std::is_compound
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) |
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