Help In code template

Question

Given this code

template <typename T>
typename T::ElementT at (T const &a , T const &b)
{
        return a[i] ;
}

what do

typename T::ElementT 

and

a[i]

mean?

Answer 1

typename T::ElementT 

Since T:ElementT is a dependent name, that is why you see the keyword typename before it. It tells the compiler that ElementT is a tested type, not value.

And as for a[i], it seems that T is a class that has defined operator[] which is being called when you write a[i]. For example, T could be sample as (partially) defined here:

class sample
{
 public:
      typedef int ElementT; //nested type!

      //...

      ElementT operator[](int i) 
      {
          return m_data[i];
      }

      ElementT *m_data;
      //...
};

Now, if T is sample, then you can write T::ElementT as well as a[i] which is of type T. In this case when T is sample, I'm assuming that the type of index i is int.

Answer 2

i guess in that code T is always a class which has operator [] overloaded and has a subclass defenition ElementT any other class which doesn't have these two qualities will make an error while compile.

Answer 3

typename T::ElementT 

This is explained exhaustingly by Johannes in this entry to the C++ FAQ.

a[i]

This operation is usually called "subscription" and accesses the i-th element in a. In order to do that, a must either be an array or some class that overloads the subscription operator (like std::vector or std::map).
However, as Nawaz pointed out in his comment, since a is of type T, and since T is expected to have a nested type ElementAt, in this case a cannot be an array.