当派生定义了析构函数时，使用复制 ctor 而不是移动 ctor

Question

当我向派生类添加析构函数时，当它尝试使用复制 ctor 而不是定义的移动 ctor 时出现编译器错误（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());
}

这段代码在 gcc-4.5 和 gcc-4.6 上编译得很好。 gcc-4.7 的错误是：

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>&)’

我在这里遗漏了什么还是应该在 gcc 4.7 上也能正常编译？当我在 Derived 类上注释掉析构函数时，一切都很好。

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) 

Answer 1

错误是正确的；只有当类没有用户定义的析构函数时，才会隐式生成移动构造函数。如果添加析构函数，那么您将禁止默认移动构造函数，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被抑制，就像你的那样，所以Derived 既不可复制也不可移动。

解决方案是简单地将移动构造函数添加到Derived。