I know this question would be bette

[chipperMDW]

I know this question would be better suited to one of the more general C++ forums, but I didn't get any answers there, so I'm trying my luck here.

Say I have:

template< class T >
class SmartPtr
{
  public:
    ...
    T *operator->() { return ptr; }

  private:
    T *ptr;
};

This works (and is apparently the/a correct way of doing it), but I don't understand the logic behind having the operator return a pointer.

Given:

SmartPtr< FooStruct > p = &someFooStruct;

that means:

p->fooStructMember;

roughly translates to:

( &someFooStruct ) fooStructMember;

which looks like it's missing a -> . Does another operator-> get called behind the scenes for the return value of operator-> ? If not, what exactly is going on here?

 

[williamu]

A pointer has to be returned since the template class doesn't know in advance what it's going to deal with.

For all it knows it could be dealing with a CString or CWnd so a pointer is returned that allows you to access the appropriate object.

William
Software Engineer