what is the convention for calling overloaded = operator in C++ using pointers?

When overloading a "=" operator for a class, the usual syntax is:

ClassA ObjA, ObjB;

ObjA = ObjB;

But what about pointers ??

Class *PtrA, *PtrB;

PtrA->operator=(PtrB)

Is the above calli开发者_C百科ng convention correct assuming the = operator in ClassA is defined for pointer on class (ClassA* operator=(const ClassA* obj)) ??

Usually the assignment operator will be defined only for the class, and not the pointers, and then you can do:

*ptrA = *ptrB;

Not that in your example

Class *PtrA, *PtrB;

PtrA->operator=(PtrB)

You are not passing in an instance of PtrB, you are passing in a pointer to a PtrB;

Actually of course your code doesnt really make sense. You cannot say

Class *PtrA, *PtrB;

I think you mean

Class A
{
}
Class B
{
  B operator=(const A &rhs){...}
}

A *ptra = ...
B *ptrb = ...

In this case you can go

*ptrb = *ptra;

or

ptrb->operator=(*ptra);

you certainly cannot go

ptrb->operator=(ptra)

or (the equivalent simple syntax)

*ptrbb = ptra

If the assignement operator in ClassA accepts an argument of type ClassA * (or const ClassA *), then you can call it as either

 PtrA->operator =(PtrB);

or as

*PtrA = PtrB;

The latter is more natural way to use the assigment operator for obvious reasons.

However, defining the assignment operator to accept an argument of type ClassA * is a rather bizzare thing to do. Normally, assignment operator of ClassA would accept an argument of type const ClassA & and be invoked as

*PtrA = *PtrB;