Overloading '->' member access operator

Question

While overloading operator* we do as follows:

T & operator * () {  return *ptr; }

That means if I have:

SmartPtr<int> obj(new int());
*Obj = 20;
Cout << *obj;

Then the *obj part is replaced by *ptr so its as good as:

*Ptr=20
cout<<*ptr

Now while overloading operator-> we do the following:

T * operator -> () { return ptr; }

And it can be accessed as:

Obj->myfun();

What I don't understand here is after evaluating the obj-> is replaced by ptr so it should look as follows:

ptrmyfun(); // and this is syntax error

Where am I going wrong?

Answer 1

C++ does not work like search/replace in a text editor. "obj1->" is not replaced by ptr verbatim, as if this was a line of text in a text file.

With:

Obj->myfun();

The fact that the -> operator invokes a custom operator that returns ptr :

T * operator -> () { return ptr; }

That doesn't mean that the original statement somehow becomes

ptrmyfun();

As if "Obj->" was replaced by "ptr". C++ simply does not work this way, like search/replace in a text editor.

The -> operator results in the custom operator->() getting invoked. That part, of how you understand operator overloading works, is correct.

But what happens is that, simply, the return value from the operator-> () becomes the actual pointer value to which the actual pointer dereference gets applied to.

So this becomes, essentially:

ptr->myfun();

That's all.