Operator overloading for lambdas?

Question

I have asked a similar question before overloading operator >> for lambdas
But i did not explained what i really wanted .

I am writing a simple wrapper around

Answer 1

What you need is a retrospective cast. A way to compose a correct function object type from passing it only a lambda (and nothing else, no template arguments, not return type specification).

A way to do it without dependencies from other libraries would be the following

#include <iostream>
#include<functional>
#include<vector>


using namespace std;


template<typename T>
struct memfun_type 
{
    using type = void;
};

template<typename Ret, typename Class, typename... Args>
struct memfun_type<Ret(Class::*)(Args...) const>
{
    using type = std::function<Ret(Args...)>;
};

template<typename F>
typename memfun_type<decltype(&F::operator())>::type
FFL(F const &func) 
{ // Function from lambda !
    return func;
}

Then you can be aware of your lambas' return type and write the following

int main() 
{
    database_bind dbb;

    dbb >> FFL([](int i, string s) { cout << i << ' ' << s << endl; });
    dbb >> FFL([](int i) { cout << i << endl; });
    dbb >> FFL([](string s,double d) { cout << s << ' ' << d << endl; });
}

Answer 2

Based on this question i wrote the fallowing code.
I did not completely understood function_traits ! but i was able to overload the >> operator for lambdas with different number of arguments . I know it's not the best possible solution , but i wrote it so someone can take it as a starting point ( a variadic template implementation will be awesome !) .

#include<iostream>
#include<string>
#include<tuple>
using namespace std;

template <typename T>
struct function_traits
: public function_traits<decltype(&T::operator())>
{};

template <typename ClassType, typename ReturnType, typename... Args>
struct function_traits<ReturnType(ClassType::*)(Args...) const>
// we specialize for pointers to member function
{
    enum { arity = sizeof...(Args) };
    // arity is the number of arguments.

    typedef ReturnType result_type;

    template <size_t i>
    struct arg
    {
        typedef typename std::tuple_element<i, std::tuple<Args...>>::type type;
        // the i-th argument is equivalent to the i-th tuple element of a tuple
        // composed of those arguments.
    };
};

// a place holder class
class database_bind {};

template<int N>
class A {
    template<typename F>
    static void run(F l);
};

template<>
struct A<1> {
    template<typename F>
    static void run(F l) {
        typedef function_traits<decltype(l)> traits;
        typedef typename traits::arg<0>::type type_1;

        type_1 col_1;
        get_from_db(0, col_1);

    l(col_1);
    }
};
template<>
struct A<2> {
    template<typename F>
    static void run(F l) {
        typedef function_traits<decltype(l)> traits;
        typedef typename traits::arg<0>::type type_1;
        typedef typename traits::arg<1>::type type_2;

        type_1 col_1;
        type_2 col_2;
        get_from_db(0, col_1);
        get_from_db(1, col_2);

        l(col_1, col_2);
    }
};


void get_from_db(int col_inx, string& str) {  str = "string"; }
void get_from_db(int col_inx, int& i) {i = 123;}
void get_from_db(int col_inx, double& d) { d = 123.456; }


template<typename F>
void operator>>(database_bind dbb, F l)
{
    typedef function_traits<decltype(l)> traits;
    A<traits::arity>::run(l);
}

And finally :

int main() {
    database_bind dbb;

    dbb >> [](int i, string s) { cout << i << ' ' << s << endl; };
    dbb >> [](int i) { cout << i << endl; };
    dbb >> [](string s,double d) { cout << s << ' ' << d << endl; };
   }