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]; but have to do this instead:
MyClass *a = new MyClass(); (*a)[1]; 
4 Answers
It's because you can't overload operators for a pointer type; you can only overload an operator where at least one of the parameters (operands) is of class type or enumeration type.
Thus, if you have a pointer to an object of some class type that overloads the subscript operator, you have to dereference that pointer in order to call its overloaded subscript operator.
In your example, a has type MyClass*; this is a pointer type, so the built-in operator[] for pointers is used. When you dereference the pointer and obtain a MyClass, you have a class-type object, so the overloaded operator[] is used.
1 Comment
earlNameless
Just like
-> and . are differentiated.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; } 
Comments
Good news. You can also do...
a->operator[](1);
To add on to the preferred answer, think of operator overloading as overloading functions.
When overloading member function of a class, you remember that the pointer is not of that class type.
Comments
Simply put, with a[1] the a pointer is treated as memory containing array, and you're trying to access the 2nd element in the array (which doesn't exist).
The (*a)[1] forces to first get the actual object at the pointer location, (*a), and then call the [] operator on it.
a[0][1];-)