Author: Tony Werten
Date: 00:22:55 04/16/04
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On April 16, 2004 at 02:48:28, Tord Romstad wrote: >On April 15, 2004 at 17:47:40, Sune Fischer wrote: > >>On April 15, 2004 at 15:40:30, Gerd Isenberg wrote: >> >>>>But since the relation between bb and 0x88 is: >>>>0x88 = bb shl 1-(bb and 7), you can also use it for bb. >>>> >>>>Tony >>> >>>Nice trick to save the additional and of the rank ;-) >> >>Same trick, only on the upper 3 bits: >>0x88 = bb + (bb & ~7) > >I use the following two macros: > >#define Expand(x) (((x)<<1)-((x)&7)) >#define Compress(x) (((x)&7)|(((x)&120)>>1)) > >The Expand() macro is, as far as I can see, identical to Tony's suggestion. >But is there a better way to write the Compress() macro? My solution looks >quite complicated, and it is hard to believe that it is optimal. bb = (x88+(x88 and 15)) shr 1 Tony > >Tord
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