I have this setup:
class DontUse;
template<class T,class U = DontUse, class V = SomeStandardType>
class Foo
{
public:
void bar(U &uh);
};
When U is set to DontUse, I want bar to be an empty function. In all other cases, I want bar to have some implementation. I tried doing this using specialization, but this code (which I realize is somehow incorrect) doesn't compile:
template<class T,class V> void Foo<T,DontUse,V>::bar(DontUse &none){}
template<class T,class U,class V> void Foo<T,U,V>::bar(U &uh)
{
//do something here
}
The error message is this (MSVC10):
1>path_to_project: error C2244: 'Foo<T,U,V>::bar' : unable to match function definition to an existing declaration
and it points to the line of the first template specialization.
How do I do this correctly?
Here's the actual code, although it's reduced to the minimalist part that's relevant:
struct DontUse;
template<typename Derived, typename Renderer = DontUse, typename TimeType = long>
class Gamestate
{
public:
void Render(Renderer &r);
};
template<typename Derived, typename TimeType> void Gamestate<Derived, DontUse,TimeType>::Render( DontUse){}
template<typename Derived, typename Renderer, typename TimeType> void Gamestate<Derived,Renderer,TimeType>::Render(Renderer &r)
{
static_cast<Derived*>(this)->Render(r);
}
You cannot specialize individual members of a template. You have to specialize the class itself:
class DontUse;
template<class T, class V>
class Foo<T, DontUse, V>
{
public:
void bar(DontUse)
{ }
};
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