std::is_compound
From cppreference.com
Defined in header <type_traits>
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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.
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
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bool
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type
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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
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
See also
(C++11) |
checks if a type is fundamental type (class template) |
(C++11) |
checks if a type is scalar type (class template) |
(C++11) |
checks if a type is object type (class template) |
(C++11) |
checks if a type is an array type (class template) |