C ++重载运算符->（）不被调用？

Question

Can anyone please explain why the overloaded operator -> for class Foo is not being called from a pointer of type Foo *?

#include <iostream>
using namespace std;

class Foo {
public:
Foo * operator -> () {
    cout << "calling Foo * operator -> ()\n";
    return this;
}
    int x;
};


void main() {
    Foo f;
    Foo * pF = &f;
    pF->x;                  // Why is overloaded operator-> not being called here?
    (pF->operator->())->x;  // This works.

    cout << "End test.\n";
}

Answer 1

Why is overloaded operator-> not being called here?

Because pF is a pointer, not a Foo instance. You have overloaded operator-> for Foo , not Foo* . You may call your overloaded operator-> directly on f .

f->x;

You may not overload operator-> for Foo* or any other pointer type.

Answer 2

You overloaded it for Foo , not for Foo* (which is impossible).

This would have worked:

(f.operator->())->x;

Or, of course:

f->x;

Being able to use that short syntax is the entire purpose of the overload, no? Although it's quite confusing to just return a pointer to the same object you invoked the operator on.

Answer 3

You have overloaded "point-to" operator of the type Foo , not type Foo* .

class Foo {
public:
    // Overload 'point-to' operator of 'Foo'
    Foo * operator -> () {
        cout << "calling Foo * operator -> ()\n";
        return this;
    }
    int x;
};

Example of usage:

Foo f;
f.operator->();