subscript operator on pointers

Question

If I have a pointer to an object that has an overloaded subscript operator ([]) why can\'t I do this:

 MyClass *a = new MyClass();
 a[1];
         


        

Answer 1

Because a is type pointer to a MyClass and not a MyClass. Changing the language to support your desired use would make many other language semantics break.

You can get the syntactic result you want from:

struct foo {
    int a[10];
    int& operator [](int i) { return a[i]; }
};

main() {
    foo *a = new foo();
    foo &b = *a;
    b[2] = 3;
}