Simplification of Template Class Specification - c++

Question

I am very interested in the possibilities of template classes. Now, I am wondering how I can achieve following:
Const<5>::getValue();
At the moment I can do this:
Const<int, 5>::getValue());
And that's the implementation of the class:
template <typename T, T value>
class Const {
public:
typedef T Type;
static T getValue() {
return value;
}
};
I know this is just a very silly example, but once I can do that I could simplify following line:
Getter<int, TestClass, &TestClass::get> getter;
to just:
Getter<&TestClass::get> getter;
That would be TestClass :
class TestClass {
private:
int _value;
public:
int get() {
return _value;
}
};
Thanks for your help!
[EDIT-1]
In regards to J.N. yes, C++11 would be fine.
In regards to Xeo, I tried to use #define AUTO_ARG(x) decltype(x), x but that doesn't work within TestClass.
[EDIT-2]
In regards to KennyTM, when I declare the Getter<...> g within TestClass it doesn’t work with Getter<AUTO_ARG(&TestClass::get)> it only works with Getter<int (TestClass::*)() const, &TestClass::get>.
Now I am wondering if this is just a bug in Visual Studio???

Answer 1

You could still use AUTO_ARG(x) as given by #Xeo's link. If you need to get the return type or the class type, just use pattern matching (i.e. template specialization):
template <typename T, T mem_fun_ptr>
struct Getter;
template <typename R, typename C, typename... A, R (C::*mem_fun_ptr)(A...) const>
struct Getter<R (C::*)(A...) const, mem_fun_ptr>
{
typedef R return_type;
typedef C class_type;
};
#define AUTO_ARG(x) decltype(x),(x)
class TestClass {
private:
int _value;
public:
int get() const {
return _value;
}
};
Getter<AUTO_ARG(&TestClass::get)> g;