Author: Vincent Diepeveen
Date: 12:09:57 01/03/03
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On January 03, 2003 at 14:42:19, Dieter Buerssner wrote: >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(); At his beloved P4 the >> instruction is very very very very slow. >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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