template class - member function specialization

Question

Here is an example code:

template<class T>
class A
{
public:
   A(T t): x(t){}
   T getX();
private:
   T x;
};

template<class T>
T A<T>::getX()
{
   return x;
}

// member function specialization
template<> // works with and without template<> 
long A<long>::getX()
{
   return 1000L;
}

The above code works with and without template<> before a member function specialization. Why ? What difference does it make in such case ?

Edit1: I use that template this way (VS 2012 compiler):

A<int> a1(1);
cout<<a1.getX()<<endl;
A<long> a2(1);
cout<<a2.getX()<<endl;

Answer 1

Not compliantly, it doesn't.

FWIW, GCC 4.8 rejects your code without the template <>.

Your compiler is either buggy or has an extension to support this; I can confirm that MSVS 2012 accepts the code. I'm told that MSVS 2013 November CTP also eats it up without complaint. To be fair, Visual Studio was always fairly lenient about template specifications.

[C++11: 14.7/3]: An explicit specialization may be declared for a function template, a class template, a member of a class template or a member template. An explicit specialization declaration is introduced by template<>. [..]

The only exception to this rule is:

[C++11: 14.7.3/5]: [..] Members of an explicitly specialized class template are defined in the same manner as members of normal classes, and not using the template<> syntax. [..]

… but that does not apply here.