A Neat Trick

[Michael Sherwin]

Just for completeness is this code now at least correct?

Code: Select all

sq = ts - ((epbit[t][ply[t] - 1] & (one << ts)) >> (ts - 3));
          target = board[t][sq];

More simple pseudo code.

Code: Select all

sq = ts - ((epbit & (1ull << ts)) >> (ts - 3));
          target = board[sq];

[Michael Sherwin]

@hgm, Benchmarks of shift instructions comparing Intel vs AMD reveals that even ten year old AMD chips run circles around even the latest Intel chips. So my question about your solution is wouldn't it be better to use a shift instruction rather than a multiply.

Code: Select all

sq = ts ^ ((u64)(epbitboard == one<<ts) << 3);

[Terje]

@hgm, Benchmarks of shift instructions comparing Intel vs AMD reveals that even ten year old AMD chips run circles around even the latest Intel chips. So my question about your solution is wouldn't it be better to use a shift instruction rather than a multiply.

Code: Select all

sq = ts ^ ((u64)(epbitboard == one<<ts) << 3);

If you multiply by a constant that's a power of 2 the compiler will output shifts instead.

[Michael Sherwin]

@hgm, Benchmarks of shift instructions comparing Intel vs AMD reveals that even ten year old AMD chips run circles around even the latest Intel chips. So my question about your solution is wouldn't it be better to use a shift instruction rather than a multiply.

Code: Select all

sq = ts ^ ((u64)(epbitboard == one<<ts) << 3);

If you multiply by a constant that's a power of 2 the compiler will output shifts instead.

Oh okay, I think I knew that, duh (to me). Thanks for the reminder!