如何用 c++ 20 中的概念声明通用高阶函数的签名？

Question

The following code declares that the function returned by the function Foo must model a function that takes 1 int and returns an int.

#include <functional>
#include <concepts>

std::convertible_to<std::function<int(int)>> auto Foo(){
    return [](int x){
        return 1;
    }; 
}

int main(){
    return Foo()(1);
}

If I now change the code and declare the returned function to be templated with auto like so

#include <functional>
#include <concepts>

std::convertible_to<std::function<int(auto)>> auto Foo(){
    return [](auto x){
        return 1;
    }; 
}

int main(){
    return Foo()(1);
}

I can't simply change the concept and insert auto as in std::convertible_to<std::function<int(auto)>>

Is there a workaround or special notation for this?

https://godbolt.org/z/sKf3Eb4MM

Answer 1

This may work under some circumstances though certainly not everything.

#include <concepts>

/// A type that converts to anything as a stand in for the
/// unknown type.
struct Auto
{
    template <typename T>
    operator T();

    template <typename T>
    operator T&();

    template <typename T>
    operator T&&();
};

template <typename Fn, typename R, typename ...Args>
concept callable = requires (Fn fn, Args ... args){
    {fn(args...)}->std::same_as<R>;

};

callable<int,Auto> auto Foo(){
    return [](auto x)->int{
        return (int)(x+1);
    }; 
}

int main(){
    return Foo()(1);
}

https://godbolt.org/z/xzbx55bG7