Question

#ifndef __TEST__
#define __TEST__

namespace std
{
    template<typename T>
    class list;
}

template<typename T>
void Pop(std::list<T> * l)
{
    while(!l->empty())
        l->pop();
}

#endif

and used that function in my main. I get errors. Of course, I know that there are more template params for std::list (allocator I think). But, that is beside the point. Do I have to know the full template declaration of a template class to be able to forward declare it?

EDIT: I wasn’t using a pointer before – it was a reference. I’ll try it out with the pointer.

Answer 1

The problem is not that you can’t forward-declare a template class. Yes, you do need to know all of the template parameters and their defaults to be able to forward-declare it correctly:

namespace std {
  template<class T, class Allocator = std::allocator<T>>
  class list;
}

But to make even such a forward declaration in namespace std is explicitly prohibited by the standard: the only thing you’re allowed to put in std is a template specialisation, commonly std::less on a user-defined type. Someone else can cite the relevant text if necessary.

Just #include <list> and don’t worry about it.

Oh, incidentally, any name containing double-underscores is reserved for use by the implementation, so you should use something like TEST_H instead of __TEST__. It’s not going to generate a warning or an error, but if your program has a clash with an implementation-defined identifier, then it’s not guaranteed to compile or run correctly: it’s ill-formed. Also prohibited are names beginning with an underscore followed by a capital letter, among others. In general, don’t start things with underscores unless you know what magic you’re dealing with.