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

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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) {}.

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