CODEWITHSUNDEEP
c++operator-overloading
wyfeizj
wyfeizj

Reputation: 83

Why does operator-> overloading not work in VC2010?

class A
{
public:
    A* operator->() const
    {
    }
    void Test() {}
};

then call it like this.

A* a = new A;
a->Test();

The code builds and runs successfully in VC2010. It seems very strange. I am wondering it is by design or it is a bug of VC2010?

Thanks

Upvotes: 2

Views: 117

Answers (2)

John Humphreys
John Humphreys

Reputation: 39294

Your code is:

A* a = new A;
a->Test();

The "a" is a pointer to an A. It is NOT an A object itself, it is a memory address of the A object on the heap.

When you call a->Test() you invoke the pointer's -> operator (built in for all pointer types in C++). You would have to do this to invoke your operator:

//Create an A (NOT a pointer).
A a;

//Invoke the A's operator ->
a->Test();

Which is how STL iterators work - they're class types, not pointers to class types. Note that the return type of operator -> must make sense for the operation/member you're trying to invoke.

So, here's an example that would call test via the ->:

#include <iostream>

class A
{
public:
    A* operator->()
    {
        return this;
    }
    void Test() { std::cout << "Hello World!"; }
};

int main()
{
    A a;
    a->Test();
}

It's strange, but it works because a->Test(); returns the current object, which Test() is then called on (see the return this; line).

See STL iterators for useful examples of why you'd actually want to do this.

Upvotes: 1

David
David

Reputation: 28178

You aren't calling your operator-> in your example, you are calling Test directly from an A*. You want:

(*a)->Test();

Or...

A a;
a->Test();

There's nothing wrong with VS2010 concerning operator-> (that I know of).

Using -> on a pointer named a effectively executes: (*a).. Using -> on a variable by value will invoke your operator-> if present, or be a syntax error if there is no operator->.

Upvotes: 3

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