Author: Matt Taylor
Date: 17:23:59 01/03/03
Go up one level in this thread
On January 03, 2003 at 13:21:40, Robert Hyatt wrote:
>On January 03, 2003 at 12:03:45, Uri Blass wrote:
>
>>code B is slightly faster than code A.
>>I know that side can get only 0 or 1(something that the compiler does not know)
>>and B is eqvivalent to A if you assume that side gets only 0 or 1.
>>
>>Is it possible to write a third code that will be even faster than B?
>>
>>I think that if the compiler can know that side is or 0 or 1 it can do B even
>>faster.
>>
>>code A:
>>
>>if (side==LIGHT)
>>{
>> if (to>=56)
>> {
>> gen_promote(from,to,bits);
>> return;
>> }
>>}
>>else
>>{
>> if (to<=7)
>> {
>> gen_promote(from,to,bits);
>> return;
>> }
>>}
>>
>>code B:
>>if ((to+side*(63-2*to))>=56)
>>{
>> gen_promote(from,to,bits);
>> return;
>>}
>>
>>Uri
>
>
>Why not something like this:
>
>rank=to>>3; (now we know the rank and only care if it is zero or seven, and
>it can't be zero for white or seven for black so side to move is not important).
>
>So, you end up with one line:
>
>
>if (to>>3==0 || to>>3==7) gen_promote();
>
>The compiler will only compute to>>3 once, although you end up with two
>branches.
>However with the newer pentium's clever "pattern matching branch prediction" it
>will
>tend to predict correctly most of the time since the above is not commonly going
>to be
>happening...
>
>I would always try the shortest/simplest approach first to let the compiler and
>the hardware
>try to do the right thing.
Or simplify:
if ((to >> 3) == (7 & side)) gen_promote();
The two expressions get computed in parallel and you end up with 1 branch.
-Matt
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.