如何修复类"Invalid operands to binary expression "类" to "类" "错误 (repl.it)

how to fix "Invalid operands to binary expression "class" to "class"" error (repl.it)

本文关键字：to repl it 错误 expression Invalid operands binary 何修复      更新时间：2023-10-16

我目前正在编写一个简单的 2 人国际象棋游戏。我目前的点是根据类的成员变量查找用户尝试移动的对象(类(。我创建了一个函数来基于变量返回类的实例，但是当我尝试将返回的值(==(与类的对象进行比较时，出现"二进制表达式的操作数无效"错误。

// this is the important class
class King{
private:
const char m_color;
int m_x;
int m_y;
public:
void setStartBlack(){
    m_color = 'B';
    m_x = 4;
    m_y = 0;
    board[m_y][m_x] = 'K';
}
void setStartWhite(){
    m_color = 'W';
    m_x = 4;
    m_y = 7;
    board[m_y][m_x] = 'K';
}
int getMX(){
    return m_x;
}
int getMY(){
    return m_y;
}
};
// this is the function I made to return the class instance
King checkPieceK(int x, int y){ // x and y is the column and row
if (blackKing.getMX() == x && blackKing.getMY() == y){
    return blackKing; // I should note here that blackKing is an 
// object of the king class and so is whiteKing
}
else if (whiteKing.getMX() == x && whiteKing.getMY() == y){
    return whiteKing;
}
else{
    return failureCondK; // this is what should be returned if the 
// piece doesn't exist at the location checked
}
}
// and here's what's happening at main()
while (GAMEOVER == false){
std::cout << " enter the row, column and type of the piece you 
want to move(e.g. "G1P" means "G1," Pawn): ";
row = getchar();
col = getchar();
typ = getchar();
getchar(); // catches newline (n or ) char
row = toupper(row);
int newRow = detYval(row);
typ = toupper(typ);
if (typ == 'K'){
    if (checkPieceK(col - 1, newRow) == failureCondK){ // this is 
// the statement is where the error occurs
    }
}
GAMEOVER = true;
}
编译器

不知道在两个King对象上使用 == 运算符时该怎么做。您必须定义操作。这涉及到编写一个具有特殊名称(operator==(的函数来比较King对象。

class King {
public:
  bool operator==(const King &rhs) const {
    return m_color == rhs.m_color && m_x == rhs.m_x && m_y == rhs.m_y;
  }
  // whenever you define operator==, you should also define operator!=
  bool operator!=(const King &rhs) const {
    return !(*this == rhs);
  }
  // ...
};

如果你想比较国王，你可以做king.operator==(otherKing)或king == otherKing.如果您愿意，还可以将operator==定义为非成员函数(在类外部(。

class King {
  // ...
};
bool operator==(const King &lhs, const King &rhs) {
  return lhs.m_color == rhs.m_color && lhs.m_x == rhs.m_x && lhs.m_y == rhs.m_y;
}
bool operator!=(const King &lhs, const King &rhs) {
  return !(lhs == rhs);
}

现在您可以将国王与operator==(king, otherKing)或king == otherKing进行比较.

您可能需要对operator==的定义进行一些思考。我们真的需要比较颜色，x和y吗？在这种情况下，您可能只能比较颜色(因为有一个白色国王，一个黑人国王和一个无效的国王(或只是比较位置(因为你不能让国王占据同一个图块(。

我刚刚演示的内容称为运算符重载，它可以用于该语言中的大多数运算符。

您需要重载该类的 == 运算符。 

class King{
private:
  bool operator==(const King& k) const
  {
    return m_x == k.m_x && m_y == k.m_y;
  }
  /* snip */
};
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