How do I determine if a type is callable with only const references?

Question

I want to write a C++ metafunction is_callable that defines value to be true, if and only if the type F has the function cal

Answer 1

Here's something I hacked up which may or may not be what you need; it does seem to give true (false) for (const) int &...

#include <utility>

template <typename F, typename Arg>
struct Callable
{
private:
  typedef char                      yes;
  typedef struct { char array[2]; } no;

  template <typename G, typename Brg>
  static yes test(decltype(std::declval<G>()(std::declval<Brg>())) *);

  template <typename G, typename Brg>
  static no test(...);

public:
  static bool const value = sizeof(test<F, Arg>(nullptr)) == sizeof(yes);
};

struct Foo
{
  int operator()(int &) { return 1; }
  // int operator()(int const &) const { return 2; } // enable and compare
};

#include <iostream>
int main()
{
  std::cout << "Foo(const int &): " << Callable<Foo, int const &>::value << std::endl
            << "Foo(int &):       " << Callable<Foo, int &>::value << std::endl
    ;
}

Answer 2

(with apologies to Kerrek for using his answer as a starting point)

EDIT: Updated to handle types without any operator() at all.

#include <utility>

template <typename F, typename Arg>
struct Callable
{
private:
  static int tester[1];
  typedef char                      yes;
  typedef struct { char array[2]; } no;

  template <typename G, typename Brg>
  static char sfinae(decltype(std::declval<G>()(std::declval<Brg>())) (G::*pfn)(Brg)) { return 0; }

  template <typename G, typename Brg>
  static char sfinae(decltype(std::declval<G>()(std::declval<Brg>())) (G::*pfn)(Brg) const) { return 0; }

  template <typename G, typename Brg>
  static yes test(int (&a)[sizeof(sfinae<G,Brg>(&G::operator()))]);

  template <typename G, typename Brg>
  static no test(...);

public:
  static bool const value = sizeof(test<F, Arg>(tester)) == sizeof(yes);
};

struct Foo
{
  int operator()(int &) { return 1; }

};

struct Bar
{
  int operator()(int const &) { return 2; }
};

struct Wazz
{
  int operator()(int const &) const { return 3; }
};

struct Frob
{
  int operator()(int &) { return 4; }
  int operator()(int const &) const { return 5; }
};

struct Blip
{
  template<typename T>
  int operator()(T) { return 6; }
};

struct Boom
{

};

struct Zap
{
  int operator()(int) { return 42; }
};

#include <iostream>
int main()
{
  std::cout << "Foo(const int &):  " << Callable<Foo,  int const &>::value << std::endl
            << "Foo(int &):        " << Callable<Foo,  int &>::value << std::endl
            << "Bar(const int &):  " << Callable<Bar,  const int &>::value << std::endl
            << "Bar(int &):        " << Callable<Bar,  int &>::value << std::endl
            << "Zap(const int &):  " << Callable<Zap , const int &>::value << std::endl
            << "Zap(int&):         " << Callable<Zap , int &>::value << std::endl
            << "Wazz(const int &): " << Callable<Wazz, const int &>::value << std::endl
            << "Wazz(int &):       " << Callable<Wazz, int &>::value << std::endl
            << "Frob(const int &): " << Callable<Frob, const int &>::value << std::endl
            << "Frob(int &):       " << Callable<Frob, int &>::value << std::endl
            << "Blip(const int &): " << Callable<Blip, const int &>::value << std::endl
            << "Blip(int &):       " << Callable<Blip, int &>::value << std::endl
            << "Boom(const int &): " << Callable<Boom, const int &>::value << std::endl
            << "Boom(int&):        " << Callable<Boom, int &>::value << std::endl;
}

Demo: http://ideone.com/T3Iry

Answer 3

Here is a possible solution that utilizes an extra test to see if your template is being instantiated with a const T&:

#include <memory>
#include <iostream>

using namespace std;

template<typename F, typename Arg>
struct is_callable {
private:

  template<typename>
  static char (&test(...))[2];

  template<bool, unsigned value>
  struct helper {};

  template<unsigned value>
  struct helper<true, value> {
    typedef void *type;
  };

  template<typename T>
  struct is_const_ref {};

  template<typename T>
  struct is_const_ref<T&> {
    static const bool value = false;
  };

  template<typename T>
  struct is_const_ref<const T&> {
    static const bool value = true;
  };

  template<typename UVisitor>
  static char test(typename helper<is_const_ref<Arg>::value, 
                                   sizeof(std::declval<UVisitor>()(std::declval<Arg>()), 0)>::type);
public:
  static const bool value = (sizeof(test<F>(0)) == sizeof(char));
};

struct foo {
  void operator()(const int &) {}
};

int main(void)
{
  cout << is_callable<foo, int &>::value << "\n";
  cout << is_callable<foo, const int &>::value << "\n";

  return 0;
}

Answer 4

Something like this maybe? It's a bit round about to make it work on VS2010.

template<typename FPtr>
struct function_traits_impl;

template<typename R, typename A1>
struct function_traits_impl<R (*)(A1)>
{
    typedef A1 arg1_type;
};

template<typename R, typename C, typename A1>
struct function_traits_impl<R (C::*)(A1)>
{
    typedef A1 arg1_type;
};

template<typename R, typename C, typename A1>
struct function_traits_impl<R (C::*)(A1) const>
{
    typedef A1 arg1_type;
};

template<typename T>
typename function_traits_impl<T>::arg1_type arg1_type_helper(T);

template<typename F>
struct function_traits
{
    typedef decltype(arg1_type_helper(&F::operator())) arg1_type;
};

template<typename F, typename Arg>
struct is_callable : public std::is_same<typename function_traits<F>::arg1_type, const Arg&>
{
}

Answer 5

Ran across this while doing something else, was able to adapt my code to fit. It has the same features (and limitations) as @Xeo, but does not need sizeof trick/enable_if. The default parameter takes the place of needing to do the enable_if to handle template functions. I tested it under g++ 4.7 and clang 3.2 using the same test code Xeo wrote up

#include <type_traits>
#include <functional>

namespace detail {
  template<typename T, class Args, class Enable=void>
  struct call_exact : std::false_type {};

  template<class...Args> struct ARGS { typedef void type; };

  template<class T, class ... Args, class C=T>
  C * opclass(decltype(std::declval<T>()(std::declval<Args>()...)) (C::*)(Args...)) { }
  template<class T, class ... Args, class C=T>
  C * opclass(decltype(std::declval<T>()(std::declval<Args>()...)) (C::*)(Args...) const) { }

  template<typename T, class ... Args>
  struct call_exact<T, ARGS<Args...>,
    typename ARGS<
       decltype(std::declval<T&>()(std::declval<Args>()...)),
       decltype(opclass<T, Args...>(&T::operator()))
     >::type
   > : std::true_type {};
}

template<class T, class ... Args>
struct Callable : detail::call_exact<T, detail::ARGS<Args...>> { };

template<typename R, typename... FArgs, typename... Args>
struct Callable<R(*)(FArgs...), Args...>
 : Callable<std::function<R(FArgs...)>, Args...>{};

Answer 6

After hours of playing around and some serious discussions in the C++ chat room, we finally got a version that works for functors with possibly overloaded or inherited operator() and for function pointers, based on @KerrekSB's and @BenVoigt's versions.

#include <utility>
#include <type_traits>

template <typename F, typename... Args>
class Callable{
  static int tester[1];
  typedef char yes;
  typedef yes (&no)[2];

  template <typename G, typename... Brgs, typename C>
  static typename std::enable_if<!std::is_same<G,C>::value, char>::type
      sfinae(decltype(std::declval<G>()(std::declval<Brgs>()...)) (C::*pfn)(Brgs...));

  template <typename G, typename... Brgs, typename C>
  static typename std::enable_if<!std::is_same<G,C>::value, char>::type
      sfinae(decltype(std::declval<G>()(std::declval<Brgs>()...)) (C::*pfn)(Brgs...) const);

  template <typename G, typename... Brgs>
  static char sfinae(decltype(std::declval<G>()(std::declval<Brgs>()...)) (G::*pfn)(Brgs...));

  template <typename G, typename... Brgs>
  static char sfinae(decltype(std::declval<G>()(std::declval<Brgs>()...)) (G::*pfn)(Brgs...) const);

  template <typename G, typename... Brgs>
  static yes test(int (&a)[sizeof(sfinae<G,Brgs...>(&G::operator()))]);

  template <typename G, typename... Brgs>
  static no test(...);

public:
  static bool const value = sizeof(test<F, Args...>(tester)) == sizeof(yes);
};

template<class R, class... Args>
struct Helper{ R operator()(Args...); };
 
template<typename R, typename... FArgs, typename... Args>
class Callable<R(*)(FArgs...), Args...>
  : public Callable<Helper<R, FArgs...>, Args...>{};

Live example on Ideone. Note that the two failing tests are overloaded operator() tests. This is a GCC bug with variadic templates, already fixed in GCC 4.7. Clang 3.1 also reports all tests as passed.

If you want operator() with default arguments to fail, there is a possible way to do that, however some other tests will start failing at that point and I found it as too much hassle to try and correct that.

Edit: As @Johannes correctly notes in the comment, we got a little inconsistency in here, namely that functors which define a conversion to function pointer will not be detected as "callable". This is, imho, pretty non-trivial to fix, as such I won't bother with it (for now). If you absolutely need this trait, well, leave a comment and I'll see what I can do.

---

Now that all this has been said, IMHO, the idea for this trait is stupid. Why whould you have such exact requirements? Why would the standard is_callable not suffice?

(Yes, I think the idea is stupid. Yes, I still went and built this. Yes, it was fun, very much so. No, I'm not insane. Atleast that's what I believe...)