Question

Could anyone explain me the reason why is in this code defined MyClass ( const MyClass & src ) { m_X = src . m_X; } ? Without this line of code it works fine as well and gives the same output which is 32 in this case. What's the difference between having it in and not? I read that copy constructor is made automatically but when you have defined pointer in the class you should define it but i don't get why in this case.

Code below:

#include <iostream>
using namespace std;

class MyClass
{
public:
    MyClass ( int x ) : m_X ( new int ( x ) ) {}
    MyClass ( const MyClass &src ) 
    {
        m_X = src.m_X; 
    }

    void  print ( void ) const 
    { 
        cout << *m_X; 
    }

private:
    int * m_X;
};


int main ( void )
{
    MyClass a ( 32 ), c = a;
    c.print ();

    return 0;
}

Answer 1

默认的复制构造函数以及复制构造函数执行成员明智的复制（或浅复制）。在这种情况下，您有一个指针，其值是数据所在的地址。这解决了复制的内容。

正如其他帖子所述，您需要更好地管理指针。

Answer 2

编译器生成一个（和你的）的问题是它复制指针。您现在有两个指针c.m_x和a.m_x都指向同一个堆分配的对象。谁拥有它？如果你删除a会发生什么，它应该删除它分配的内存（你错误地不这样做），但是c仍然在蠢蠢欲动。

出于这个原因，有std::shared_ptr。它专为这种情况而设计，它执行Right Thing By Magic

将int * m_X替换为std::shared_ptr<int> m_X

Answer 3

The reason you get the same output is you are doing a shallow copy just like the compiler would do. It is not enough just to copy a pointer. You need to create a new pointer and then set its value to be a copy of what you are copying from. A proper constructor would look like

MyClass ( const MyClass & src ) : m_X(new int(*src.m_X) {}

Now the copy has its own independent pointer that has the same value.

Also note that you need a destructor to delete the memory and a proper copy assignment operator as the default copy assignment operator will do the same thing as the default copy constructor.

Copy constructor confusion

3 个答案: