C++ 为什么在我的类中添加析构函数会使我的类无法 move ？

Question

编译器提醒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;

如果我删除析构函数，我的代码可以正常编译。 https://ideone.com/UFB18P

Answer 1

因为 that's what the standard says .当您开始添加特殊成员函数时，您会禁止自动生成其他一些成员函数。这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属于同一类规则。

添加这个:

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