Author: Uri Blass
Date: 10:39:13 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();
Not exactly because in one case I need side=0 and in another case I need side=1
I think that I will simply do one function for white and one function for black.
I already have
#define rank0(i) ((i)>>3)
Does the computer do faster if rank0(to)==0 and not if to<=7?
I assumed that if to<=7 is faster(only one comparison when to>>3=0 has one
comparison and one shift).
I used rank0(i) in case that it is not 0 so I can avoid things like
if (i>=8&&i<=15).
When I think about it a smart compiler should translate
if (i>=8&&i<=15) to if ((i>>3)==1) but I do not know if they do it.
Uri
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