Simplify hashfull calculation.

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.

diff --git a/src/tt.cpp b/src/tt.cpp
index 080d3a6bd480006e4c821661d2ec9800a4b7523b..46860fe900df5b9157d3cbafa5e932c688e0de83 100644 (file)
--- a/src/tt.cpp
+++ b/src/tt.cpp
@@ -148,9 +148,9 @@ TTEntry* TranspositionTable::probe(const Key key, bool& found) const {
 int TranspositionTable::hashfull() const {
 
   int cnt = 0;
 int TranspositionTable::hashfull() const {
 
   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)
           cnt += (table[i].entry[j].genBound8 & 0xF8) == generation8;
 
       for (int j = 0; j < ClusterSize; ++j)
           cnt += (table[i].entry[j].genBound8 & 0xF8) == generation8;
 

-  return cnt * 1000 / (ClusterSize * (1000 / ClusterSize));
+  return cnt / ClusterSize;

 }
 }