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# std::is_compound

From cppreference.com
< 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.

## 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)

### Member types

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

### [edit] Notes

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

### [edit] Possible implementation

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

### [edit] Example

Run this code

    #include <iostream>
    #include <type_traits>

    int main() {
        class cls {};
        std::cout << (std::is_compound<cls>::value
                         ? "T is compound"
                         : "T is not a compound") << '\n';
        std::cout << (std::is_compound<int>::value
                         ? "T is compound"
                         : "T is not a compound") << '\n';
    }

Output:

    T is compound
    T is not a compound

### [edit] See also

is_fundamental (C++11) | checks if a type is fundamental type (class template) [edit]
is_scalar (C++11) | checks if a type is scalar type (class template) [edit]
is_object (C++11) | checks if a type is object type (class template) [edit]
is_array (C++11) | checks if a type is an array type (class template) [edit]

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