Author: Dieter Buerssner
Date: 11:42:19 01/03/03
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On January 03, 2003 at 13:21:40, Robert Hyatt wrote: >Why not something like this: > >rank=to>>3; (now we know the rank and only care if it is zero or seven, and >it can't be zero for white or seven for black so side to move is not important). > >So, you end up with one line: > > >if (to>>3==0 || to>>3==7) gen_promote(); Why should this be faster than if (to <=7 || to >= 56) ? With some bit fiddling tricks (which I would not suggest), one test will be enough and also include the side. For example if ((((to ^ ~((unsigned)side-1))) & 0x38) == 0x38) Note, that this is untested, and probably more clever ideas are possible. It assumes, that side is 0 for white and 1 for black. Then side-1 will be -1 or 0. -1 is all bits set in unsigned. Now reverse the bits by not. So for white, the ~((unsigned)side-1)) will be 0. 0x38 is binary 111000 (decimal 56). So, only when to goes to the 8th rank (56-63), the 111000 bits will be set. And the binary and gives 0x38. For side = 1, ~((unsigned)side-1)) will be all bits set. The xor will only set the 111000 bits, when they were cleared before (so to <= 7). If I haven't totally failed until here, the binary and could even be avoided, by if ((to ^ ~((unsigned)side-1)) >= ~7U) Uri, why do you need side? Will you movegenerator generate to to squares >= 56 for black pawns? Regards, Dieter
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