std::alignment_of
From cppreference.com
Defined in header <type_traits>
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template< class T > struct alignment_of; |
(since C++11) | |
Provides the member constant value
equal to the alignment requirement of the type T
, as if obtained by an alignof expression. If T
is an array type, returns the alignment requirements of the element type. If T
is a reference type, returns the alignment requirements of the type referred to.
If alignof(T) is not a valid expression, the behavior is undefined.
Helper variable template
template< class T > inline constexpr std::size_t alignment_of_v = alignment_of<T>::value; |
(since C++17) | |
Inherited from std::integral_constant
Member constants
value [static] |
alignof(T) (public static member constant) |
Member functions
operator std::size_t |
converts the object to std::size_t, returns value (public member function) |
operator() (C++14) |
returns value (public member function) |
Member types
Type | Definition |
value_type
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std::size_t
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type
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std::integral_constant<std::size_t, value> |
Possible implementation
template< class T > struct alignment_of : std::integral_constant< std::size_t, alignof(T) > {}; |
Notes
This type trait predates the alignof keyword, which can be used to obtain the same value with less verbosity.
Example
Run this code
#include <iostream> #include <type_traits> class A {}; int main() { std::cout << std::alignment_of<A>::value << '\n'; std::cout << std::alignment_of<int>() << '\n'; // alt syntax std::cout << std::alignment_of_v<double> << '\n'; // c++17 alt syntax }
Output:
1 4 8
See also
alignof operator | queries alignment requirements of a type (since C++11) |
(C++11) |
defines the type suitable for use as uninitialized storage for types of given size (class template) |
(C++11) |
defines the type suitable for use as uninitialized storage for all given types (class template) |
(C++11) |
POD type with alignment requirement as great as any other scalar type (typedef) |