当 T 包含一个 const 数据成员时，为什么要删除 std::optional<T>::operator=？

Question

以下代码会导致编译错误：

#include <optional>

class A {
};
class B {
private:
    const A a;
};

int main()
{
    B b;
    std::optional<B> bo1;
    bo1 = b;
}

例如，在 gcc 上，错误显示为：

main.cpp: In function 'int main()':
main.cpp:12:7: error: uninitialized const member in 'class B'
12 |     B b;
   |       ^
main.cpp:7:13: note: 'const A B::a' should be initialized
7 |     const A a;
  |             ^
main.cpp:14:11: error: use of deleted function 'std::optional<B>& std::optional<B>::operator=(std::optional<B>&&)'
14 |     bo1 = b;
   |           ^
In file included from main.cpp:1:
/lib/gcc-head/include/c++/12.0.0/optional:663:11: note: 'std::optional<B>& std::optional<B>::operator=(std::optional<B>&&)' is implicitly deleted because the default definition would be ill-formed:
663 |     class optional
    |           ^~~~~~~~
/lib/gcc-head/include/c++/12.0.0/optional:663:11: error: use of deleted function 'std::_Enable_copy_move<true, false, true, false, _Tag>& std::_Enable_copy_move<true, false, true, false, _Tag>::operator=(std::_Enable_copy_move<true, false, true, false, _Tag>&&) [with _Tag = std::optional<B>]'
In file included from /lib/gcc-head/include/c++/12.0.0/optional:43,
             from main.cpp:1:
/lib/gcc-head/include/c++/12.0.0/bits/enable_special_members.h:248:5: note: declared here
248 |     operator=(_Enable_copy_move&&) noexcept                         = delete;
    |     ^~~~~~~~

在 MSVC 上，作为另一个示例，错误显示为：

main.cpp
<source>(14): error C2280: 'std::optional<B> &std::optional<B>::operator =(const std::optional<B> &)': attempting to reference a deleted function
C:/data/msvc/14.29.29917-Pre/include\optional(445): note: compiler has generated 'std::optional<B>::operator =' here
C:/data/msvc/14.29.29917-Pre/include\optional(445): note: 'std::optional<B> &std::optional<B>::operator =(const std::optional<B> &)': function was implicitly deleted because a base class invokes a deleted or inaccessible function 'std::_Deleted_move_assign<std::_Optional_construct_base<_Ty>,_Ty> &std::_Deleted_move_assign<std::_Optional_construct_base<_Ty>,_Ty>::operator =(const std::_Deleted_move_assign<std::_Optional_construct_base<_Ty>,_Ty> &)'
        with
        [
            _Ty=B
        ]
C:/data/msvc/14.29.29917-Pre/include\xsmf_control.h(131): note: 'std::_Deleted_move_assign<std::_Optional_construct_base<_Ty>,_Ty> &std::_Deleted_move_assign<std::_Optional_construct_base<_Ty>,_Ty>::operator =(const std::_Deleted_move_assign<std::_Optional_construct_base<_Ty>,_Ty> &)': function was implicitly deleted because a base class invokes a deleted or inaccessible function 'std::_Deleted_copy_assign<std::_Optional_construct_base<_Ty>,_Ty> &std::_Deleted_copy_assign<std::_Optional_construct_base<_Ty>,_Ty>::operator =(const std::_Deleted_copy_assign<std::_Optional_construct_base<_Ty>,_Ty> &)'
        with
        [
            _Ty=B
        ]
C:/data/msvc/14.29.29917-Pre/include\xsmf_control.h(92): note: 'std::_Deleted_copy_assign<std::_Optional_construct_base<_Ty>,_Ty> &std::_Deleted_copy_assign<std::_Optional_construct_base<_Ty>,_Ty>::operator =(const std::_Deleted_copy_assign<std::_Optional_construct_base<_Ty>,_Ty> &)': function was explicitly deleted
        with
        [
            _Ty=B
        ]
Compiler returned: 2

只要我们去掉const A a;中的const关键字，错误就消失了。根据 referenced standard on std::optional，根据 std::is_...able_v 测试，有些情况下 = 运算符的某些重载将被删除。在这种情况下，为什么 const 关键字会影响其中一些测试？

Answer 1

const 数据成员在构造对象后不能修改。因此，正常工作 operator= 是不可能的。

Answer 2

optional 使用对象的 =.

无法分配给具有 const 数据成员的 class。只能建造。

试试这个：

B b0;    
B b1;
b0=b1;

optional 不起作用，因为 B 不起作用。

此外，尝试 static_assert(!std::is_copy_assignable_v<B>);，它通过了。

现在，std::optional这里有个“后门”。

B b;
std::optional<B> bo1;
bo1.emplace(b);

构造而不是赋值。

从理论上讲，可选的可以依赖于此；但如果内容存在，则需要销毁它然后放置一个新对象。这是有风险的，会扰乱语义，而且是一个默默地做的糟糕计划。

当 T 包含一个 const 数据成员时，为什么要删除 std::optional<T>::operator=？

Why is `std::optional<T>::operator=` deleted when T contains a `const` data member?

c++

assignment-operator

c++17

stdoptional