overloading operator -> to access data member

Question

I have a question about the following sample program about overloading -> operator, same across in a C++ tutorial:

 5     class myclass
 6     {
 7         public:
 8         int i;
 9
10         myclass *operator->()
11         {return this;}
12     };
13
14     int main()
15     {
16         myclass ob;
17
18         ob->i = 10;
19         cout << ob.i << " " << ob->i << endl;
20
21         return 0;
22     }

$ ./a.out
10 10

I am trying to understand how line 18 works. I understand that "ob" is not a pointer, but since "class myclass" has defined the operator "->", "ob->i" is valid (syntactically), so far good. However, "ob->" returns a pointer, and I don't see how it is de-referenced to get access to member "i" and setting it.

I am assuming the above explanation will also explain how in line 19 "ob->i" is printed as an int.

Thank you, Ahmed.

Answer 1

operator->在一个链中被调用，直到它不再被调用 - 在你的情况下，它实际上被调用了两次 - 一次，你的对象上的重载操作符，它返回一个指针，第二次，内置的操作符，取消引用指针并访问该成员。

Answer 2

x->y is equivalent to x.operator->()->y if x is a class object and an overloaded member operator-> is found.

I hope it gets clearer from that.