Forward declaration of std::hash template specialization - c++

Question

Why forward declaration as follows :
template<typename T> struct std::hash;
fails to compile with gcc and clang, but compiles with Visual Studio 2015?
gcc 6.1.0 (using coliru):
main.cpp:11:34: error: invalid use of template-name 'std::hash' without an argument list
template<typename T> struct std::hash;
^~~~
clang 3.8.0 (using coliru):
main.cpp:11:29: error: forward declaration of struct cannot have a nested name specifier
template<typename T> struct std::hash;
^~~~~
it works under VS (http://webcompiler.cloudapp.net/). Which compiler is right?
btw. the same declaration is used in C++ Primer 5th edition. Well - nearly the same it uses class instead of struct: template <class T> class std::hash; which is wrong.
full code:
#include <unordered_map>
/*
// compiles with gcc,clang,VS
namespace std {
template<typename T>
struct hash;
}*/
// Compiles only with VS
template<typename T> struct std::hash;
struct MyData {
MyData() {}
MyData(int d1, int d2) : data1(d1), data2(d2) {}
bool operator==(const MyData& rop) const {
return rop.data1 == data1 && rop.data2 == data2;
}
friend struct std::hash<MyData>;
private:
int data1;
int data2;
};
namespace std {
template<>
struct hash<MyData> {
typedef MyData argument_type;
typedef size_t result_type;
size_t operator()(const argument_type& data) const noexcept;
};
size_t hash<MyData>::operator()(const argument_type& data) const noexcept {
return hash<unsigned>()(data.data1) ^ hash<unsigned>()(data.data2);
}
}
int main() {
std::unordered_map<MyData, std::string> mm;
mm[MyData(1,1)] = "test1";
mm[MyData(2,2)] = "test1";
}

Answer 1

The reason, seems to be largely because a forward declaration has to function much like a regular declaration. i.e. encased in a namespace, not prefixed by one. I guess this would allow the same parser to be used for declarations and forward declarations which makes sense.