运算符重载：仅涉及相同类型对象时的成员与非成员

Question

这个问题很好地回答了为什么要将运算符重载定义为非成员：Operator overloading : member function vs. non-member function?

If you define your operator overloaded function as member function, then compiler translates expressions like s1 + s2 into s1.operator+(s2). That means, the operator overloaded member function gets invoked on the first operand. That is how member functions work!

But what if the first operand is not a class? There's a major problem if we want to overload an operator where the first operand is not a class type, rather say double. So you cannot write like this 10.0 + s2. However, you can write operator overloaded member function for expressions like s1 + 10.0.

现在我遇到了需要超载的情况operator==。在我的例子中，只会比较 (a) (b) 相同 类型的对象。

是否还有理由将 operator== 定义为非成员，或者在这种情况下我应该将其实现为成员？

Answer 1

因为 operator== 的 LHS 和 RHS 参数具有对称语义，推荐的方法是始终根据其操作数的 public 接口将其实现为非成员（或者如果需要私有数据，以在 class).

中将其声明为好友

所以

class Bla
{
public:
    // complete interface to data required for comparison
    auto first();
    auto second();
    // ... more
private:
    // data goes here
};

bool operator==(Bla const& L, Bla const& R)
{
    return 
        std::forward_as_tuple(L.first(), L.second() /*, ... */) == 
        std::forward_as_tuple(R.first(), R.second() /*, ... */)
    ;
}

这样，L 和 R 参数都考虑了到 Bla 的隐式转换（我并不是说隐式转换是个好主意，但如果你有这些，最好避免出现意外情况它们仅被考虑用于 RHS 参数。

运算符重载：仅涉及相同类型对象时的成员与非成员

Operator overload: Member vs. non-member when only same type objects can be involved

c++

operator-overloading