C++为什么在我的类中添加析构函数会使我的类无法移动？

Question

编译器提醒我,我正在使用已删除的函数. https://ideone.com/3YAIlA

#include <memory>
using namespace std;
class foo
{
public:
    unique_ptr<int> p;
    ~foo()
    {

    }
};
int main()
{
    foo a, b;
    a = move(b);
    return 0;
}

编译信息

prog.cpp: In function 'int main()':
prog.cpp:15:4: error: use of deleted function 'foo& foo::operator=(const foo&)'
  a = move(b);


prog.cpp:3:7: note: 'foo& foo::operator=(const foo&)' is implicitly deleted because the default definition would be ill-formed:
 class foo


prog.cpp:3:7: error: use of deleted function 'std::unique_ptr<_Tp, _Dp>& std::unique_ptr<_Tp, _Dp>::operator=(const std::unique_ptr<_Tp, _Dp>&) [with _Tp = int; _Dp = std::default_delete<int>]'

In file included from /usr/include/c++/5/memory:81:0,
                 from prog.cpp:1:


/usr/include/c++/5/bits/unique_ptr.h:357:19: note: declared here
       unique_ptr& operator=(const unique_ptr&) = delete;

如果我删除了析构函数,我的代码编译得很好. https://ideone.com/UFB18P

Answer 1

因为这就是标准所说的.当您开始添加特殊成员函数时,您将禁止自动生成其他一些函数.与编写非默认构造函数意味着不会自动为您生成默认构造函数的规则属于同一类规则.

添加这个:

foo& operator=(foo&&) = default;