当派生函数定义了析构函数时，使用复制函数而不是移动函数

Using copy ctor instead of move ctor when derived has destructor defined

本文关键字：函数 复制 移动 定义 派生 析构函数      更新时间：2023-10-16

当我向派生类添加析构函数时，当它试图使用复制函数而不是定义的移动函数时，我得到编译器错误(使用gcc 4.7):

#include <utility>
#include <iostream>
template <typename T>
struct Base 
{
  T value;
  Base(T&& value) :
    value(value)
  {
    std::cout << "base ctor" << std::endl;
  }
  Base& operator=(const Base&) = delete;
  Base(const Base&) = delete;
  Base& operator=(Base&& rhs)
  {
    value = rhs.value;
    std::cout << "move assignment" << std::endl;
  }
  Base(Base&& other) :
    value(other.value)
  {
    std::cout << "move ctor" << std::endl;
  }
  virtual ~Base()
  {
    std::cout << "base dtor" << std::endl;
  }
};
template <typename T>
struct Derived : public Base<T>
{
  Derived(T&& value) :
    Base<T>(std::forward<T>(value))
  {
    std::cout << "derived ctor" << std::endl;
  }
  ~Derived()
  {
    std::cout << "derived dtor" << std::endl;
  }
};
template <typename T>
Derived<T> MakeDerived(T&& value)
{
  return Derived<T>(std::forward<T>(value));
}
struct Dummy {};
int main()
{
  auto test = MakeDerived(Dummy());
}

此代码在gcc-4.5和gcc-4.6上可以很好地编译。来自gcc-4.7的错误如下:

test.cpp: In function ‘int main()’:
test.cpp:61:34: error: use of deleted function ‘Derived<Dummy>::Derived(const Derived<Dummy>&)’
test.cpp:37:8: note: ‘Derived<Dummy>::Derived(const Derived<Dummy>&)’ is implicitly deleted because the default definition would be ill-formed:
test.cpp:37:8: error: use of deleted function ‘Base<T>::Base(const Base<T>&) [with T = Dummy; Base<T> = Base<Dummy>]’
test.cpp:16:3: error: declared here
test.cpp: In instantiation of ‘Derived<T> MakeDerived(T&&) [with T = Dummy]’:
test.cpp:61:34:   required from here
test.cpp:54:43: error: use of deleted function ‘Derived<Dummy>::Derived(const Derived<Dummy>&)’

我在这里错过了什么，或者这也应该在gcc 4.7上编译好吗?当我注释掉派生类的析构函数时，一切都好了。

gcc version 4.5.3 (Ubuntu/Linaro 4.5.3-12ubuntu2) 
gcc version 4.6.3 (Ubuntu/Linaro 4.6.3-1ubuntu5) 
gcc version 4.7.2 (Ubuntu/Linaro 4.7.2-11precise2)

错误是正确的;Move构造函数仅在类没有用户定义析构函数时隐式生成。如果添加析构函数，则会抑制默认的move-构造函数，并尝试复制构造函数。当一个类具有不可复制的基类时，默认复制构造函数的生成将被禁止，因此Derived既不可复制也不可移动。

解决方案是简单地添加一个移动构造函数到Derived。

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