Array subscript operator overloading

I am having trouble understanding the difference between Array obj; and Array* obj = new Array; while overloading the array index operator []. When I have a pointer to the object, I get these error messages on VS 2010.

error C2679: bina开发者_JS百科ry '=' : no operator found which takes a right-hand operand of type 'int' (or there is no acceptable conversion)

could be 'Array &Array::operator =(const Array &)' while trying to match the argument list '(Array, int)'

#include <iostream>
class Array
{
    int arr[10] ;

    public:
       int& operator[]( int index )
       {
           return arr[index] ;
       }
};

int main()
{
    //Array* obj = new Array; Error

    Array obj;   // Correct
    for( int i=0; i<10; ++i )
        obj[i] = i;

    getchar();
    return 0;
}

Can some one explain the rationale between the two kind of instances for operator overloading? Thanks.

In case of Array *obj, obj[i] is the equivalent of *(obj+i), so it evaluates into an Array object.

You would have to do

int main()
{
    Array* obj = new Array;

    for( int i=0; i<10; ++i )
        (*obj)[i] = i;

    getchar();
    return 0;
}

You defined operator[] for Array, not for Array*. In the commented-out code, you create an Array*. In fact, you cannot overload an operator for any pointer type. Applying [] to a pointer treats it as an array, converting the array indexing into pointer arithmetic. Thus, applying [] to an Array* yields an Array (really an Array&). You can't assign an int to an Array because you didn't define that (and don't want to, either).

Modern, well-written C++ uses the keyword new very seldom. You should not be suspicious that your C++ code doesn't contain the keyword new. You should be suspicious every time it does!