Question

Hya,

Can anyone please tell me how this thing is working?

template <typename T,
template <typename ELEM> class CONT = std::deque >
class Stack {
private:
CONT<T> elems; // elements

public:
void push(T const&); // push element
void pop(); // pop element
T top() const; // return top element
bool empty() const { // return whether the stack is empty
return elems.empty();
}
};

What i don’t understand is this :
template class V or say this “template class CONT = std::deque”

i visualize this as

template <class>
class CONT = std::deque // here CONT is templatized class declaration.

but what pesters me is , how can we assign something to class name CONT , rather than writing its definition (which i’ve done till this time):

template <class>
class CONT{
//def
}

one more thing :

template <class> // why its only class written in angle bracket there should be also be name
like : template<class ty>

Thanks a lot , any help is very appreciated)

Answer 1

What i don’t understand is this : template class V

There is no such line in your question, so I can’t help with that.

template< template <typename ELEM> class CONT = std::deque >
class Stack

This is a declaration of a template template parameter. You pass a template into the Stack template, and then Stack can use it internally.

The = std::deque part is a default value, in case you leave the CONT parameter unspecified. (std::deque is a predefined template.)

However, this will not work, because std::deque takes two arguments. This will work:

template< template <typename ELEM, typename ALLOC> class CONT = std::deque >
class Stack

However ELEM and ALLOC do not actually name anything; they exist merely to clarify what the parameter list of the required template is. So, you can omit them:

template< template <typename, typename> class CONT = std::deque >
class Stack