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

Given a lambda, is it possible to figure out it's parameter type and return type? If yes, how?

Basically, I want lambda_traits which can be used in following ways:

auto lambda = [](int i) { return long(i*10); };

lambda_traits<decltype(lambda)>::param_type  i; //i should be int
lambda_traits<decltype(lambda)>::return_type l; //l should be long

The motivation behind is that I want to use lambda_traits in a function template which accepts a lambda as argument, and I need to know it's parameter type and return type inside the function:

template<typename TLambda>
void f(TLambda lambda)
{
   typedef typename lambda_traits<TLambda>::param_type  P;
   typedef typename lambda_traits<TLambda>::return_type R;

   std::function<R(P)> fun = lambda; //I want to do this!
   //...
}

For the time being, we can assume that the lambda takes exactly one argument.

Initially, I tried to work with std::function as:

template<typename T>
A<T> f(std::function<bool(T)> fun)
{
   return A<T>(fun);
}

f([](int){return true;}); //error

But it obviously would give error. So I changed it to TLambda version of the function template and want to construct the std::function object inside the function (as shown above).

Funny, I've just written a function_traits implementation based on Specializing a template on a lambda in C++0x which can give the parameter types. The trick, as described in the answer in that question, is to use the decltype of the lambda's operator().

template <typename T>
struct function_traits
    : public function_traits<decltype(&T::operator())>
{};
// For generic types, directly use the result of the signature of its '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.
    };
};

// test code below:
int main()
{
    auto lambda = [](int i) { return long(i*10); };

    typedef function_traits<decltype(lambda)> traits;

    static_assert(std::is_same<long, traits::result_type>::value, "err");
    static_assert(std::is_same<int, traits::arg<0>::type>::value, "err");

    return 0;
}

Note that this solution does not work for generic lambda like [](auto x) {}.

Though I'm not sure this is strictly standard conforming, ideone compiled the following code:

template< class > struct mem_type;

template< class C, class T > struct mem_type< T C::* > {
  typedef T type;
};

template< class T > struct lambda_func_type {
  typedef typename mem_type< decltype( &T::operator() ) >::type type;
};

int main() {
  auto l = [](int i) { return long(i); };
  typedef lambda_func_type< decltype(l) >::type T;
  static_assert( std::is_same< T, long( int )const >::value, "" );
}

However, this provides only the function type, so the result and parameter types have to be extracted from it. If you can use boost::function_traits, result_type and arg1_type will meet the purpose. Since ideone seems not to provide boost in C++11 mode, I couldn't post the actual code, sorry.

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

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