Explicitly choose which template overload should be used - c++

Question

Can we chose which function template overload should be used in this case?
struct X { };
struct A { A(X){} };
struct B { B(X){} };
template<class T>
void fun(T, A) { }
template<class T>
void fun(T, B) { }
int main() {
/* explicitly choose overload */ fun(1, X());
}
Error:
error: call of overloaded 'fun(int, X)' is ambiguous
/* explicitly choose overload */ fun(1, X());
^
candidate: void fun(T, A) [with T = int]
void fun(T, A) { }
^~~
candidate: void fun(T, B) [with T = int]
void fun(T, B) { }
^~~
For normal function it looks like this:
void fun(A){}
void fun(B){}
int main() {
((void(*)(A))(fun))(X());
}
Is it possible?

Answer 1

Improving your example, you can try with
((void(*)(int, A))(fun))(1, X());

Answer 2

If you choose not to explicitly specify the first parameter type but still want to specify the second you could go along with lambda dedicated for casting purpose (c++14 solution):
struct X { };
struct A { A(X){} };
struct B { B(X){} };
template<class T>
void fun(T, A) { }
template<class T>
void fun(T, B) { }
int main() {
[](auto v, auto x){ static_cast<void(*)(decltype(v), A)>(fun)(v, x); }(1, X());
}
[live demo]

Answer 3

A low-tech solution to this problem would be to add one extra level of indirection. Add a function like funA, whose sole purpose is to give an explicit name to the first version of fun:
struct X { };
struct A { A(X){} };
struct B { B(X){} };
template<class T>
void fun(T, A) { }
template<class T>
void fun(T, B) { }
template <class T>
void funA(T t, A a) { fun(t, a); }
int main() {
/* explicitly choose overload */ funA(1, X());
}
However, I wonder why you cannot just change the argument to A(X()). You will have to change the calling code anyway, so what's the problem?