When would I need to implement operator [] ?

given the following template function :

template <class T>
void DoSomething(T &obj1, T &obj2)
{
      if(obj1 > obj2)
        cout<<"obj1 bigger: "<<obj1;
      else if(obj1 == obj2)
    开发者_如何学编程    cout<<"equal";
      else cout<<"obj2 bigger: "<<obj2;

      T tmp(3);
      T array[2];
      array[0]=obj1;
      array[1]=obj2;
}

I need to define a class called MyClass (declarations only , i.e. just the .h file) , that would be able to work with that template function . I defined the next declarations :

class MyClass
{
public:
    MyClass();    // default ctor
    MyClass(int x);  // for ctor with one argument
    bool operator ==(const MyClass& myclass) const;
    bool operator >(const MyClass& myclass) const;
    friend ostream& operator<<(ostream &out,const MyClass& myclass);  // output operator

};

What I don't understand is why there is no need to define operator [] for the lines:

array[0]=obj1; array[1]=obj2; 

? When would I need to define operator []? thanks ,Ron

You declared an array for your type:

T array[2];

But your are talking about implementing operator[] for T, which is totally different concept.

If you need

T t;
t[1] = blah

Then you need to implement operator[]

Because

T array[2];

Isn't a T object, its an array of T. So

array[0];

Is indexing an array, not one of your objects, therefore you don't need an operator[].

Assuming you call DoSomething with a couple of MyClass objects, you have declared array to be a normal array of MyClass objects. You did not need a [] operator for MyClass because array is not an instance of MyClass; it is just an array.

You will want to overload the [] operator in your own classes when it makes sense to, or is convenient. A good example is a collection (such as a map). Another example is a custom string class, where you might want to index by a regex object to find matches for your pattern inside your string.

If your class was an implementation of a dynamic array, for example, you would want to access the (single) object as though it was an array - you this by overloading the [] operator.