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Simplify hashfull calculation.

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We can simplify the calculation of the hashfull info by looping over exact 1,000 entries,
and then divide the result by ClusterSize. Somewhat memory accesses, somewhat more accurate.

Passed non-regression LTC
https://tests.stockfishchess.org/tests/view/5e30079dab2d69d58394fd5d
LLR: 2.94 (-2.94,2.94) {-1.50,0.50}
Total: 30125 W: 3987 L: 3926 D: 22212
Ptnml(0-2): 177, 2504, 9558, 2642, 141

closes https://github.com/official-stockfish/Stockfish/pull/2523

No functional change.
This commit is contained in:
joergoster 2020-01-27 18:53:25 +01:00 • committed by Joost VandeVondele
parent 71e0b5385e
commit a910ba71ee
1 changed files with 2 additions and 2 deletions
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4
src/tt.cpp
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@ -148,9 +148,9 @@ TTEntry* TranspositionTable::probe(const Key key, bool& found) const {
int TranspositionTable::hashfull() const { int TranspositionTable::hashfull() const {
int cnt = 0; int cnt = 0;
for (int i = 0; i < 1000 / ClusterSize; ++i) for (int i = 0; i < 1000; ++i)
for (int j = 0; j < ClusterSize; ++j) for (int j = 0; j < ClusterSize; ++j)
cnt += (table[i].entry[j].genBound8 & 0xF8) == generation8; cnt += (table[i].entry[j].genBound8 & 0xF8) == generation8;
return cnt * 1000 / (ClusterSize * (1000 / ClusterSize)); return cnt / ClusterSize;
} }