Author: Robert Hyatt
Date: 10:05:57 11/03/98
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On November 03, 1998 at 09:44:36, John Stoneham wrote: >On November 02, 1998 at 21:45:05, Robert Hyatt wrote: > >>a couple of tricks... >> >>1. you can use square&56 as the shift amount for the 90 degree rotated bit map, >>since that will get you exactly the right shift count for any rank... >> >>2. you may use a bishop shift value, but you can get away by always using 8 >>bits from the rotated diagonal bitmaps, if you simply set up the array so that >>the "unused bits" still produce the right answer. This avoids having a >>different mask for each diagonal length... > >The reason I pre-calculate even the simple shift and mask amounts, is that I >believe an operation such as ((bitboard & mask[sq]) >> shift[sq]) is probably >faster than, for example, ((bitboard >> (sq & 56)) & 0x00ff). I may be wrong, >and I should probably set up a simple test program to see what the actual >differnce is. It should be much slower. mask[sq] has *two* memory referencess. sq&56 has one. If you do that a lot, it might not matter as you should get a L1 cache hit (or at least a L2 hit). However, if you look at the above, on a non-alpha it gets even worse, because bitboard & mask is a 64 bit and operation. the bitboard>>is a 64 bit shift only... so on a 32 bit architecture the sq&56 ought to be signficantly faster... on the PC's you count memory references *not* instructions. Because the typical memory delay is 20+ clock cycles...
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