How to make a variadic is_same? - c++

How can I make a class template that returns whether any of its variadic types are equal to the first type. I want to be able to do this:
is_same<T, A, B, C>::value; // true if T is one of A, B or C
And if T is equal to any one of those types, its static value member will be true, otherwise false. How can I do this?

Nice and concise with C++17:
template <class T, class... Ts>
struct is_any : std::disjunction<std::is_same<T, Ts>...> {};
And the dual:
template <class T, class... Ts>
struct are_same : std::conjunction<std::is_same<T, Ts>...> {};
A variation that uses fold expressions:
template <class T, class... Ts>
struct is_any : std::bool_constant<(std::is_same_v<T, Ts> || ...)> {};
template <class T, class... Ts>
struct are_same : std::bool_constant<(std::is_same_v<T, Ts> && ...)> {};

Use template recursion:
template<typename T, typename... Rest>
struct is_any : std::false_type {};
template<typename T, typename First>
struct is_any<T, First> : std::is_same<T, First> {};
template<typename T, typename First, typename... Rest>
struct is_any<T, First, Rest...>
: std::integral_constant<bool, std::is_same<T, First>::value || is_any<T, Rest...>::value>
{};
static_assert(is_any<int, char, double, int>::value, "error 1"); // OK
static_assert(is_any<int, char, double, short>::value, "error 2"); // error

In C++17 you have an even nicer solution, using template variables and fold expressions:
template<class T, class... Rest>
inline constexpr bool are_all_same = (std::is_same_v<T, Rest> && ...);
And the usage is also simpler than all other examples:
are_all_same<T, A, B, C>
No ::value, no parentheses!

Something like this. First, a small metaprogramming library, because it adds like 2 lines to do it generically:
template<template<typename,typename>class checker, typename... Ts>
struct is_any_to_first : std::false_type {};
template<template<typename,typename>class checker, typename T0, typename T1, typename... Ts>
struct is_any_to_first<checker, T0, T1, Ts...> :
std::integral_constant< bool, checker<T0, T1>::value || is_any_to_first<checker, T0, Ts...>::value>
{};
Then a 2 line implementation of is_any_same_to_first:
template<typename... Ts>
using is_any_same_to_first = is_any_to_first< std::is_same, Ts... >;
And for completeness, the original is_all, which may also prove useful:
template<template<typename,typename>class checker, typename... Ts>
struct is_all : std::true_type {};
template<template<typename,typename>class checker, typename T0, typename T1, typename... Ts>
struct is_all<checker, T0, T1, Ts...> :
std::integral_constant< bool, checker<T0, T1>::value && is_all<checker, T0, Ts...>::value>
{};
template<typename... Ts>
using is_all_same = is_all< std::is_same, Ts... >;
Live example of the is_all_same.
Note that calling is_any_same_to_first anything less explicit is asking for trouble. 2/3 people who tried to answer this question, including me, assumed that is_same<A,B,C> is true iff all three are the same type!

Using the relaxed C++14 constexpr functions, these kinds of things are much easier to code, and probably much faster to compile as well, so you could write:
template <class T, class ... Candidates>
constexpr bool is_all_same() {
bool pairs[] = {std::is_same<T,Candidates>::value...};
for(bool p: pairs) if(!p) return false;
return true;
}
template <class T, class ... Candidates>
constexpr bool is_any_same() {
bool pairs[] = {std::is_same<T,Candidates>::value...};
for(bool p: pairs) if(p) return true;
return false;
}
This is enabled by the fact that in C++14 constexpr functions can have for loops.

The most generic version that works :
since C++11
without dependencies (no #include <type_traits> needed)
only one name is_same works for n-types (no is_same_all needed)
template <typename First, typename Second, typename ... Next>
struct is_same {
template <typename A, typename B>
struct is_same_min {
enum { value = false };
};
template <typename A>
struct is_same_min<A,A> {
enum { value = true };
};
template <typename X, typename Y>
constexpr static bool check() {
return is_same_min<X,Y>::value;
};
template <typename X, typename Y, typename Z, typename ... K>
constexpr static bool check() {
return is_same_min<X,Y>::value and check<Y, Z, K...>();
};
enum { value = check<First, Second, Next...>() };
};
Just use is_same<T1,T2,T3...>::value

Related

Implementation of concepts in C++17

I'd seen the Bjarne stroustrup's talk named "Concepts the future of generic programming" and I thought maybe I can write "Concepts" in C++17...
But I'm a C programmer I've start with C++ about 6 months ago... so I don't know why but I've got e trouble with this code. Somebody can help me?
I'd implemented the Addable's concept with a shortly tmp code...
The idea is that will be in C++20 will be something like this:
template<typename T>
concept Summable = requires(T x) { x + x; };
template<typename T> requires Summable<T>
T sum(T n1, T n2) {
return n1 + n2;
}
And this code is what actually have in C++17:
template <typename, typename, typename = void>
struct Addable : std::false_type {};
template <typename T, typename U>
struct Addable < T, U,
std::void_t<decltype(std::declval<T&>()+std::declval<U&>())> >
: std::true_type {};
The top's code compile and works, but when I whrite the following code doesn't compile...
template<typename Head, typename ... Tail>
auto sum(Head h, Tail ... args) {
static_assert(Addable<Head, Tail...>::value, "Requires addable objects");
return h+(args+...);
}
You might change you assertion to
template <typename Head, typename ... Tail>
auto sum(Head h, Tail ... args) {
static_assert((Addable<Head, Tail>::value || ...), "Requires addable objects");
return (h + ... + args);
}
as your traits only works for 2 types.
Notice that SFINAE looks more like concept than static_assert.
Your code does work as long as you only provide two arguments to sum.
The problem is that when you call something like sum(1, 2, 3), Addable<Head, Tail...> becomes Addable<int, int, int>, which doesn't match your Addable<T, U, void> partial specialization. If you call sum with more than three arguments then there's no Addable template that matches at all, since Addable only accepts three template parameters.
If you want to be able to deal with more than two arguments, you'll need a different approach. This is what I came up with. It uses a similar approach to what you used, just with an extra layer indirection to support more than two arguments:
template <typename T, typename U, typename = void>
struct Helper : std::false_type {};
template <typename T, typename U>
struct Helper<T, U, std::void_t<decltype(std::declval<T>() + std::declval<U>())>> : std::true_type {};
template <typename, typename...>
struct Addable : std::false_type {};
template <typename T, typename U, typename... V>
struct Addable<T, U, V...> : std::conditional_t<Addable<T, U>::value,
Addable<U, V...>,
std::false_type> {};
template <typename T, typename U>
struct Addable<T, U> : Helper<T, U> {};
template<typename Head, typename ... Tail>
auto sum(Head h, Tail ... args) {
static_assert(Addable<Head, Tail...>::value, "Requires addable objects");
return h+(args+...);
}
Live Demo

A variadic template method to accept a given number of doubles?

template <unsigned int N> class myclass
{
public:
template <typename... Args> void mymethod(Args... args)
{
// Do interesting stuff
}
};
I want mymethod to be called only with exactly N doubles. Is that possible? That is, say that I have:
myclass <3> x;
x.mymethod(3., 4., 5.); // This works
x.mymethod('q', 1., 7.); // This doesn't work
x.mymethod(1., 2.); // This doesn't work
How can I get this done?
For the number of arguments constraint you can easily check if sizeof...(Args) == N but for checking if all the arguments are doubles you need to build a recursive type trait that checks std::is_same for each of the arguments.
template<typename...>
struct are_same : std::true_type
{};
template<typename T>
struct are_same<T> : std::true_type
{};
template<typename T, typename U, typename... Types>
struct are_same<T, U, Types...> :
std::integral_constant<bool, (std::is_same<T, U>::value && are_same<T, Types...>::value)>
{};
Notice are_same is first declared and then specialized.
Then just implement the constraint in your method return type using std::enable_if by taking advantage of SFINAE.
template <unsigned int N> class myclass
{
public:
template <typename... Args>
typename std::enable_if<(are_same<double, Args...>::value && sizeof...(Args) == N), void>::type
/* void */ mymethod(Args... args)
{
// Do interesting stuff
}
};
Can try something like following :
#include <type_traits>
template<class T, class...>
struct all_same : std::true_type
{};
template<class T, class U, class... TT>
struct all_same<T, U, TT...>
: std::integral_constant<bool, std::is_same<T,U>{} && all_same<T, TT...>{}>
{};
template <unsigned int N> class myclass
{
public:
template <typename... Args>
typename std::enable_if<sizeof...(Args) == N, void >::type mymethod(Args... args)
{
static_assert(all_same<double, Args...>{},
"Not all args as Double");
}
};
<Demo>

Do static_assert on every other type in a template

How can I do static_asserts (or other checks) on every other type in a template?
template<typename... Ts> //T1,T2,T3,...
struct foo {
//How can I
//for T1,T3,T5,T7,...
//do some checks, for example:
//static_assert(std::is_default_constructible<Tn>::value,"invalid type");
//static_assert(std::is_copy_constructible<Tn>::value,"invalid type");
};
Please try this:
#include <type_traits>
template <typename... Ts>
struct default_constructible;
template <typename T>
struct default_constructible<T>
{
static constexpr bool value = std::is_default_constructible<T>::value;
};
template <>
struct default_constructible<>
{
static constexpr bool value = true;
};
template <typename T, typename U, typename... Ts>
struct default_constructible<T, U, Ts...>
{
static constexpr bool value = std::is_default_constructible<T>::value && default_constructible<Ts...>::value;
};
template <typename... Ts>
struct foo
{
static_assert(default_constructible<Ts...>::value, "");
};
class A { A() = delete; };
template class foo<int, bool>; // Compiles
template class foo<int, bool, A>; // Does not compile
Demo
You could try this:
template <template <typename...> class Pred, typename ...Args> struct check_odd;
template <template <typename...> class Pred>
struct check_odd<Pred> : std::true_type { };
template <template <typename...> class Pred, typename T>
struct check_odd<Pred, T> : Pred<T> { };
template <template <typename...> class Pred, typename T1, typename T2, typename ...Args>
struct check_odd<Pred, T1, T2, Args...>
{
static constexpr bool value = Pred<T1>::value && check_odd<Pred, Args...>::value;
};
Usage:
static_assert(check_odds<std::is_arithmetic, T1, T2, T3, T4, T5, T6>::value,
"Not all odd types are arithmetic.");
Obvious routes of generalization are parametrizing the combiner (currently hardcoded as "AND", or &&), and giving the predicate additional parameters.

Avoiding creating a variable in this template code

I have a template:
template<typename... Ts> //T1,T2,T3,...
struct foo {
//my struct
};
I want to do static_assert checks on T1,T3,T5,... (the "odd types") and on T2,T4,T6,... (the "even types") separately.
I have found this simple solution:
template<size_t N, typename... Ts>
struct perform_checks {};
template<size_t N, typename T, typename U, typename... Ts>
struct perform_checks<N, T, U, Ts...> : perform_checks<N, Ts...>
{
//check for odd types
static_assert(std::is_default_constructible<T>::value,"failure");
//check for even types
static_assert(std::is_copy_constructible<U>::value,"failure");
};
The N parameter allows it to end. I use it like this:
template<typename... Ts>
struct foo {
perform_checks<0,Ts...> hello;
};
This seems to be working fine. But is it possible to avoid the hello variable? I never use it for any other purpose.
Derive foo from perform_checks<> privately:
template <typename... Ts> struct foo : private perform_checks<Ts...> {
// stuff
};
Oh, and get rid of the N parameter that you don't need:
template <typename... Ts> struct perform_checks {};
template <typename T> struct perform_checks<T> {
template <typename U> struct dependent_name_hack : std::false_type {};
static_assert(dependent_name_hack<T>::value,
"Odd number of parameters not acceptable.");
};
template <typename T, typename U, typename... Ts>
struct perform_checks<T, U, Ts...> : perform_checks<Ts...> {
//check for odd types
static_assert(std::is_default_constructible<T>::value,"failure");
//check for even types
static_assert(std::is_copy_constructible<U>::value,"failure");
};
You can use enable_if1 (and boost::mpl) in more-or-less the following way:
#include <boost/mpl/and.hpp>
template<size_t N, typename... Ts>
struct perform_checks {};
template<size_t N, typename T, typename U, typename... Ts>
struct perform_checks<N, T, U, Ts...> : perform_checks<N, Ts...>
{
typedef boost::mpl::and_<std::is_default_constructible<T>::type,
std::is_copy_constructible<U>::type> type;
};
template < class... Ts,
class = typename std::enable_if<perform_checks<0, Ts...>::type>
struct foo {
//my struct
};
The only purpose of foo in the OP is triggering the check when it's instantiated. That's why you need the variable hello: it's an instantiation of foo.
I would rather follow the approach of traits in <type_traits>. More precisely, I would turn perform_checks into class (or struct) that has public static constexpt bool member called value which is true or false depending on whether the given types pass the test or not. Then I would use a single static_assert to stop compilation if value is false.
My solution, which assumes that the number of template type arguments is even, follows:
#include <type_traits>
template<typename First, typename Second, typename... Others>
struct perform_checks :
std::integral_constant<bool,
perform_checks<First, Second>::value && // Checks First and Second
perform_checks<Others...>::value // Recursively "calls" itself on Others
> {
};
// This specialization finishes the recursion and effectively performs the test
template<typename First, typename Second>
struct perform_checks<First, Second> :
std::integral_constant<bool,
std::is_default_constructible<First>::value && // Checks First
std::is_copy_constructible<Second>::value // Checks Second
> {
};
Here is a simple test:
struct NonDefaultConstructible {
NonDefaultConstructible() = delete;
};
struct NonCopyConstructible {
NonCopyConstructible(const NonCopyConstructible&) = delete;
};
int main() {
static_assert(perform_checks<int, double>::value, "Failure");
static_assert(perform_checks<int, int, double, double>::value, "Failure");
static_assert(!perform_checks<NonDefaultConstructible, int>::value, "Failure");
static_assert(!perform_checks<int, NonCopyConstructible>::value, "Failure");
static_assert(!perform_checks<int, int, double, NonCopyConstructible>::value, "Failure");
}
Notice that no variable was created.

Find out whether a C++ object is callable

Is it possible to write a type trait, say is_callable<T> which tells if an object has an operator() defined?
It is easy if the arguments to the call operator are known in advance, but not in the general case.
I want the trait to return true if and only if there is at least one overloaded call operator defined.
This question is related and has a good answer, but it doesn't work on all types (only on int-convertible types). Also, std::is_function works, but only on proper C++ functions, not on functors. I'm looking for a more general solution.
I think this trait does what you want. It detects operator() with any kind of signature even if it's overloaded and also if it's templatized:
template<typename T>
struct is_callable {
private:
typedef char(&yes)[1];
typedef char(&no)[2];
struct Fallback { void operator()(); };
struct Derived : T, Fallback { };
template<typename U, U> struct Check;
template<typename>
static yes test(...);
template<typename C>
static no test(Check<void (Fallback::*)(), &C::operator()>*);
public:
static const bool value = sizeof(test<Derived>(0)) == sizeof(yes);
};
The principle is based on Member Detector idiom. As it is, it will fail to compile if you pass it a non-class type, but that shouldn't be hard to fix, I just left it out for brevity. You can also extend it to report true for functions.
Of course it doesn't give you any info about the signature(s) of operator() whatsoever, but I believe that's not what you asked for, right?
EDIT for Klaim:
It's simple enough to make it work (return false) with non-class types. If you rename the above class to is_callable_impl, you can write this, for example:
template<typename T>
struct is_callable
: std::conditional<
std::is_class<T>::value,
is_callable_impl<T>,
std::false_type
>::type
{ };
The answers here were helpful but I came here wanting something that could also spot whether something was callable regardless of whether it happened to be an object or a classic function. jrok's answer to this aspect of the problem, alas, didn't work because std::conditional actually evaluates the types of both arms!
So, here's a solution:
// Note that std::is_function says that pointers to functions
// and references to functions aren't functions, so we'll make our
// own is_function_t that pulls off any pointer/reference first.
template<typename T>
using remove_ref_t = typename std::remove_reference<T>::type;
template<typename T>
using remove_refptr_t = typename std::remove_pointer<remove_ref_t<T>>::type;
template<typename T>
using is_function_t = typename std::is_function<remove_refptr_t<T>>::type;
// We can't use std::conditional because it (apparently) must determine
// the types of both arms of the condition, so we do it directly.
// Non-objects are callable only if they are functions.
template<bool isObject, typename T>
struct is_callable_impl : public is_function_t<T> {};
// Objects are callable if they have an operator(). We use a method check
// to find out.
template<typename T>
struct is_callable_impl<true, T> {
private:
struct Fallback { void operator()(); };
struct Derived : T, Fallback { };
template<typename U, U> struct Check;
template<typename>
static std::true_type test(...);
template<typename C>
static std::false_type test(Check<void (Fallback::*)(), &C::operator()>*);
public:
typedef decltype(test<Derived>(nullptr)) type;
};
// Now we have our final version of is_callable_t. Again, we have to take
// care with references because std::is_class says "No" if we give it a
// reference to a class.
template<typename T>
using is_callable_t =
typename is_callable_impl<std::is_class<remove_ref_t<T>>::value,
remove_ref_t<T> >::type;
But in the end, for my application, I really wanted to just know whether you could say f() (i.e., call it with no arguments), so I instead went with something much simpler.
template <typename T>
constexpr bool noarg_callable_impl(
typename std::enable_if<bool(sizeof((std::declval<T>()(),0)))>::type*)
{
return true;
}
template<typename T>
constexpr bool noarg_callable_impl(...)
{
return false;
}
template<typename T>
constexpr bool is_noarg_callable()
{
return noarg_callable_impl<T>(nullptr);
}
In fact, I went even further. I knew the function was supposed to return an int, so rather than just check that I could call it, I checked the return type, too, by changing the enable_if to:
typename std::enable_if<std::is_convertible<decltype(std::declval<T>()()),
int>::value>::type*)
Hope this helps someone!
Here is a possible solution using C++11 that works without requiring to know the signature of the call operator for functors, but only as long the functor does not have more than one overload of operator ():
#include <type_traits>
template<typename T, typename = void>
struct is_callable : std::is_function<T> { };
template<typename T>
struct is_callable<T, typename std::enable_if<
std::is_same<decltype(void(&T::operator())), void>::value
>::type> : std::true_type { };
This is how you would use it:
struct C
{
void operator () () { }
};
struct NC { };
struct D
{
void operator () () { }
void operator () (int) { }
};
int main()
{
static_assert(is_callable<C>::value, "Error");
static_assert(is_callable<void()>::value, "Error");
auto l = [] () { };
static_assert(is_callable<decltype(l)>::value, "Error");
// Fires! (no operator())
static_assert(is_callable<NC>::value, "Error");
// Fires! (several overloads of operator ())
static_assert(is_callable<D>::value, "Error");
}
C++17 brings std::is_invocable and friends.
This answer also given a solution on how to emulate it with C++14.
There are several other answers already, of course, and they are useful, but none of them seem to cover every use case AFAICT. Borrowing from those answers and this possible implementation of std::is_function, I created a version that covers every possible use case of which I could think. It's kind of lengthy, but very feature complete (Demo).
template<typename T, typename U = void>
struct is_callable
{
static bool const constexpr value = std::conditional_t<
std::is_class<std::remove_reference_t<T>>::value,
is_callable<std::remove_reference_t<T>, int>, std::false_type>::value;
};
template<typename T, typename U, typename ...Args>
struct is_callable<T(Args...), U> : std::true_type {};
template<typename T, typename U, typename ...Args>
struct is_callable<T(*)(Args...), U> : std::true_type {};
template<typename T, typename U, typename ...Args>
struct is_callable<T(&)(Args...), U> : std::true_type {};
template<typename T, typename U, typename ...Args>
struct is_callable<T(Args......), U> : std::true_type {};
template<typename T, typename U, typename ...Args>
struct is_callable<T(*)(Args......), U> : std::true_type {};
template<typename T, typename U, typename ...Args>
struct is_callable<T(&)(Args......), U> : std::true_type {};
template<typename T, typename U, typename ...Args>
struct is_callable<T(Args...)const, U> : std::true_type {};
template<typename T, typename U, typename ...Args>
struct is_callable<T(Args...)volatile, U> : std::true_type {};
template<typename T, typename U, typename ...Args>
struct is_callable<T(Args...)const volatile, U> : std::true_type {};
template<typename T, typename U, typename ...Args>
struct is_callable<T(Args......)const, U> : std::true_type {};
template<typename T, typename U, typename ...Args>
struct is_callable<T(Args......)volatile, U> : std::true_type{};
template<typename T, typename U, typename ...Args>
struct is_callable<T(Args......)const volatile, U> : std::true_type {};
template<typename T, typename U, typename ...Args>
struct is_callable<T(Args...)&, U> : std::true_type {};
template<typename T, typename U, typename ...Args>
struct is_callable<T(Args...)const&, U> : std::true_type{};
template<typename T, typename U, typename ...Args>
struct is_callable<T(Args...)volatile&, U> : std::true_type{};
template<typename T, typename U, typename ...Args>
struct is_callable<T(Args...)const volatile&, U> : std::true_type{};
template<typename T, typename U, typename ...Args>
struct is_callable<T(Args......)&, U> : std::true_type {};
template<typename T, typename U, typename ...Args>
struct is_callable<T(Args......)const&, U> : std::true_type{};
template<typename T, typename U, typename ...Args>
struct is_callable<T(Args......)volatile&, U> : std::true_type{};
template<typename T, typename U, typename ...Args>
struct is_callable<T(Args......)const volatile&, U> : std::true_type{};
template<typename T, typename U, typename ...Args>
struct is_callable<T(Args...)&&, U> : std::true_type{};
template<typename T, typename U, typename ...Args>
struct is_callable<T(Args...)const&&, U> : std::true_type{};
template<typename T, typename U, typename ...Args>
struct is_callable<T(Args...)volatile&&, U> : std::true_type{};
template<typename T, typename U, typename ...Args>
struct is_callable<T(Args...)const volatile&&, U> : std::true_type{};
template<typename T, typename U, typename ...Args>
struct is_callable<T(Args......)&&, U> : std::true_type{};
template<typename T, typename U, typename ...Args>
struct is_callable<T(Args......)const&&, U> : std::true_type{};
template<typename T, typename U, typename ...Args>
struct is_callable<T(Args......)volatile&&, U> : std::true_type{};
template<typename T, typename U, typename ...Args>
struct is_callable<T(Args......)const volatile&&, U> : std::true_type{};
template<typename T>
struct is_callable<T, int>
{
private:
using YesType = char(&)[1];
using NoType = char(&)[2];
struct Fallback { void operator()(); };
struct Derived : T, Fallback {};
template<typename U, U>
struct Check;
template<typename>
static YesType Test(...);
template<typename C>
static NoType Test(Check<void (Fallback::*)(), &C::operator()>*);
public:
static bool const constexpr value = sizeof(Test<Derived>(0)) == sizeof(YesType);
};
This works correctly with non-class types (returns false, of course), function types (<T()>), function pointer types, function reference types, functor class types, bind expressions, lambda types, etc. This works correctly even if the class constructor is private and/or non-defaulted, and even if operator() is overloaded. This returns false for member function pointers by design because they are not callable, but you can use bind to create a callable expression.
Note: These assume that the default constructor is valid for the type your checking. Not sure offhand how to get around that.
The following seems to work if it's callable with 0 arguments. Is there something in is_function's implementation that might help to extend this to 1 or more argument callables?:
template <typename T>
struct is_callable {
// Types "yes" and "no" are guaranteed to have different sizes,
// specifically sizeof(yes) == 1 and sizeof(no) == 2.
typedef char yes[1];
typedef char no[2];
template <typename C>
static yes& test(decltype(C()())*);
template <typename>
static no& test(...);
// If the "sizeof" the result of calling test<T>(0) would be equal to the sizeof(yes),
// the first overload worked and T has a nested type named foobar.
static const bool value = sizeof(test<T>(0)) == sizeof(yes);
};
If you know the type of the argument (even if it's a template parameter), the following would work for 1 argument, and I imagine one could extend pretty easily from there:
template <typename T, typename T2>
struct is_callable_1 {
// Types "yes" and "no" are guaranteed to have different sizes,
// specifically sizeof(yes) == 1 and sizeof(no) == 2.
typedef char yes[1];
typedef char no[2];
template <typename C>
static yes& test(decltype(C()(T2()))*);
template <typename, typename>
static no& test(...);
// If the "sizeof" the result of calling test<T>(0) would be equal to the sizeof(yes),
// the first overload worked and T has a nested type named foobar.
static const bool value = sizeof(test<T>(0)) == sizeof(yes);
};
Edit
here is a modification that handles the case where default constructor isn't available.
This is a neat and short trick for finding if T is callable. It goes along the lines originally proposed by Walter E. Brown at CPPCON'14 in his talk on modern template metaprogramming.
template <class... >
using void_t = void;
template <class T>
using has_opr_t = decltype(&T::operator());
template <class T, class = void>
struct is_callable : std::false_type { };
template <class T>
struct is_callable<T, void_t<has_opr_t<typename std::decay<T>::type>>> : std::true_type { };
Here is another implementation.
It makes use of std::is_function template for free functions.
For classes, it uses something similar to the Member Detector Idiom. If T has a call operator, callable_2 will contain more than one operator(). This will cause the first can_call function to be discarded (through SFINAE) due to ambiguity failure in decltype(&callable_2<T>::operator()) and the second can_call function will return true. If T does not have a call operator, the first can_call function will be used (due to overload resolution rules).
namespace impl
{
struct callable_1 { void operator()(); };
template<typename T> struct callable_2 : T, callable_1 { };
template<typename T>
static constexpr bool can_call(decltype(&callable_2<T>::operator())*) noexcept { return false; }
template<typename>
static constexpr bool can_call(...) noexcept { return true; }
template<bool is_class, typename T>
struct is_callable : public std::is_function<T> { };
template<typename T> struct is_callable<false, T*> : public is_callable<false, T> { };
template<typename T> struct is_callable<false, T* const> : public is_callable<false, T> { };
template<typename T> struct is_callable<false, T* volatile> : public is_callable<false, T> { };
template<typename T> struct is_callable<false, T* const volatile> : public is_callable<false, T> { };
template<typename T>
struct is_callable<true, T> : public std::integral_constant<bool, can_call<T>(0)> { };
}
template<typename T>
using is_callable = impl::is_callable<std::is_class<std::remove_reference_t<T>>::value,
std::remove_reference_t<T>>;