Author: Gerd Isenberg
Date: 01:18:07 04/16/04
Go up one level in this thread
On April 16, 2004 at 03:22:55, Tony Werten wrote: >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 > Sune's expand to 0x88 only needs two instructions, because ~7 is a compile time constant. But for Compress there seems no way to get rid of the third "expensive" shift or div 2 ;-) x88 = bb + (bb&~7); bb = (x88 + (x88&7)) >> 1; Gerd
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