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what is the difference between overloading an operator inside or outside a class?

开发者 https://www.devze.com 2023-02-21 16:01 出处:网络
In C++, i know there are two ways to overload. We can overload it inside (like class a) or outside (like class b). But, the question is, is there any difference between these two either in compile tim

In C++, i know there are two ways to overload. We can overload it inside (like class a) or outside (like class b). But, the question is, is there any difference between these two either in compile time or runtime or not?

class a
{
public:
    int x;
    a operator+(a p) // operator is overloaded inside class
    {
        a temp;
    开发者_StackOverflow中文版    temp.x = x;
        temp.x = p.x;
        return temp;
    }
};

class b
{
public:
    friend b operator+(b, b);
    int x;
};

b operator+(b p1, b p2) // operator is overloaded outside class
{
    p1.x += p2.x;
    return p1;
}


The member operator+ requires the LHS to be an a - The free operator requires LHS or RHS to be a b and the other side to be convertible to b

struct Foo {
    Foo() {}
    Foo(int) {}
    Foo operator+(Foo const & R) { return Foo(); }
};


struct Bar {
    Bar() {}
    Bar(int) {}
};

Bar operator+(Bar const & L, Bar const & R) {
    return Bar();
}


int main() {
    Foo f;
    f+1;  // Will work - the int converts to Foo
    1+f;  // Won't work - no matching operator
    Bar b;
    b+1;  // Will work - the int converts to Bar
    1+b;  // Will work, the int converts to a Bar for use in operator+

}
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