One cannot explicitly specialize member templates. Consider:
template <class T>
struct X
{
template <class U> struct Y;
};
...Now (imagine we could do this):
template <class T>
template <>
struct X<T>::Y<int>
{};
...For X of which T are we explicitly specializing?
What if, after the point of definition of our explicit specialization, someone does this in one compilation unit...
void foo()
{
X<int>::Y<int> xy;
}
... and then this in another...(valid code, btw).
template <>
template<>
struct X<int>::Y<int>
{};
void foo()
{
X<int>::Y<int> xy;
}
... which would imply multiple definitions of the same class???
As mentioned previously, this is treated well here
Now, considering that the default value actually depends on the type T, perhaps one can get it from the type T.
template <class T>
struct X
{
static T defaultValue(){ return T::defaultValue(); }
};
or better yet, one could change the behavior of defaultValue based on whether T has the member defaultValue.