How to have a function pointer to a generic lambda?

Question

As the title says, how can I have the following intent expressed in code ? The requirements being the function pointer takes any type of arguments. Variadic doesn't work because of std::string.

https://godbolt.org/z/1E1szT

Note that I cannot directly use auto fp = <the lambda> because fp is a class member variable.

#include <iostream>

template<typename T1, typename T2>
void(*fp)(T1 t1, T2 t2) = [](auto a, auto b){
                              std::cout << a << "--"
                              << b << std::endl;
                          };

int main(){
    fp(1, 2);
    fp('a', 'b');
}

Answer 1

A variable template is perfectly fine.
Even the way you initialize them is fine.

Just be aware it is a template for variables, not a variable storing a template (which cannot exist).

Thus when finally using it, things fall apart:

How should the compiler know which of the myriad potential instances you mean?
Simple, you have to actually tell it:

int main(){
    fp<int, int>(1, 2);
    fp<char, char>('a', 'b');
}

Of course, manually desugaring the lambda and getting an instance of it would be more practical:

struct {
    template <class T1, class T2>
    void operator()(T1 a, T2 b) const {
        std::cout << a << "--" << b << std::endl;
    };
} fp;

A shame we cannot simply give a lambda a name and let the compiler figure it out.
Brainstorming candidates for appropriate syntax:

struct A = []...;
struct B : []... {
};
using C = []...;