Determine type from non-type template parameter

Question

I wish I could do this:

template <typename T x>
struct Test {
    T val{x};
};

int main() {
    Test<3> test;
    return test.val;
}

But I can't. Right?

---

I was answering a question here and I use the following template:

template <typename T, typename V, typename VP, V(T::*getf)(), void (T::*setf)(VP)>

Each of the types are manually specified. But that is a duplication because T, V and VP are already contained in the pointers to member functions getf and setf's types.

---

But if I try a template with only

template <V(T::*getf)(), void (T::*setf)(VP)>

or

template <V(T::*getf)(), void (T::*setf)(VP), typename T, typename V, typename VP>

then the types cannot be determined.

---

Next I tried specialization:

template <typename T, typename T2>
struct Accessor;

template <typename V, typename T, typename VP>
struct Accessor <V(T::*)(), void (T::*)(VP)>

which will determine all of the types if use

typedef Accessor<
    decltype(&TargetClass::GetFoo), 
    decltype(&TargetClass::SetFoo)> fooAcessor;

But now I don't have the pointers anymore, only the types.

---

Is there a way to write a template so that the types can be determined automatically from the non-type template parameter?

Answer 1

Is there a way to write a template so that the types can be determined automatically from the non-type template parameter?

In C++17, yes, thanks to declaring non-type template parameters with auto:

template <auto x>
struct Test {
    decltype(x) val{x};
};

Before C++17, no. You'd have to write:

template <class T, T x>
struct Test {
    T val{x};
};