When first condition is met then second one is
always true.
Spotted by Ralph Stoesser.
No functional change.
Signed-off-by: Marco Costalba <mcostalba@gmail.com>
// value if the other side has a rook or queen.
if (square_file(s) == FILE_A || square_file(s) == FILE_H)
{
- if ( pos.non_pawn_material(Them) <= KnightValueMidgame
- && pos.piece_count(Them, KNIGHT) <= 1)
+ if (pos.non_pawn_material(Them) <= KnightValueMidgame)
ebonus += ebonus / 4;
else if (pos.pieces(ROOK, QUEEN, Them))
ebonus -= ebonus / 4;