There is no function template that matches the parameter list

Question

include <iostream>

using namespace std;
template <class T>
class A
{
private:
T abc;
public:
void(A<T>::*oP)(T k);
typedef void (A<T>::P)(T k);
A(T g) {
this->abc = g;
}
template<class To>
To show(To o) {
std::cout << "hi";
};
template<class L>
Afun(P o) {
o =&show;
return nullptr;
};
void func(int p) {};
};
namespace ko {
int s = 90;
}
typedef void (A<int>::*oP)(int);
int main() {
oP cd;
cd= &A<int>::func;
A<int> p(90);
A<int> o(0);

auto l=p.fun(cd);

}

Answer 1

fun is a function template. Its template parameter L doesn't appear anywhere in the function signature, and so cannot be deduced from the arguments. The only way to call fun is to provide the template argument explicitly, as in p.fun<SomeType>(cd)

Now, it's unclear why fun is made template, seeing as it doesn't use L at all.