std:: is_compound

std::is_compound is a UnaryTypeTrait .

If T is a compound type (that is, array, function, object pointer, function pointer, member object pointer, member function pointer, reference, class, union, or enumeration, including any cv-qualified variants), provides the member constant value equal true . For any other type, value is false .

If the program adds specializations for std::is_compound or std::is_compound_v , the behavior is undefined.

Template parameters

Helper variable template

Notes

Compound types are the types that are constructed from fundamental types. Any C++ type is either fundamental or compound.

Possible implementation

template<class T>
struct is_compound : std::integral_constant<bool, !std::is_fundamental<T>::value> {};

Example

#include <type_traits>
#include <iostream>
 
static_assert(not std::is_compound_v<int>);
static_assert(std::is_compound_v<int*>);
static_assert(std::is_compound_v<int&>);
 
void f();
static_assert(std::is_compound_v<decltype(f)>);
static_assert(std::is_compound_v<decltype(&f)>);
 
static_assert(std::is_compound_v<char[100]>);
 
class C {};
static_assert(std::is_compound_v<C>);
 
union U {};
static_assert(std::is_compound_v<U>);
 
enum struct E { e };
static_assert(std::is_compound_v<E>);
static_assert(std::is_compound_v<decltype(E::e)>);
 
struct S
{
    int i : 8;
    int j;
    void foo();
};
static_assert(not std::is_compound_v<decltype(S::i)>);
static_assert(not std::is_compound_v<decltype(S::j)>);
static_assert(std::is_compound_v<decltype(&S::j)>);
static_assert(std::is_compound_v<decltype(&S::foo)>);
 
int main()
{
    std::cout << "All checks have passed\n";
}

See also