std::is_nothrow_assignable

Defined in header <type_traits>
template< class T, class U > struct is_assignable;	(1)	(since C++11)
template< class T, class U > struct is_trivially_assignable;	(2)	(since C++11)
template< class T, class U > struct is_nothrow_assignable;	(3)	(since C++11)

1%29如果表达式std::declval<T>()=std::declval<U>()在未计算的上下文中格式良好，则提供成员常量。value平等true.否则，value是false...存取检查执行时，就好像是从与任何类型无关的上下文中执行的。

2%29与%281%29相同，但赋值表达式的计算不会调用任何不简单的操作。为了这张支票的目的，打电话给std::declval被认为是琐碎的，而不是ODR-使用成std::declval...

3%29与%281%29相同，但赋值表达式的计算不会调用除“否”以外的任何操作。

T和U每一个都是一个完整的类型，%28可能是cv-合格%29void，或者一系列未知的界限。否则，行为就没有定义。

辅助变量模板

template< class T, class U > inline constexpr bool is_assignable_v = is_assignable<T, U>::value;		(since C++17)
template< class T, class U > inline constexpr bool is_trivially_assignable_v = is_trivially_assignable<T, U>::value;		(since C++17)
template< class T, class U > inline constexpr bool is_nothrow_assignable_v = is_nothrow_assignable<T, U>::value;		(since C++17)

继承自STD：积分[医]常量

成员常数

value static	true if T is assignable from U , false otherwise (public static member constant)

成员函数

operator bool	converts the object to bool, returns value (public member function)
operator() (C++14)	returns value (public member function)

成员类型

Type	Definition
value_type	bool
type	std::integral_constant<bool, value>

注记

此特性不检查赋值表达式的直接上下文之外的任何内容：如果使用T或U将触发模板的专门化，生成隐式定义的特殊成员函数等，并且这些函数都有错误，即使在以下情况下，实际分配也可能不会编译。std::is_assignable<T,U>::value编译并计算为true...

例

二次

#include <iostream>
#include <string>
#include <type_traits>
struct Ex1 { int n; };
int main() {
    std::cout << std::boolalpha
              << "int is assignable from int? "
              << std::is_assignable<int, int>::value << '\n' // 1 = 1; wouldn't compile
              << "int& is assignable from int? "
              << std::is_assignable<int&, int>::value << '\n' // int a; a = 1; works
              << "int is assignable from double? "
              << std::is_assignable<int, double>::value << '\n'
              << "int& is nothrow assignable from double? "
              << std::is_nothrow_assignable<int&, double>::value << '\n'
              << "string is assignable from double? "
              << std::is_assignable<std::string, double>::value << '\n'
              << "Ex1& is trivially assignable from const Ex1&? "
              << std::is_trivially_assignable<Ex1&, const Ex1&>::value << '\n';
}

二次

产出：

二次

int is assignable from int? false
int& is assignable from int? true
int is assignable from double? false
int& is nothrow assignable from double? true
string is assignable from double? true
Ex1& is trivially assignable from const Ex1&? true

二次

另见

is_copy_assignableis_trivially_copy_assignableis_nothrow_copy_assignable (C++11)(C++11)(C++11)	checks if a type has a copy assignment operator (class template)
is_move_assignableis_trivially_move_assignableis_nothrow_move_assignable (C++11)(C++11)(C++11)	checks if a type has a move assignment operator (class template)

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最后更新于：2017-12-18