std::is_compound

< cpp‎ | types
Defined in header <type_traits>
template< class T >
struct is_compound;
 (since C++11)

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.

The behavior of a program that adds specializations for is_compound or is_compound_v (since C++17) is undefined.

Template parameters

T - a type to check

Helper variable template

template< class T >
inline constexpr bool is_compound_v = is_compound<T>::value;
 (since C++17)

Inherited from std::integral_constant

Member constants

value
[static]
 true if T is a compound type , false otherwise
(public static member constant)

Member functions

operator bool
 converts the object to bool, returns value
(public member function)
operator()
(C++14)
 returns value
(public member function)

Member types

Type Definition
value_type bool
type std::integral_constant<bool, value>

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 <iostream>
#include <type_traits>
 
int main() {
    class cls {};
    std::cout << (std::is_compound<cls>::value
                     ? "`cls` is compound"
                     : "`cls` is not a compound") << '\n';
    std::cout << (std::is_compound_v<int>
                     ? "`int` is compound"
                     : "`int` is not a compound") << '\n';
}

Output:

`cls` is compound
`int` is not a compound

See also

(C++11)
 checks if a type is a fundamental type
(class template)
(C++11)
 checks if a type is a scalar type
(class template)
(C++11)
 checks if a type is an object type
(class template)
(C++11)
 checks if a type is an array type
(class template)