Question

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)
        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

Answer 1

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[]