我正在写一个国际象棋引擎,我有一个看起来像这样的函数:
U64 find_moves(Piece type, Team side, uint8_t square, U64 occupied) {
switch (type) {
case PAWN: {
U64 result = 0;
result |= occupied & bb_normal_moves::pawn_caps[side][square];
if (!(occupied & bb_normal_moves::pawn_moves_x1[side][square])) {
result |= bb_normal_moves::pawn_moves_x1[side][square];
if (!(occupied & bb_normal_moves::pawn_moves_x2[side][square])) {
result |= bb_normal_moves::pawn_moves_x2[side][square];
}
}
return result;
}
case KNIGHT:
return bb_normal_moves::knight_moves[square];
case BISHOP:
return bb_magics::bishop_moves(square, occupied);
case ROOK:
return bb_magics::rook_moves(square, occupied);
case QUEEN:
return bb_magics::bishop_moves(square, occupied) | bb_magics::rook_moves(square, occupied);
case KING:
return bb_normal_moves::king_moves[square];
}
return 0; // Can't happen
}
它本质上根据类型参数委托另一个函数调用.在程序周围的许多地方,在循环遍历不同的Piece值之后调用此函数,这恰好是枚举.
不幸的是,这意味着每次在该循环中调用此函数,因此在此函数分支中浪费了大量CPU时间.
我想做的是更改此函数以允许编译器优化调用:
template <Piece type> U64 find_moves(Team side, uint8_t square, U64 occupied)
但是我的循环不能编译,因为在编译时无法解析函数调用的目标.
有没有办法优化这个功能,而无需手动展开我的所有循环?
编辑:这是一个调用find_moves的循环的示例:
for (uint8_t piece = 1; piece < 6; piece++) {
move.info.piece = piece;
U64 bb_piece = board.bb_pieces[team][piece];
while (bb_piece) {
uint8_t from = pop_bit(team, bb_piece);
move.info.from = from;
U64 bb_targets = find_moves((Piece) piece, team, from, board.bb_all) & mask;
while (bb_targets) {
uint8_t to = pop_bit(x_team, bb_targets);
move.info.to = to;
buf[buf_size++] = move;
}
}
}
最佳答案 鉴于Piece枚举的值是1到6,您可以使用模板,std :: make_index_sequence,std :: index_sequence来展开.
对不起,但我只能准备一个最小的例子(没有移动,没有董事会等).
如果你称之为
foo(std::make_index_sequence<6U>{});
在foo()中,您可以使用单个值调用另一个函数(您标记为C 17,因此您可以使用模板折叠)
template <std::size_t ... Ps>
void foo (std::index_sequence<Ps...> const &)
{ (bar<Ps>(), ...); }
我的想法是在bar()中你可以放置你的例子的for(uint8_t piece = 1; piece< 6; piece)循环体的内容;我只调用一个简单的(没有其他参数)find_moves()函数.
template <std::size_t Ps>
void bar ()
{ find_moves<pieces(1+Ps)>(); }
现在你可以使用完全特化来开发六个find_moves()模板函数(我只编写std :: cout消息;你使用其他参数,可以将案例的内容放在你的交换机中.
template <pieces P>
void find_moves ();
template <>
void find_moves<Pawn> ()
{ std::cout << "Case Pawn" << std::endl; }
template <>
void find_moves<Knight> ()
{ std::cout << "Case Knight" << std::endl; }
template <>
void find_moves<Bishop> ()
{ std::cout << "Case Bishop" << std::endl; }
template <>
void find_moves<Rook> ()
{ std::cout << "Case Rook" << std::endl; }
template <>
void find_moves<Queen> ()
{ std::cout << "Case Queen" << std::endl; }
template <>
void find_moves<King> ()
{ std::cout << "Case King" << std::endl; }
以下是完整的编译示例
#include <iostream>
#include <utility>
#include <type_traits>
enum pieces { Pawn = 1, Knight, Bishop, Rook, Queen, King };
template <pieces P>
void find_moves ();
template <>
void find_moves<Pawn> ()
{ std::cout << "Case Pawn" << std::endl; }
template <>
void find_moves<Knight> ()
{ std::cout << "Case Knight" << std::endl; }
template <>
void find_moves<Bishop> ()
{ std::cout << "Case Bishop" << std::endl; }
template <>
void find_moves<Rook> ()
{ std::cout << "Case Rook" << std::endl; }
template <>
void find_moves<Queen> ()
{ std::cout << "Case Queen" << std::endl; }
template <>
void find_moves<King> ()
{ std::cout << "Case King" << std::endl; }
template <std::size_t Ps>
void bar ()
{ find_moves<pieces(1+Ps)>(); }
template <std::size_t ... Ps>
void foo (std::index_sequence<Ps...> const &)
{ (bar<Ps>(), ...); }
int main ()
{
foo(std::make_index_sequence<6U>{});
}