Syntax using the return value from an overloaded -> operator

Question

So I think I might be overthinking this but I wanted to know if someone could clarify why the following statement works in the given code

f->hello();

This is the code

struct bar
{
    void hello()
    {
        std::cout << "Hello World";
    }
};

struct foo
{
    bar* f;
    foo() {
        f = new bar();
    }

    ~foo() {
        delete f;
    }

    bar* operator->() {
        return f;
    }
};

int main()
{
    foo f;
    f->hello(); //Works
}

Since the following code from above returns a pointer

bar* operator->() {
        return f;
    }

should'nt

f->hello(); actually be f->->hello();

why does f->hello() work and f->->hello() fails ? The reason i am thinking that f->->hello() should work is because

f->ptrReturned->hello();

If we overload the -> operator are we required to return a ptr type ? Form what I have tried it seems returning an object type is not allowed

->-> is not a correct syntax. f->hello() is handled by the compiler. — hamidi
– hamidi, Commented Jul 29, 2018 at 3:45
Yeah, the compiler fudges this and re-applies the -> operator to the returned pointer so you don't have the awkward syntax problems. — Galik
– Galik, Commented Jul 29, 2018 at 4:04

songyuanyao · Accepted Answer · 2018-07-29 04:45:22Z

1

Your understanding is correct basically, but ->-> is not valid syntax. You can use the overloaded operator-> like

f.operator->()->hello();
//^^^^^^^^^^^^          return the pointer
//            ^^^^^^^^^ call hello() on the returned pointer

f->hello() is treated as the same of f.operator->()->hello(); that makes the usage of class with overloaded operator-> (e.g. smart pointers) consistent with built-in pointers. The usage would be more natural, and could be used with raw pointers in more general context like templates.

And,

If we overload the -> operator are we required to return a ptr type ?

There're some restrictions:

The overload of operator -> must either return a raw pointer, or return an object (by reference or by value) for which operator -> is in turn overloaded.

edited Jul 29, 2018 at 4:45

answered Jul 29, 2018 at 3:52

songyuanyao

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Comments

user9254539 · Accepted Answer · 2018-07-29 03:49:15Z

1

Apparently, that's just how an overloaded operator-> works. The part before the -> (not including the operator itself) gets replaced with the return value of the overloaded operator. For example, std::unique_ptr::operator-> returns a pointer, which then gets dereferenced by the ->.

answered Jul 29, 2018 at 3:49

user9254539

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