How arrow-> operator overloading works internally in c++?

Question

I understand the normal operator overloading. Compiler can translate them to method call directly. I am not very clear about the -> operator. I was writing my first custom i

Answer 1

myClassIterator->APublicMethodInMyClass()

is nothing but the following:

myClassIterator.operator->()->APublicMethodInMyClass()

The first call to the overloaded operator-> gets you a pointer of some type which has an accessible (from your call-site) member function called APublicMethodInMyClass(). The usual function look-up rules are followed to resolve APublicMethodInMyClass(), of course, depending on whether it is a virtual or not.

There is not necessarily a temporary variable; the compiler may or may not copy the pointer returned by &(m_iterator->second). In all probability, this will be optimized away. No temporary objects of type MyClass will be created though.

The usual caveats also do apply to m_iterator -- make sure that your calls do not access an invalidated iterator (i.e. if you are using vector for example).

Answer 2

The operator-> has special semantics in the language in that, when overloaded, it reapplies itself to the result. While the rest of the operators are applied only once, operator-> will be applied by the compiler as many times as needed to get to a raw pointer and once more to access the memory referred by that pointer.

struct A { void foo(); };
struct B { A* operator->(); };
struct C { B operator->(); };
struct D { C operator->(); };
int main() {
   D d;
   d->foo();
}

In the previous example, in the expression d->foo() the compiler will take the object d and apply operator-> to it, which yields an object of type C, it will then reapply the operator to get an instance of B, reapply and get to A*, after which it will dereference the object and get to the pointed data.

d->foo();
// expands to:
// (*d.operator->().operator->().operator->()).foo();
//   D            C            B           A*