C++ auto_ptr and copy construction

Question

If I have a class

template <typename T>
struct C {
...

private:
  auto_ptr<T> ptr;
};

How do I define the copy constructor for C:

It cannot be

template <typename T>
C<T>::C(const C& other) 

because I want to if I copy the auto_ptr from other, I have changed other by removing ownership. Is it legal to define a copy constructor as

template <typename T>
C<T>::C(C& other) {}

Answer 1

如果要防止所有权转让和/或复制，可以定义复制构造函数和赋值运算符private ，从而禁止类的用户复制或分配对象。

Answer 2

Do you actually want to copy your class's state or transfer it? If you want to copy it then you do it just like any other class with pointers in it:

template < typename T >
C<T>::C(C const& other) : ptr(new T(*other.ptr)) {} // or maybe other.ptr->clone()

If you actually want to transfer ownership of the pointer you could do it with a non-const "copy" constructor but I'd recommend you do something that's more obvious at the call site; something that tells people reading the code that ownership has transfered.

Answer 3

There's no such thing as a "standard" copy constructor, but people do expect them to behave a certain way. If your object is going to do any transferring of ownership on copy, then it's a bad idea to make a copy constructor that obscures this fact. A better approach would be to create a method that makes this clear. (You could call it CloneAndTransfer or something similar.)