Why is partial specialization of a nested class template allowed, while complete isn't?
template<int x> struct A {
template<int y> struct B {};.
template<int y, int unused> struct C {};
};
template<int x> template<>
struct A<x>::B<x> {}; // error: enclosing class templates are not explicitly specialized
template<int x> template<int unused>
struct A<x>::C<x, unused> {}; // ok
So why is the explicit specialization of a inner, nested class (or function) not allowed, if the outer class isn't specialized too? Strange enough, I can work around this behaviour if I only partially specialize the inner class with simply adding a dummy template parameter. Makes things uglier and more complex, but it works.
I would consider complete specializations as a subset of partial specializations - especially because you can express every complete specialization as a partial with adding a dummy parameter. So this disambiguation between partial and full specialization doesn't really make sense for me.
Unfortunatly nobody at comp.std.c++ dared to answer, so I am putting it up here again with a bounty.
Note: I need this feature for recursive templates of the inner class for a set of the outer class and the specialization of the inner parameter does depend on the outer template parameter.
Solution 1:
My guess as to why this happens: complete specializations are no longer "template classes/functions", they are are "real" classes/methods, and get to have real (linker-visible) symbols. But for a completely-specialized template inside a partially-specialized one, this would not be true. Probably this decision was taken just to simplify the life of compiler-writers (and make life harder for coders, in the process :P ).
Solution 2:
C++ Standard explicitly prohibits full specialization of member template classes in first case. According to 14.7.3/18:
In an explicit specialization declaration for a member of a class template or a member template that appears in namespace scope, the member template and some of its enclosing class templates may remain unspecialized, except that the declaration shall not explicitly specialize a class member template if its enclosing class templates are not explicitly specialized as well.
Solution 3:
You can work around this behavior by delegating the real work to another structure though:
namespace detail
{
template <class T, class U>
struct InnerImpl {};
}
template <class T>
struct Outer
{
template <class U>
struct Inner: detail::InnerImpl<T,U>
{
};
};
Now you can specialize InnerImpl
as you wish
Solution 4:
Backing up Virgil's argument (he was faster than I posting the same rationale), consider this:
template<typename T1>
class TOuter
{
public:
template<typename T2>
class TInner
{
public:
T1 m_1;
T2 m_2;
};
};
template<typename T1>
template<>
class TOuter<T1>::TInner<float>
{
public:
T1 m_1;
float m_2;
};
Is TInner fully specialized, or partially specialized because of T1?
Edit:
After considering some of the other comments - it seems that you want to have a full specialization based on the template parameter in the outer class. If you nest the inner class's implementation, that seems to work in Visual Studio 2005:
template<typename T1>
class TOuter
{
public:
template<typename T2>
class TInner
{
public:
std::string DoSomething() { return "Inner - general"; }
T2 m_2;
};
template<>
class TInner<T1>
{
public:
std::string DoSomething() { return "Inner - special"; }
T1 m_1;
};
};
TOuter::TInner will correctly be the specialization of TInner. I could not get it to compile with the implementation outside of the template.