Is an is_functor C++ trait class possible? - c++

Question

How can I deduce statically if an argument is a C++ function object (functor)?
template <typename F>
void test(F f) {}
I tried is_function<F>::value, but this doesn't work. It also seems there is no is_functor trait, so perhaps it's not possible. I appear to be only looking for a specific member function, in this case the function call operator: F::operator().

Answer 1

It is possible to create such a trait, with two restrictions:
For the compiler, a free function is something fundamentally different from a class functor that overloads operator(). Thus we have to treat both cases seperately when implementing. This is not a problem for usage though, we can hide this implementation detail from the user.
We need to know the signature of the function you want to call. This is usually not a problem, and it does have the nice side effect that our trait is able to handle overloads pretty natively.
Step one: Free functions
Let's start with free functions, because they are little easier to detect. Our task is, when given a function pointer, to determine whether the signature of that function pointer matches the signature passed as the second template argument. To be able to compare those, we either need to get a grasp of the underlying function signature, or create a function pointer of our signature. I arbitrarily chose the latter:
// build R (*)(Args...) from R (Args...)
// compile error if signature is not a valid function signature
template <typename, typename>
struct build_free_function;
template <typename F, typename R, typename ... Args>
struct build_free_function<F, R (Args...)>
{ using type = R (*)(Args...); };
Now all that's left to do is to compare and we are done with the free function part:
// determine whether a free function pointer F has signature S
template <typename F, typename S>
struct is_function_with_signature
{
// check whether F and the function pointer of S are of the same
// type
static bool constexpr value = std::is_same<
F, typename build_free_function<F, S>::type
>::value;
};
Step two: Class functors
This one is a little more involved. We could easily detect with SFINAE whether a class defines an operator():
template <typename T>
struct defines_functor_operator
{
typedef char (& yes)[1];
typedef char (& no)[2];
// we need a template here to enable SFINAE
template <typename U>
static yes deduce(char (*)[sizeof(&U::operator())]);
// fallback
template <typename> static no deduce(...);
static bool constexpr value = sizeof(deduce<T>(0)) == sizeof(yes);
};
but that does not tell us whether one exists for our desired function signature! Luckily, we can use a trick here: pointers are valid template parameters. Thus we can first use the member function pointer of our desired signature, and check whether &T::operator() is of that type:
template <typename T, T> struct check;
Now check<void (C::*)() const, &C::operator()> will only be a valid template instantiation if C does indeed have a void C::operator()() const. But to do this we first have to combine C and the signature to a member function pointer. As we already have seen, we need to worry about two extra cases we did not have to care about for free functions: const and volatile functions. Besides that it's pretty much the same:
// build R (C::*)(Args...) from R (Args...)
// R (C::*)(Args...) const from R (Args...) const
// R (C::*)(Args...) volatile from R (Args...) volatile
// compile error if signature is not a valid member function signature
template <typename, typename>
struct build_class_function;
template <typename C, typename R, typename ... Args>
struct build_class_function<C, R (Args...)>
{ using type = R (C::*)(Args...); };
template <typename C, typename R, typename ... Args>
struct build_class_function<C, R (Args...) const>
{ using type = R (C::*)(Args...) const; };
template <typename C, typename R, typename ... Args>
struct build_class_function<C, R (Args...) volatile>
{ using type = R (C::*)(Args...) volatile; };
Putting that and our findings concerning the check helper struct together, we get our check metafunction for functor objects:
// determine whether a class C has an operator() with signature S
template <typename C, typename S>
struct is_functor_with_signature
{
typedef char (& yes)[1];
typedef char (& no)[2];
// helper struct to determine that C::operator() does indeed have
// the desired signature; &C::operator() is only of type
// R (C::*)(Args...) if this is true
template <typename T, T> struct check;
// T is needed to enable SFINAE
template <typename T> static yes deduce(check<
typename build_class_function<C, S>::type, &T::operator()> *);
// fallback if check helper could not be built
template <typename> static no deduce(...);
static bool constexpr value = sizeof(deduce<C>(0)) == sizeof(yes);
};
Step three: Putting the pieces together
We are almost done. Now we only need to decide when to use our free function, and when the class functor metafunctions. Luckily, C++11 provides us with a std::is_class trait that we can use for this. So all we have to do is specialize on a boolean parameter:
// C is a class, delegate to is_functor_with_signature
template <typename C, typename S, bool>
struct is_callable_impl
: std::integral_constant<
bool, is_functor_with_signature<C, S>::value
>
{};
// F is not a class, delegate to is_function_with_signature
template <typename F, typename S>
struct is_callable_impl<F, S, false>
: std::integral_constant<
bool, is_function_with_signature<F, S>::value
>
{};
So we can finally add the last piece of the puzzle, being our actual is_callable trait:
// Determine whether type Callable is callable with signature Signature.
// Compliant with functors, i.e. classes that declare operator(); and free
// function pointers: R (*)(Args...), but not R (Args...)!
template <typename Callable, typename Signature>
struct is_callable
: is_callable_impl<
Callable, Signature,
std::is_class<Callable>::value
>
{};
Now we clean up our code, put implementation details into anonymous namespaces so they are not acessible outside of our file, and have a nice is_callable.hpp to use in our project.
Full code
namespace // implementation detail
{
// build R (*)(Args...) from R (Args...)
// compile error if signature is not a valid function signature
template <typename, typename>
struct build_free_function;
template <typename F, typename R, typename ... Args>
struct build_free_function<F, R (Args...)>
{ using type = R (*)(Args...); };
// build R (C::*)(Args...) from R (Args...)
// R (C::*)(Args...) const from R (Args...) const
// R (C::*)(Args...) volatile from R (Args...) volatile
// compile error if signature is not a valid member function signature
template <typename, typename>
struct build_class_function;
template <typename C, typename R, typename ... Args>
struct build_class_function<C, R (Args...)>
{ using type = R (C::*)(Args...); };
template <typename C, typename R, typename ... Args>
struct build_class_function<C, R (Args...) const>
{ using type = R (C::*)(Args...) const; };
template <typename C, typename R, typename ... Args>
struct build_class_function<C, R (Args...) volatile>
{ using type = R (C::*)(Args...) volatile; };
// determine whether a class C has an operator() with signature S
template <typename C, typename S>
struct is_functor_with_signature
{
typedef char (& yes)[1];
typedef char (& no)[2];
// helper struct to determine that C::operator() does indeed have
// the desired signature; &C::operator() is only of type
// R (C::*)(Args...) if this is true
template <typename T, T> struct check;
// T is needed to enable SFINAE
template <typename T> static yes deduce(check<
typename build_class_function<C, S>::type, &T::operator()> *);
// fallback if check helper could not be built
template <typename> static no deduce(...);
static bool constexpr value = sizeof(deduce<C>(0)) == sizeof(yes);
};
// determine whether a free function pointer F has signature S
template <typename F, typename S>
struct is_function_with_signature
{
// check whether F and the function pointer of S are of the same
// type
static bool constexpr value = std::is_same<
F, typename build_free_function<F, S>::type
>::value;
};
// C is a class, delegate to is_functor_with_signature
template <typename C, typename S, bool>
struct is_callable_impl
: std::integral_constant<
bool, is_functor_with_signature<C, S>::value
>
{};
// F is not a class, delegate to is_function_with_signature
template <typename F, typename S>
struct is_callable_impl<F, S, false>
: std::integral_constant<
bool, is_function_with_signature<F, S>::value
>
{};
}
// Determine whether type Callable is callable with signature Signature.
// Compliant with functors, i.e. classes that declare operator(); and free
// function pointers: R (*)(Args...), but not R (Args...)!
template <typename Callable, typename Signature>
struct is_callable
: is_callable_impl<
Callable, Signature,
std::is_class<Callable>::value
>
{};
Ideone example with some tests
http://ideone.com/7PWdiv

Answer 2

Although this does not work for overloaded functions, for all other cases (free functions, classes implementing operator(), and lambdas) this short solutions works in C++11:
template <typename T, typename Signature>
struct is_callable: std::is_convertible<T,std::function<Signature>> { };
Note: std::is_invocable is available since C++17.

Answer 3

You can use the following concept in c++20
template<typename F>
concept FunctionObject = requires (F) {
&F::operator();
};

Answer 4

Is an is_functor C++ trait class possible?
Yes, it is possible to manually implement a validation for functors.
I tried is_function::value, but this doesn't work.
You are on the right path, it is possible to implement using std::function
Remember that std::function also accepts functions, function pointers and functor instances in its constructor.
Example:
struct Test {
public:
bool operator()(int){
return true;
}
};
void test(int){
}
void example(std::function<void(int)> ex){
cout << "Called" << endl;
};
int main()
{
example(test);
example(&test);
example(Test{});
}
That said, the logic that will be used to validate if a class has a function call overload operator (functor) is similar to the code above.
That is, if the std::function<void(int)> accepts an instance of the class Test{} means the class has a functor, otherwise it doesn't.
Example of an possible solution
Here is the source code:
//F: Test class
//Args...: The arguments, ex: int or none
template <typename F, typename... Args>
struct is_functor :
is_constructible <
function<void(Args ...)>, F
>
{};
Example usage:
is_functor<Test, int> -> True result
is_functor -> False result
Info about std::is_constructible
Is a trait class that identifies whether T is a constructible type with the set of argument types specified by Arg.
For this class, a constructible type is a type that can be constructed using a particular set of arguments.
is_constructible inherits from integral_constant as being either true_type or false_type, depending on whether T is constructible with the list of arguments Args.
In short, it checks if a given class has a constructor, for example:
struct Test2 {
Test(bool, int){}
};
std::is_constructible<Test2, bool, int> -> True result
std::is_constructible<Test2, float> -> False result
Example of implementation:
template <typename, typename, typename ...Args>
struct is_constructible_impl : false_type {};
template <typename T, typename ...Args>
struct is_constructible_impl <
void_t<decltype(T(std::declval<Args>()...))>, T, Args...
> : true_type {};
template <typename T, typename ...Args>
struct is_constructible : is_constructible_impl<void_t<>, T, Args...> {};
Final explanation
In the implementation of is_functor it was checked if std::function<void(int)> accepts an instance of Test{}, which is true.
References:
How is std::is_constructible<T, Args> implemented?
Can std::is_invocable be emulated within C++11?
https://replit.com/#LUCASP6/RowdyAlphanumericCode#main.cpp

Answer 5

template<typename T, typename Sign>
struct is_functor
{
typedef char yes[1];
typedef char no [2];
template <typename U, U> struct type_check;
template <typename _1> static yes &chk(type_check<Sign, &_1::operator()>*);
template <typename > static no &chk(...);
static bool const value = sizeof(chk<T>(nullptr)) == sizeof(yes);
};
Altered from this answer.
It could be used like...
template<typename T>
typename std::enable_if<is_functor<T, void(T::*)()>::value>::type func()
{
}