At Thu, 23 May 2019 17:08:42 +0300, Valery Ushakov wrote: > > +#if defined(__GNUC__) > > +/* gcc defines '>>' as ASR. */ > > +#define AUDIO_ASR(value, shift) ((value) >> (shift)) > > +#else > > +#define AUDIO_ASR(value, shift) ((value) / (1 << (shift))) > > +#endif > > This feels inverted. The mathematically correct operation required > here is symmetric division (that rounds towards zero). Arithmetic > shift is flooring division (that rounds towards minus infinity).
Both are not correct but are acceptable. # I also used to feel that division is the correct operation. # But I don't think so now. For example, let's consider to convert 16bits to 15bits. In case of round-to-zero, 32767, 32766 => 16383 : 3, 2 => 1 1, 0 => 0 <---- *1 -1 => 0 <---- *1 -2, -3 => -1 : -32766, -32767 => -16383 -32768 => -16384 <---- *2 In case of round-to-minus-inf, 32767, 32766 => 16383 : 3, 2 => 1 1, 0 => 0 -1, -2 => -1 : -32767, -32768 => -16384 As above, in round-to-zero, there are three inputs that output 0 (*1) and there are only one input that output -16384 (*2). In contrast, in round-to-minus-inf, all outputs correspond to two input values. Which is "correct"? The correct operation is not exist whenever rounding to integer occurs. And in audio area, we need to understand that both rounding are not correct but are acceptable. Furthermore, we have to consider that this operation is performed in the bottom of loop in realtime mixing. If both rounding are acceptable, I'd like to use faster one. Of course, unless it is "undefined". > So > it feels confusing and wrong to *name* the operation the opposite of > what it really is. What I want to do here is arithmetic right shift operation and 2nd argument in this macro indicates shift count. So I named it ASR. Thanks, --- Tetsuya Isaki <is...@pastel-flower.jp / is...@netbsd.org>
