is_assignable< T, U > Struct Template Reference

If the expression etl::declval<T>() = etl::declval<U>() is well-formed in unevaluated context, provides the member constant value equal true. Otherwise, value is false. Access checks are performed as if from a context unrelated to either type. More...

#include <is_assignable.hpp>

Inheritance diagram for is_assignable< T, U >:

Additional Inherited Members
Public Types inherited from integral_constant< bool, B >
using	type
using	type
using	value_type
using	value_type
Public Member Functions inherited from integral_constant< bool, B >
constexpr	operator value_type () const noexcept
constexpr	operator value_type () const noexcept
constexpr auto	operator() () const noexcept -> value_type
constexpr auto	operator() () const noexcept -> value_type
Static Public Attributes inherited from integral_constant< bool, B >
static constexpr bool	value
static constexpr bool	value

Detailed Description

template<typename T, typename U>
struct etl::is_assignable< T, U >

If the expression etl::declval<T>() = etl::declval<U>() is well-formed in unevaluated context, provides the member constant value equal true. Otherwise, value is false. Access checks are performed as if from a context unrelated to either type.

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The documentation for this struct was generated from the following file:

• include/etl/_type_traits/is_assignable.hpp