std::is_standard_layout

template< class T >
struct is_standard_layout;

If T is a standard-layout type, provides the member constant value equal to 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.

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

Template parameters

Helper variable template

template< class T >
inline constexpr bool is_standard_layout_v = is_standard_layout<T>::value;

Notes

A pointer to a standard-layout class may be converted (with reinterpret_cast) to a pointer to its first non-static data member and vice versa.

If a standard-layout union holds two or more standard-layout structs, it is permitted to inspect the common initial part of them.

The macro offsetof is only guaranteed to be usable with standard-layout classes.

Example

#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_standard_layout<A>::value << '\n';
    std::cout << std::is_standard_layout<B>::value << '\n';
    std::cout << std::is_standard_layout<C>::value << '\n';
}

true
false
false

See also