First off, many thanks. > + const int inv_1 = l_ptwo << ((4 - b_ptwo) & 3); > + const int inv_2 = 0xeeeeeeef & ((1U << b_ptwo) - 1);
It would be nice to add a comment here that the expression for inv_1 is (2^b_ptwo)^-1 mod 15 and inv_2 is 15^-1 mod 2^b_ptwo. (A general PFA FFT would need to use extended Euclidean algorithm here, but because both cases are fixed, it simplifies to these expressions. I have a sketch of a proof (basically solving the relevant diophantine equation you get) in case anyone is nervous, though it's easy to verify by hand for 1 < b_ptwo < 18, which are all the cases that ffmpeg's power-of-two FFT currently supports). Rest of patch seems good. -Peter _______________________________________________ ffmpeg-devel mailing list ffmpeg-devel@ffmpeg.org http://ffmpeg.org/mailman/listinfo/ffmpeg-devel