Simplify pawn moves (#1900)

If we define dcCandidates with & pawnsNotOn7,
we don't have to & it both times.

This seems more clear to me as well.

Tested for no regression.
STC
LLR: 2.96 (-2.94,2.94) [-3.00,1.00]
Total: 44042 W: 9663 L: 9585 D: 24794
http://tests.stockfishchess.org/tests/view/5c21d9120ebc5902ba12e84d

No functional change.

diff --git a/src/movegen.cpp b/src/movegen.cpp
index 5ed248933ce132b6ad820b6dc0ed0fb466266bfd..2f7c55c8640bda0f93923c5b480241ac84471386 100644 (file)
--- a/src/movegen.cpp
+++ b/src/movegen.cpp
@@ -93,10 +93,10 @@ namespace {
             // if the pawn is not on the same file as the enemy king, because we
             // don't generate captures. Note that a possible discovery check
             // promotion has been already generated amongst the captures.
             // if the pawn is not on the same file as the enemy king, because we
             // don't generate captures. Note that a possible discovery check
             // promotion has been already generated amongst the captures.

-            Bitboard dcCandidates = pos.blockers_for_king(Them);
-            if (pawnsNotOn7 & dcCandidates)
+            Bitboard dcCandidateQuiets = pos.blockers_for_king(Them) & pawnsNotOn7;
+            if (dcCandidateQuiets)

             {
             {

-                Bitboard dc1 = shift<Up>(pawnsNotOn7 & dcCandidates) & emptySquares & ~file_bb(ksq);
+                Bitboard dc1 = shift<Up>(dcCandidateQuiets) & emptySquares & ~file_bb(ksq);

                 Bitboard dc2 = shift<Up>(dc1 & TRank3BB) & emptySquares;
 
                 b1 |= dc1;
                 Bitboard dc2 = shift<Up>(dc1 & TRank3BB) & emptySquares;
 
                 b1 |= dc1;