From cppreference:

The overload of operator -> must either return a raw pointer, or return an object (by reference or by value) for which operator -> is in turn overloaded.

However, I tested the following example, and it is accepted by GCC, Clang and MSVC:

struct A {
  int operator->();
};

According to cppreference, the return type int is neither a pointer nor a type that overloads operator->.

Why is this example correct? Where is this situation described in the standard? What is the intent of the standard to allow this?

CodePudding user response：

While the standard doesn't directly place restrictions on the return type of operator->, it does describe what happens when the overloaded -> operator is used:

12.6.5 Class member access [over.ref]

A class member access operator function is a function named operator-> that is a non-static member function taking no parameters. For an expression of the form
    postfix-expression -> template(opt) id-expression
the operator function is selected by overload resolution (12.4.1.2), and the expression is interpreted as
     ( postfix-expression . operator -> () ) -> template(opt) id-expression.

So when using your class, you can call the overloaded operator like a normal function by writing a.operator->(). But if you attempt to use the operator with a->name, this is interpreted as (a.operator->())->name which won't compile because it attempts to use -> on an int.