当派生 class 在基 class 中被删除时，它是否会拥有隐式复制构造函数或赋值运算符？

Question

Qt defines Q_DISABLE_COPY如下：

#define Q_DISABLE_COPY(Class) \
    Class(const Class &) = delete;\
    Class &operator=(const Class &) = delete;

Q_DISABLE_COPY 是 used in the QObject class, but the documentation for it 说它也应该在它的所有子class 中使用：

when you create your own subclass of QObject (director or indirect), you should not give it a copy constructor or an assignment operator. However, it may not enough to simply omit them from your class, because, if you mistakenly write some code that requires a copy constructor or an assignment operator (it's easy to do), your compiler will thoughtfully create it for you. You must do more.

但是考虑这个程序：

struct Base {
    Base() = default;

private:
    Base(const Base &) = delete;
    Base &operator=(const Base &) = delete;
};

struct Derived : Base {};

int main() {
    Derived d1;
    Derived d2(d1); // error: call to implicitly-deleted copy constructor of 'Derived'
    Derived d3;
    d3 = d1; // error: object of type 'Derived' cannot be assigned because its copy assignment operator is implicitly deleted
}

尝试编译该程序时出现的错误似乎表明编译器不会在派生的 classes 中创建复制构造函数或赋值运算符，而它们在基 classes 中被删除。 Qt 的文档在这方面是错误的，还是在创建它们时存在一些极端情况？

相关但不重复：Repeating Q_DISABLE_COPY in QObject derived classes。它给出了为什么在 class 中使用 Q_DISABLE_COPY 可能有用的原因，即使它无论如何都不可复制，但不能确认没有它它实际上永远不会被复制。

Answer 1

由于基 class 复制构造函数被删除，派生 class 无法知道如何复制基 class 对象。这将禁用编译器提供的任何隐式复制构造函数。

来自 cppreference:

The implicitly-declared or defaulted copy constructor for class T is defined as deleted if any of the following conditions are true:

T has direct or virtual base class that cannot be copied (has deleted, inaccessible, or ambiguous copy constructors)

T has direct or virtual base class with a deleted or inaccessible destructor;

当用户从删除默认复制构造函数的 class 继承时，继承 Q_DISABLE_COPY 会很有用，但会提供默认实现来覆盖它。

struct Base {
    Base() = default;

private:
    Base(const Base &) = delete;
    Base &operator=(const Base &) = delete;
};

struct Derived : Base {
    Derived() = default;
    Derived(const Derived&) : Derived() {}
    Derived &operator=(const Derived&) {
        return *this;
    }
};

struct MoreDerived : Derived {};


int main() {
    Derived d1;
    Derived d2(d1); // Works fine!
    Derived d3;
    d3 = d1; // Works fine!
    MoreDerived md1;
    MoreDerived md2(md1); // Works fine!
    MoreDerived md3;
    md3 = md1; // Works fine!!
}

编辑：正如@SR_ 正确指出的那样，在 Derived 的上述实现中，Base 没有被复制构造。我只是想说明一个事实，即在继承层次结构中修改另一个 class 时很容易引入无意的复制构造函数。

Answer 2

在提交 a2b38f6, QT_DISABLE_COPY was instead defined like this (credit to Swift - Friday Pie 之前在评论中指出这一点）：

#define Q_DISABLE_COPY(Class) \
    Class(const Class &) Q_DECL_EQ_DELETE;\
    Class &operator=(const Class &) Q_DECL_EQ_DELETE;

和Q_DECL_EQ_DELETE like this:

#ifdef Q_COMPILER_DELETE_MEMBERS
# define Q_DECL_EQ_DELETE = delete
#else
# define Q_DECL_EQ_DELETE
#endif

Q_COMPILER_DELETE_MEMBERS 如果 C++11 支持（或至少有足够新的草案支持 = delete）可用，则得到定义。

因此，如果您当时针对 C++03 编译器编译 Qt，它会编译如下内容：

struct Base {
    Base() {};

private:
    Base(const Base &);
    Base &operator=(const Base &);
};

struct Derived : Base {};

int main() {
    Derived d1;
    Derived d2(d1);
    Derived d3;
    d3 = d1;
}

用 g++ -std=c++03 编译会出现这些错误：

<source>: In copy constructor 'Derived::Derived(const Derived&)':
<source>:9:8: error: 'Base::Base(const Base&)' is private within this context
    9 | struct Derived : Base {};
      |        ^~~~~~~
<source>:5:5: note: declared private here
    5 |     Base(const Base &);
      |     ^~~~
<source>: In function 'int main()':
<source>:13:18: note: synthesized method 'Derived::Derived(const Derived&)' first required here
   13 |     Derived d2(d1);
      |                  ^
<source>: In member function 'Derived& Derived::operator=(const Derived&)':
<source>:9:8: error: 'Base& Base::operator=(const Base&)' is private within this context
    9 | struct Derived : Base {};
      |        ^~~~~~~
<source>:6:11: note: declared private here
    6 |     Base &operator=(const Base &);
      |           ^~~~~~~~
<source>: In function 'int main()':
<source>:15:10: note: synthesized method 'Derived& Derived::operator=(const Derived&)' first required here
   15 |     d3 = d1;
      |          ^~

那时，“您的编译器会为您精心创建它”在技术上是正确的，但实际上并非如此，因为创建它的编译器会导致编译失败，只是出现不同的（并且可能不太清楚）错误。我现在确信，现在 = delete 被无条件使用，这根本不再是真的，所以我打算请 Qt 的维护者 remove/reword 他们文档的那一部分。

当派生 class 在基 class 中被删除时，它是否会拥有隐式复制构造函数或赋值运算符？

Would a derived class ever have an implicit copy constructor or assignment operator when it's deleted in the base class?

c++

derived-class

copy-constructor

copy-assignment

deleted-functions