C++ std::function cannot find correct overload

Question

Consider the following case:

void Set(const std::function &fn);
void Set(const std::function &fn);

Answer 1

I would suggest this solution. It should work with lambdas as well as with function-objects. It can be extended to make it work for function pointer as well (just go through the link provided at the bottom)

Framework:

template 
struct function_traits : public function_traits
{};

template 
struct function_traits
{
    enum { arity = sizeof...(Args) };
};

template
struct count_arg : std::enable_if::arity==NArgs, int>
{};

Usage:

template
typename count_arg::type Set(Functor f) 
{
    std::function fn = f;
    std::cout << "f with one argument" << std::endl;
}

template
typename count_arg::type Set(Functor f)
{
    std::function fn = f;
    std::cout << "f with two arguments" << std::endl;
}

int main() {
        Set([](int a){});
        Set([](int a, int b){});
        return 0;
}

Output:

f with one argument
f with two arguments

I took some help from the accepted answer of this topic:

• Is it possible to figure out the parameter type and return type of a lambda?

---

Work around for Visual Studio 2010

Since Microsoft Visual Studio 2010 doesn't support variadic templates, then the framework-part can be implemented as:

template 
struct function_traits : public function_traits
{};

template 
struct function_traits { enum { arity = 1 }; };

template 
struct function_traits { enum { arity = 2 }; };

template 
struct function_traits { enum { arity = 3 }; };

//this is same as before 
template
struct count_arg : std::enable_if::arity==NArgs, ReturnType>
{};

EDIT
Now this code supports any return type.