How to instantiate a template class when constructor takes arguments

Question

I'm trying to create a new object of the class Queue<T>, with the following:

Queue<char[40]> *q = new Queue<char[40]>(100);

Then, I want to use this q on another class constructor:

Interface<char[40]> i(q);

If I just use it, I get an error saying invalid conversion from ‘Queue<char [40]>*’ to ‘int’, which I figure means I'm trying to pass the value of q, which is the pointer to it instead of the actual value. Fine. Then I redefined the Interface<T> constructor to receive a pointer instead of the value itself, so the signature is

Interface(DataStructure<T>& q);

Since Queue extends DataStructure. For some reason, now the instantiation of q fails:

undefined reference to `Queue<char [40]>::Queue(int)

But I am quite sure I have written the constructor method WITH an INT parameter

template<typename T>
Queue<T>::Queue(int size): DataStructure<T>(size) {
    std::cout << size << std::endl;
}

Why the hell am I getting this error, then? To play with pointers is just getting a mess and I could not figure out by any means what to do now.

Answer 1

This code Queue<T>::Queue(int size) suggests you put the definition in a source file. If you use templates all your code must be in the header file. (There are other approaches, but this is the easiest).