitted from "void main" and "void F_OO", as permitted in traditional C)
Thanks in advance to anyone who takes the time and effort to analyse this puzzle for me.
link: http://www.cise.ufl.edu/~manuel/obfuscate/pi.c
#define _ F-->00 || F-OO--;
long F=00,OO=00;
main(){F_OO();printf("%1.3f\n", 4.*-F/OO/OO);}F_OO()
{
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}
(The preceding white spaces in each line are deleted by the forum posting system, otherwise the program body looks like a perfect circle. Please do check the provided web link to look at the program.)
I find it amazing but don't understand. I'm pretty sure that F counts the number of "_" characters, and OO counts the number of lines, i.e. the size of the square. Therefore 4*F/OO/OO is an approximation for Pi. But I have no idea how F and OO are actually counted in the program.
(The "void" keyword is omitted from "void main" and "void F_OO", as permitted in traditional C)
Thanks in advance to anyone who takes the time and effort to analyse this puzzle for me.
Last edited on
It seems to take advantage of short-circuit evaluation.
Preprocessed and reformatted:
Preprocessed and reformatted:
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Thanks very much, but I have no problem understanding why the program looks like this after preprocessing. My problem is I that can't even understand the preprocessed code.
In particular, I'm not sure about operator precedences regarding "--", ">" and "||". I would be very grateful if someone could post a detailed anatomy of the program.
In particular, I'm not sure about operator precedences regarding "--", ">" and "||". I would be very grateful if someone could post a detailed anatomy of the program.
-F-- > 00 || F - OO--; is just an obfuscated way of saying |
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The F- after the || does nothing. It's there only to complete the "foo".
Thanks very much. Your reply was very helpful and pointed me to the right direction. I played with the code in a compiler and finally I was able to understand it. However, you made two mistakes in your explanation:
1. I found "F--" is evaluated after the "if" statement.
2. With the logical OR operator ||, the right hand side is only evaluated if the left hand side evaluated to false, in contrary to what you have said.
Therefore
On the other hand,
So the "_" character at the beginning of each line decreases both F and OO by 1, whereas every "-_" pair decreases F by 1 but leaves OO unchanged.
In this way, after the function F_OO finishes, -F is the number of "_" characters, i.e. the area of the circle. -OO is the number of lines, i.e. the size of the square. Therefore -4*F/OO/OO gives Pi.
1. I found "F--" is evaluated after the "if" statement.
2. With the logical OR operator ||, the right hand side is only evaluated if the left hand side evaluated to false, in contrary to what you have said.
Therefore
-F-- > 00 || F - OO--; is equivalent to |
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On the other hand,
F-- > 00 || F - OO--; is equivalent to |
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So the "_" character at the beginning of each line decreases both F and OO by 1, whereas every "-_" pair decreases F by 1 but leaves OO unchanged.
In this way, after the function F_OO finishes, -F is the number of "_" characters, i.e. the area of the circle. -OO is the number of lines, i.e. the size of the square. Therefore -4*F/OO/OO gives Pi.
Last edited on
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