std::is_pod
From cppreference.com
Defined in header <type_traits>
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template< class T > struct is_pod; |
(since C++11) | |
If T
is a PODType
("plain old data type"), that is, both trivial and standard-layout, provides the member constant value
equal true. For any other type, value
is false.
The behavior is undefined if std::remove_all_extents_t<T> is an incomplete type and not (possibly cv-qualified) void.
Template parameters
T | - | a type to check |
Helper variable template
template< class T > inline constexpr bool is_pod_v = is_pod<T>::value; |
(since C++17) | |
Inherited from std::integral_constant
Member constants
value [static] |
true if T is a POD 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
Objects of POD types are fully compatible with the C programming language.
Example
Run this code
#include <iostream> #include <type_traits> struct A { int m; }; struct B { int m1; private: int m2; }; struct C { virtual void foo(); }; int main() { std::cout << std::boolalpha; std::cout << std::is_pod<A>::value << '\n'; std::cout << std::is_pod<B>::value << '\n'; std::cout << std::is_pod<C>::value << '\n'; }
Output:
true false false
See also
(C++11) |
checks if a type is standard-layout type (class template) |
(C++11) |
checks if a type is trivial (class template) |