Question

Is this answer considered “good” code or is it just an ugly hack?

And I would like to know how this is forward-declared (both classes).

When I just forward-declare the class with 2 template-parameters, it just always takes this one, no matter what value flag has.

I would like to do this because I have 2 special member functions which should behave differently on flag being true and I don’t feel like reimplementing the whole class. Also, it should have the same name. According to this example, this seems to be possible.

And I have to forward-declare it because I’m creating a library in which we forward-declare everything.

Any idea?

Answer 1

In order to use specialisation its definition always has to be visible to the caller. If, for example, you have template <class Type, bool flag> struct something defined in one header and template <class Type> struct something<Type, true> : public something<Type, false> defined in the second one, to use the latter you have to include the second header. Without that you will always get the first, more generic type.

EDIT: the bit about forward-declaring got me thinking. If you want to use only type declaration, as in pointer variable, do the following:

Header

template <class Type, bool flag>
struct something;

struct Test
{
    something<int, true>* ptr; // definition not needed
    Test();
}

Source

#include "something.h" // header with template
#include "something_spec.h" // header with specialisation

Test::Test()
{
    ptr = new something<int, true>(); // specialisation used
}