Question

为了代表国际象棋棋盘，我有一些便利constexpr功能。为了记录易读的结果，我创建了一个函数，将两个表示形式的幂转换为const char*

我不确定使用constexpr const char*函数返回字符串文字是否安全，下面是否有任何问题？据我所知，编译器会替换say：

::std::cout << "Before: " << typePrint(bounds())

使用：

::std::cout << "Before: " << "Bounds"

代码：

constexpr unsigned int bounds()  { return 1;       }
constexpr unsigned int nothing() { return 1 << 1;  }
constexpr unsigned int bPawn()   { return 1 << 2;  }
constexpr unsigned int bRook()   { return 1 << 3;  }
constexpr unsigned int bKnight() { return 1 << 4;  }
constexpr unsigned int bBishop() { return 1 << 5;  }
constexpr unsigned int bQueen()  { return 1 << 6;  }
constexpr unsigned int bKing()   { return 1 << 7;  }
constexpr unsigned int wPawn()   { return 1 << 8;  }
constexpr unsigned int wRook()   { return 1 << 9;  }
constexpr unsigned int wKnight() { return 1 << 10; }
constexpr unsigned int wBishop() { return 1 << 11; }
constexpr unsigned int wQueen()  { return 1 << 12; }
constexpr unsigned int wKing()   { return 1 << 13; }

constexpr const char* typePrint(const unsigned int ac_type) {
    return ac_type == bounds()  ? "Bounds" :
           ac_type == nothing() ? "Nothing" :
           ac_type == bPawn()   ? "Black Pawn" :
           ac_type == bRook()   ? "Black Rook" :
           ac_type == bKnight() ? "Black Knight" :
           ac_type == bBishop() ? "Black Bishop" :
           ac_type == bQueen()  ? "Black Queen" :
           ac_type == bKing()   ? "Black King" :
           ac_type == wPawn()   ? "White Pawn" :
           ac_type == wRook()   ? "White Rook" :
           ac_type == wKnight() ? "White Knight" :
           ac_type == wBishop() ? "White Bishop" :
           ac_type == wQueen()  ? "White Queen" :
           ac_type == wKing()   ? "White King" :
           "Unknown";
}

编辑：引起我注意的是，执行以下操作是有效的C ++：

::std::cout << "Before: " "Bounds";

但是，返回constexpr的{{1}}不是有效替代品，因此我认为编译器不会将const char*替换为typePrint(bounds())，以下是错误：< / p>

"Bounds"

从constexpr函数返回一个const char * literal字符串

0 个答案: