Scope of operator->

Question

When overloading operator ->, one would eventually have to return a pointer. If this pointer points to an object created inside the function like:

struct Something {
  char c;
  Something(char k){
    c = k;
  }
};
struct Hello {
  int a;
  Something* operator->(){
    Something s(a);
    return &s;
  }
};

Would then dereferencing this result in undefined behavior (as Can a local variable's memory be accessed outside its scope?):

Hello h {5};
cout << h->c;

If so, can this be resolved, still using operator->?

why do you want to do that? you may add static keyword, it will work but I do not understand your SW design. — nivpeled, Jul 07 '19 at 08:47
@nivpeled It's far more complicated than this question to address. — , Jul 07 '19 at 08:50
`operator->` is a perfectly normal function. The name and the syntax are somewhat unusual but they are just sugar. — n. m. could be an AI, Jul 07 '19 at 08:51

LogicStuff · Accepted Answer · 2019-07-07T09:05:26.720

Would then dereferencing this result in undefined behavior?

Yes.

If so, can this be resolved, still using operator->?

The following applies to any other member function.

Create Something instance as a member of Hello. Then the pointer to it remains valid for the duration of Hello's lifetime. You can construct Something from the integer you are storing immediately, so you can get rid of it.

If Something takes additional resources (memory, handles, etc.), you probably want to construct it at the time of operator-> call. You can choose std::unique_ptr (dynamic) or std::optional (static) for this. Or, if Something supports an uninitialized state (e.g. default-constructed), you can initialize (e.g. move-assign) to it later.

If you cannot alter Hello class then your only option is to overload operator-> to return by Something by value directly.

But, I must say, the circumstances of this problem are starting to get weird and the way of overloading operators even more so.

Did you return `std::shared_ptr` or just raw pointer? What did not work with `std::unique_ptr`? — LogicStuff, Jul 07 '19 at 09:06

Scope of operator->

1 Answers1