Author: Wylie Garvin
Date: 17:27:37 12/11/01
Hi,
I've noticed some posts where people mention doing things like this:
BSF ecx,dword ptr [foo+4]
ADD ecx,32
BSF ecx,dword ptr [foo]
On the assumption that undefined --> unchanged for PIII, PIV. While I don't
have any evidence to contradict that assumption, it seems like a bad idea to
rely on undefined behaviour.
I point out the conditional CMOVcc instructions which have been supported
since around the time of the PPro I think:
BSF ecx,dword ptr [foo+4]
ADD ecx,32
BSF eax,dword ptr [foo]
CMOVZ eax,ecx
This technique will work correctly even if eax happens to be trashed by the
BSR instruction if no bits were set.
Also I'd like to mention something that I use to speed up my attack
calculations with BSF/BSR...I don't know if it will help you C/C++ programmers,
but it's pretty cool in assembly.
Notice that of the 8 rays originating in one half of the board, only 3 of them
can possibly cross over to the other side of the board. So you can reduce that
construction above to a single BSR/BSF in the other 5 cases. My assembly code
first decides which half of the board the source square is in, then individually
probes the ends of the ray. To ensure that rays are always terminated, before I
start my attack calculations I build a temporary bitboard from the "all pieces"
bitboard by ORing in the edges (0xFF818181,818181FF) and then clearing the bit
for the source square. Now I need about 5 BSF/BSRs for rook/queen attacks and 6
for bishop/queen attacks.
Also keep in mind that removing a conditional branch in the fashion shown
above requires doing the computation for *both* halves of the branch. In this
case it appears to be a win; but if extra computation is required to produce the
bitboard, the cost could exceed the cost of a conditional branch. In general,
if the branch is predictable (e.g. if the rays are almost always short, and so
almost always stay on the same side of the board) it's likely better not to
remove it.
wylie
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