Detect if C++ lambda can be converted to function pointer

Question

I have some code that generates assembly for a JIT idea I'm working on. I use meta-programming to generate calls by analyzing the function type and then generating the correct assembly to call it. I recently wanted to add lambda support, and lambdas have two versions, non-capturing (normal __cdecl function call) and capturing (__thiscall, member-function call with the lambda object as context).

__thiscall is slightly more expensive so I'd like to avoid it whenever possible, and I'd also like to avoid having to use different call generation functions depending on the lambda type.

I tried many ways to detect the lambda type via templates and SFINAE and all failed.

Non-capturing lambdas have an ::operator function_type* which one can use to convert them to function pointers, while capturing lambdas don't.

Relevant C++ spec: http://en.cppreference.com/w/cpp/language/lambda

Any ideas?

edit I'd like to have a solution that works for vs 2013/2015, gcc and clang

Test code follows

#include <utility>

    //this doesn't work
    template < class C, class T >
    struct HasConversion {
        static int test(decltype(std::declval<C>().operator T*, bool()) bar) {
            return 1;
        }

        static int test(...) {
            return 0;
        }
    };

    template <class C>
    void lambda_pointer(C lambda) {
        int(*function)() = lambda;

        printf("Lambda function: %p without context\n", function);
    }

    template <class C>
    void lambda_pointer_ctx(C lambda) {
        int(C::*function)() const = &C::operator();

        void* context = &lambda;

        printf("Lambda function: %p with context: %p\n", function, context);
    }

    int main() {
        int a;

        auto l1 = [] {
            return 5;
        };

        auto l2 = [a] {
            return a;
        };


        //non capturing case

        //works as expected
        lambda_pointer(l1);

        //works as expected (ctx is meaningless and not used)
        lambda_pointer_ctx(l1);



        //lambda with capture (needs context)

        //fails as expected
        lambda_pointer(l1);

        //works as expected (ctx is a pointer to class containing the captures)
        lambda_pointer_ctx(l1);

        /*
        //this doesn't work :<
        typedef int afunct() const;

        HasConversion<decltype(l1), afunct>::test(0);
        HasConversion<decltype(l2), afunct>::test(0);
        */


        return 0;
    }

Answer 1

If you know the signature of a function you want your lambda to be converted to, you can leverage the std::is_assignable trait:

auto lambda = [] (char, double) -> int { return 0; };
using signature = int(char, double);
static_assert(std::is_assignable<signature*&, decltype(lambda)>::value, "!");

DEMO

This way it can work also with generic lambdas.

---

I'd like to have a solution that works for vs 2013/2015, gcc and clang

If you don't know the signature, here's an approach that is an alternative to checking whether + triggers an implicit conversion. This one exploits the std::is_assignable test and verifies whether a lambda is assignable to a function pointer with the same signature as the lambda's function call operator. (Just like a test with unary operator plus, this doesn't work with generic lambdas. But in C++11 there are no generic lambdas).

#include <type_traits>

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

template <typename...>
using void_t = void;

template <typename F>
struct call_operator;

template <typename C, typename R, typename... A>
struct call_operator<R(C::*)(A...)> : identity<R(A...)> {};

template <typename C, typename R, typename... A>
struct call_operator<R(C::*)(A...) const> : identity<R(A...)> {};

template <typename F>
using call_operator_t = typename call_operator<F>::type;

template <typename, typename = void_t<>>
struct is_convertible_to_function
    : std::false_type {};

template <typename L>
struct is_convertible_to_function<L, void_t<decltype(&L::operator())>>
    : std::is_assignable<call_operator_t<decltype(&L::operator())>*&, L> {};

Test:

int main()
{
    auto x = [] { return 5; };
    auto y = [x] { return x(); };

    static_assert(is_convertible_to_function<decltype(x)>::value, "!");
    static_assert(!is_convertible_to_function<decltype(y)>::value, "!");
}

GCC, VC++, Clang++

Answer 2

Non-capturing lambdas have a very interesting property : they can convert to an adequate function pointer, but they can also do so implicitly when you apply unary operator + to them. Thus :

template<class...> using void_t = void;

template <class T, class = void>
struct has_capture : std::true_type {};

template <class T>
struct has_capture<T, void_t<decltype(+std::declval<T>())>> : std::false_type {};

int main() {
    auto f1 = []{};
    auto f2 = [&f1]{};

    static_assert(!has_capture<decltype(f1)>{}, "");
    static_assert( has_capture<decltype(f2)>{}, "");
}

Answer 3

The HasConversion approach you have going is a holdover from C++03. The idea there was to use the different return types of the overloads of test (have one return a char and the other a long, for instance) and check that the sizeof() of the return type matches what you expect.

Once we're on C++11 though, there are much better methods. We can, for instance, use void_t:

template <typename... >
using void_t = void;

to write a type trait:

template <typename T, typename = void>
struct has_operator_plus : std::false_type { };

template <typename T>
struct has_operator_plus<T, 
    void_t<decltype(+std::declval<T>())>>
: std::true_type { };

int main() {
    auto x = []{ return 5; };
    auto y = [x]{ return x(); };

    std::cout << has_operator_plus<decltype(x)>::value << std::endl; // 1
    std::cout << has_operator_plus<decltype(y)>::value << std::endl; // 0
}