Author: Gerd Isenberg
Date: 00:31:29 12/06/05
Go up one level in this thread
On December 05, 2005 at 23:24:52, Zappa wrote:
>I am getting really, really tired of coding all my evaluation twice (once for
>white and once for black). However, one of the things that is keeping me from
>switching to a for(i < 2) loop is that I can't do a shift!
>
>For example, if I have some pattern based on (pawns << 8) for white, than that
>is (pawns >> 8) for black, and you can't do a negative shift in IA32.
>
>My ideas:
>
>Eugene will happily point out that on the Itanium doing two shifts and selecting
>the correct value is 1 (2?) bundles.
>
>Otherwise on AMD64 I could do
>
>a) two shifts & cmov. I think 5 instructions (as compared to 1, and I have a
>LOT of shifts).
>
>b) << followed by >>. 1 extra instruction but I have twice as many loads for
>constants.
>
>c) rotate (X | 64-x) (but then I have the possibility of things ending up
>rotating around).
>
>d) your name here . . . :)
>
>I am not that concerned about latency because there would usually be alot of
>stuff around that could be rescheduled, but if I have to do 5 instructions for
>every shift my code size will triple.
>
>anthony
a) mixture of a and b
// assuming color ::= {0,1} := {white, black}
shiftCountWhite = shiftCount & (color-1);
shiftCountBlack = shiftCount & -color; // shiftCountWhite ^ shiftCount
x <<= shiftCountWhite;
x >>= shiftCountBlack;
d) conditional generalized shift.
if (color)
x >>= shiftCount;
else
x <<= shiftCount;
If the routine is inlined and color is a compile time constant (due to unrolling
color loops) the compiler will optimize the none taken branch away - otherwise
how likely is a misprediction?
Gerd
This page took 0 seconds to execute
Last modified: Thu, 15 Apr 21 08:11:13 -0700
Current Computer Chess Club Forums at Talkchess. This site by Sean Mintz.