Template specializations for inner class - c++

Question

consider the following code :
struct X
{
template <typename T>
class Y
{};
};
template<>
class X::Y<double>{
};
here we are specializing the Y class for the type double and the code works fine. the problem is that if I change the code to this:
template<typename A>
struct X
{
template <typename T>
class Y
{};
};
template<typename A>
class X<A>::Y<double>{
};
the compiler will report an error:
'X::Y': explicit specialization is using partial
specialization syntax, use template <> instead!
dose any one know how can I specialize class Y in this case?

Answer 1

You cannot specialize an inner template class without explicitly specializing the outer one as well. See this question for more details and a quote from the standard.
Workaround: create an outer class that takes both T and A in a detail namespace, and alias it inside X:
namespace impl
{
template <typename A, typename T>
struct Y { };
}
template<typename A>
struct X
{
template <typename T>
using Y = impl::Y<A, T>;
};
If you are fine with explicitly specializing both inner and outer class, you can use the following syntax:
template <>
template <>
class X<int>::Y<double>
{
// ...
};
Example on wandbox

Answer 2

The simple answer - you can't fully specialize templated inner class of templated outer class. But if you really want to achieve similar effect you could try partial specialization with dummy defaulted template parameter:
#include <iostream>
template<typename A>
struct X
{
template <typename T, T* =nullptr>
class Y{};
};
template<typename A>
template<double *Ptr>
class X<A>::Y<double, Ptr> {
public:
static constexpr int value = 1;
};
int main() {
std::cout << X<int>::Y<double>::value << std::endl;
}
[live demo]