我正在尝试编写一个有条件地禁用四个特殊成员函数(复制构造、移动构造、复制赋值和移动赋值)的包装类，下面是我用于测试目的的快速草稿:

enum class special_member : uint8_t {
    copy_ctor, move_ctor,
    copy_asgn, move_asgn
};

template
struct _disabled_wrapper {
công cộng:
    constexpr _disabled_wrapper(MemberType type) : _type(type) {}

công cộng:
    constexpr MemberType& unwrapped() { return _type; }
    constexpr const MemberType& unwrapped() const { return _type; }

riêng tư:
    MemberType _type;
};

template
struct _disabled_wrapper 
{
riêng tư:
    using _parent_t = _disabled_wrapper;

công cộng:
    constexpr _disabled_wrapper(const _parent_t& parent) : _parent(parent) {}

    constexpr _disabled_wrapper(const _disabled_wrapper&) = delete;
    constexpr _disabled_wrapper(_disabled_wrapper&&) = default;
    constexpr _disabled_wrapper& operator=(const _disabled_wrapper&) = default;
    constexpr _disabled_wrapper& operator=(_disabled_wrapper&&) = default;

công cộng:
    constexpr MemberType& unwrapped() { return _parent.unwrapped(); }
    constexpr const MemberType& unwrapped() const { return _parent.unwrapped(); }

riêng tư:
    _parent_t _parent;
};

template
struct _disabled_wrapper
{
public: //Todo: Make private post fix.
    using _parent_t = _disabled_wrapper;

công cộng:
    constexpr _disabled_wrapper(const _parent_t& parent) : _parent(parent) {}

    constexpr _disabled_wrapper(const _disabled_wrapper&) = default;
    constexpr _disabled_wrapper(_disabled_wrapper&&) = delete;
    constexpr _disabled_wrapper& operator=(const _disabled_wrapper&) = default;
    constexpr _disabled_wrapper& operator=(_disabled_wrapper&&) = default;

công cộng:
    constexpr MemberType& unwrapped() { return _parent.unwrapped(); }
    constexpr const MemberType& unwrapped() const { return _parent.unwrapped(); }

riêng tư:
    _parent_t _parent;
};

template
struct _disabled_wrapper
{
riêng tư: 
    using _parent_t = _disabled_wrapper;

công cộng:
    constexpr _disabled_wrapper(const _parent_t& parent) : _parent(parent) {}

    constexpr _disabled_wrapper(const _disabled_wrapper&) = default;
    constexpr _disabled_wrapper(_disabled_wrapper&&) = default;
    constexpr _disabled_wrapper& operator=(const _disabled_wrapper&) = delete;
    constexpr _disabled_wrapper& operator=(_disabled_wrapper&&) = default;

công cộng:
    constexpr MemberType& unwrapped() { return _parent.unwrapped(); }
    constexpr const MemberType& unwrapped() const { return _parent.unwrapped(); }

riêng tư:
    _parent_t _parent;
};

template
struct _disabled_wrapper
{
riêng tư:
    using _parent_t = _disabled_wrapper;

công cộng:
    constexpr _disabled_wrapper(const _parent_t& parent) : _parent(parent) {}

    constexpr _disabled_wrapper(const _disabled_wrapper&) = default;
    constexpr _disabled_wrapper(_disabled_wrapper&&) = default;
    constexpr _disabled_wrapper& operator=(const _disabled_wrapper&) = default;
    constexpr _disabled_wrapper& operator=(_disabled_wrapper&&) = delete;

công cộng:
    constexpr MemberType& unwrapped() { return _parent.unwrapped(); }
    constexpr const MemberType& unwrapped() const { return _parent.unwrapped(); }

riêng tư:
    _parent_t _parent;
};

除了三种情况( , Và )，以上代码运行正常。具体来说，下面代码中的断言失败了，这让我相信由于某种原因默认的移动赋值运算符没有被删除，在 move_ctor ở giữa. _disabled_wrapper 的部分特化, 即使它的非静态成员 _parent已删除其移动分配。 cppreference.com 指出:

The implicitly-declared or defaulted move assignment operator for class T is defined asdeleted if any of the following is true:

• T has a non-static data member that is const;
• T has a non-static data member of a reference type;
• T has a non-static data member that cannot be move-assigned (has deleted, inaccessible, or ambiguous move assignment operator);
• T has direct or virtual base class that cannot be move-assigned (has deleted, inaccessible, or ambiguous move assignment operator).

A deleted implicitly-declared move assignment operator is ignored by overload resolution.

这让我相信，根据标准，确实应该删除移动赋值运算符(要点三)。我缺少什么，或者，如果可能的话，我如何在不手动输入所有可能的特化的情况下使包装类按预期工作？谢谢。

int chính() {
    using type = _disabled_wrapper;

    //Passes, hence member variable has its move_assignment deleted as intended.
    static_assert(std::is_move_assignable_v == false); 
    //Fails here! Move assignment defined?
    static_assert(std::is_move_assignable_v == false); 

    trả về 0;
}

1 Câu trả lời

您引用中的最后一句话“重载决策忽略了已删除的隐式声明的移动赋值运算符。”更准确地说:

A defaulted move assignment operator that is defined as deleted is ignored by overload resolution.

class.copy.assign

kiểu 有一个默认的移动赋值运算符，它被定义为已删除，因为它在基类中被删除。

因此，重载决策会忽略此运算符。

因此，从右值赋值选择复制赋值。

Vì vậy,kiểu 是可移动赋值的，因为它是可复制赋值的。

关于c++ - 预期 move_assignment 被删除，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/72645574/