Simplify hashfull calculation.

joergoster · vondele · commit a910ba71eedd · 2020-01-28T19:25:39.000+01:00
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 #2523 No functional change.
diff --git 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;
-  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;
 
-  return cnt * 1000 / (ClusterSize * (1000 / ClusterSize));
+  return cnt / ClusterSize;
 }
