deducing template specilization based on constructor arguments - c++

Question

I'm trying to provide a small class specialization when 2 provided types are similar and I have reached following code that works:
template<typename Fn, typename U, typename V>
class K {
public:
K(Fn f, U u, V v) : u_(u), v_(v) {
std::cout << "2 args\n";
}
private:
V v_;
U u_;
};
template<typename Fn, typename U>
class K<Fn, U, U> {
public:
K(Fn f, U u) : u_(u) {
std::cout << "1 args\n";
}
private:
U u_;
};
void koo(int i, double d) {}
void moo(int i) {}
int main() {
K(koo, 3, 5.6);
K<decltype(moo), int, int>(moo, 3);
}
My only problem is that to select the specialization, I must provide <decltype(moo), int, int> manually or the specialization will not picked. I wonder if it is possible to pick this specialization based on single constructor argument so K(moo,3); works too. Any idea?

Answer 1

In C++17, provide deduction guide for CTAD:
template <typename Fn, typename U> K(Fn, U) -> K<Fn, U, U>;
Demo

Answer 2

A simple C++11-compatible solution would be to delegate the argument substitution to a helper function like make_shared for the std::shared_ptr:
template<typename Fn, typename U, typename V>
K<Fn, U, V> make_K(Fn f, U u, V v) {
return K<Fn, U, V>(f, u, v);
}
template<typename Fn, typename U>
K<Fn, U, U> make_K(Fn f, U u) {
return K<Fn, U, U>(f, u);
}
used as
auto k = make_K(koo, 3, 5.6);
auto m = make_K(moo, 3);