This Q is an extension of:
Templated check for the existence of a class member function?
Is there any utility which will help to find:
With std::experimental::is_detected and std::experimental::disjunction you could do this:
//check for a type member named foo
template <typename T>
using foo_type_t = typename T::foo;
//check for a non-type member named foo
template <typename T>
using foo_non_type_t = decltype(&T::foo);
template <typename T>
using has_foo = disjunction<is_detected<foo_type_t, T>,
is_detected<foo_non_type_t, T>>;
Then you would use has_foo<my_class>::value in whatever you want.
The above will work for more than just types and member functions, but you could easily constrain it by using traits like std::is_member_function_pointer and std::is_member_object_pointer if you like.
To supply your optional argument, you could use the std::experimental::is_detected_exact helper.
Live Demo
Note that if you take the implementations of the above traits from the pages I linked, you can use this with C++14. A minor change to the disjunction code will let you use it in C++11.
#define HasMember(NAME) \
template<class Class, typename Type = void> \
struct HasMember_##NAME \
{ \
typedef char (&yes)[2]; \
template<unsigned long> struct exists; \
template<typename V> static yes Check (exists<sizeof(static_cast<Type>(&V::NAME))>*); \
template<typename> static char Check (...); \
static const bool value = (sizeof(Check<Class>(0)) == sizeof(yes)); \
}; \
template<class Class> \
struct HasMember_##NAME<Class, void> \
{ \
typedef char (&yes)[2]; \
template<unsigned long> struct exists; \
template<typename V> static yes Check (exists<sizeof(&V::NAME)>*); \
template<typename> static char Check (...); \
static const bool value = (sizeof(Check<Class>(0)) == sizeof(yes)); \
}
Usage: Simply invoke the macro with whatever member you want to find:
HasMember(Foo); // Creates a SFINAE `class HasMember_Foo`
HasMember(i); // Creates a SFINAE `class HasMember_i`
Now we can utilize HasMember_X to check X in ANY class as below:
#include<iostream>
struct S
{
void Foo () const {}
// void Foo () {} // If uncommented then type should be mentioned in `HasMember_Foo`
int i;
};
int main ()
{
std::cout << HasMember_Foo<S, void (S::*) () const>::value << "\n";
std::cout << HasMember_Foo<S>::value << "\n";
std::cout << HasMember_i<S, int (S::*)>::value << "\n";
std::cout << HasMember_i<S>::value << "\n";
}
Catches:
class
must not have overloaded methods. If it has then this trick fails.
i.e. even though the named member is present more than once, the result will be false.B is base of S & void B::Bar () is present, then HasMember_Bar<S, void (B::*)()>::value or HasMember_Bar<S, void (S::*)()>::value or HasMember_Bar<S>::value will give false
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