c++ - C++中的运算符重载

Question

struct T
{
        int a;
        int b;
};

class Ptr
{
public:
        Ptr(int a, int b) { t_.a = a; t_.b = b; }
        T* operator->() {return &t_;}
        T& operator*() {return t_;}

private:
        T t_;
};

int main()
{
        Ptr ptr(1, 2);
        cout << "a = " << ptr->a << " b = " << ptr->b << endl;
        cout << "a = " << ptr.operator->()->a << " b = " << ptr.operator->()->b << endl;
} 

输出:

a = 1 b = 2
a = 1 b = 2

为什么ptr->a和ptr.operator->()->a是一样的，它的原理是什么？

Answer 1

因为这就是这个重载运算符的工作方式...

An expression x->m is interpreted as (x.operator->())->m for a class object x of type T if T::operator->() exists and if the operator is selected as the best match function by the overload resolution mechanism.

(C++ §[over.ref]/1)