Return to Answer

C (GCC) (44 chars)

float p(i){return i<1E6?4./++i-p(++i):0;}

float p(i){return i<1E6?4./++i-p(++i):0;}

That's 41 chars, but it also has to be compiled with -O2 to get the optimiser to eliminate the tail recursion. This also relies on undefined behaviour with respect to the order in which the ++ are executed; thanks to ugoren for pointing this out. I've tested with gcc 4.4.3 under 64-bit Linux .

Note that unless the optimiser also reorders the sum, it will add from the smallest number, so it avoids loss of significance.

Call as p().

C (GCC) (44 chars)

float p(i){return i<1E6?4./++i-p(++i):0;}

That's 41 chars, but it also has to be compiled with -O2 to get the optimiser to eliminate the tail recursion. This also relies on undefined behaviour with respect to the order in which the (Note:++ are executed; thanks to ugoren for pointing this out. I've tested with gcc 4.4.3) under 64-bit Linux .

Note that unless the optimiser also reorders the sum, it will add from the smallest number, so it avoids loss of significance.

Call as p().

C (GCC) (44 chars)

float p(i){return i<1E6?4./++i-p(++i):0;}

That's 41 chars, but it also has to be compiled with -O2 to get the optimiser to eliminate the tail recursion. (Note: tested with gcc 4.4.3).

Note that unless the optimiser also reorders the sum, it will add from the smallest number, so it avoids loss of significance.

Call as p().