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Remove backmost_sq (#2190)
This is a non-functional simplification. backmost_sq and frontmost_sq are redundant. It seems quite clear to always use frontmost_sq and use the correct color. Non functional change.
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3 changed files with 3 additions and 6 deletions
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@ -376,10 +376,7 @@ inline Square pop_lsb(Bitboard* b) {
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}
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/// frontmost_sq() and backmost_sq() return the most/least advanced square in
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/// the given bitboard relative to the given color.
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/// frontmost_sq() returns the most advanced square for the given color
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inline Square frontmost_sq(Color c, Bitboard b) { return c == WHITE ? msb(b) : lsb(b); }
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inline Square backmost_sq(Color c, Bitboard b) { return c == WHITE ? lsb(b) : msb(b); }
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#endif // #ifndef BITBOARD_H_INCLUDED
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@ -365,7 +365,7 @@ ScaleFactor Endgame<KBPsK>::operator()(const Position& pos) const {
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&& pos.count<PAWN>(weakSide) >= 1)
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{
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// Get weakSide pawn that is closest to the home rank
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Square weakPawnSq = backmost_sq(weakSide, pos.pieces(weakSide, PAWN));
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Square weakPawnSq = frontmost_sq(strongSide, pos.pieces(weakSide, PAWN));
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Square strongKingSq = pos.square<KING>(strongSide);
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Square weakKingSq = pos.square<KING>(weakSide);
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@ -190,7 +190,7 @@ void Entry::evaluate_shelter(const Position& pos, Square ksq, Score& shelter) {
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for (File f = File(center - 1); f <= File(center + 1); ++f)
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{
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b = ourPawns & file_bb(f);
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Rank ourRank = b ? relative_rank(Us, backmost_sq(Us, b)) : RANK_1;
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Rank ourRank = b ? relative_rank(Us, frontmost_sq(Them, b)) : RANK_1;
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b = theirPawns & file_bb(f);
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Rank theirRank = b ? relative_rank(Us, frontmost_sq(Them, b)) : RANK_1;
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