+ // The squares occupied by enemy pieces (not defended by pawns) will be
+ // counted two times instead of one. The shift (almost) guarantees that
+ // intersection with b is zero so when we 'or' the two bitboards togheter
+ // and count we get the correct sum of '1' in b and attacked bitboards.
+ Bitboard attacked = Us == WHITE ? ((b & pos.pieces_of_color(Them)) >> 1)
+ : ((b & pos.pieces_of_color(Them)) << 1);