C ++ copy control problem

#include <iostream>
#include <string>
#include <vector>
using namespace std;

class HasPtr{
 public:
    //constructor accepts a string
    HasPtr(const string &s = string()) : ps(new string(s)), i(0), use(new size_t(1)) {}  
    //copy constructor
    HasPtr(const HasPtr &h) : ps(h.ps), i(h.i), use(h.use) { ++*use; }
    //copy assignment operator 
    HasPtr &operator=(const HasPtr &h)
    {
        ++*h.use;
        if(--*use == 0)
        {
            delete ps;
            delete use;
        }
        ps = h.ps;
        i = h.i;
        use = h.use;

        return *this;
    }
    ~HasPtr()
    {
        if(--*use == 0)
        {
            delete ps;
            delete use;
        }
    }

//private:
    string *ps;
    int i;

    size_t *use;
};


int main()
{
    HasPtr h("hi mom");
    HasPtr h2 = h;
    HasPtr h3("hi dad");
    h2 = h3;

    cout << "h: " << *h.ps << endl;
    cout << "h2: " << *h2.ps << endl;
    cout << "h3: " << *h3.ps << endl;
}

      

Output:

h: hi mom
h2: hi dad
h3: hi dad

      

I think the output should be:

h: hi dad
h2: hi dad
h3: hi dad

      

The reason why I think the output should look like above:

h is "hi mom", h2 is h1 share, so h2 is "hi mom", h3 is "hi dad" and I think I h2 = h3

changed h1 either because h2 is divisible by h1, but it is not.

What am I doing wrong?

+3


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2 answers


To make it clearer, consider a simplified example

int x = 10;
int y = 20;

int *h = &x;
int *h2 = h;

      

Now h

they h2

point to x

. You can check this output of the object pointed to by pointers. for example

std::cout << "*h = " << *h << ", *h2 = " << *h2 << std::endl;

      

Then you use the third pointer

int *h3 = &y;
h2 = h3;

      

Now h2

they h3

point to y

. h

has not been changed. You can also check this

std::cout << "*h2 = " << *h2 << ", *h3 = " << *h3 << std::endl;
std::cout << "*h = " << *h << std::endl;

      



The same happens with your objects in statements

HasPtr h("hi mom");
HasPtr h2 = h;
HasPtr h3("hi dad");
h2 = h3;

      

only the relevant pointers (ps) are wrapped in class objects.

Here is a program that demonstrates the explanation

#include <iostream>
#include <string>


int main()
{
    std::string hi_mom( "hi mom" );
    std::string hi_dad( "hi dad" );

    std::string *h  = &hi_mom;
    std::string *h2 = h;

    std::cout << "*h = " << *h << ", *h2 = " << *h2 << std::endl;

    std::string *h3 = &hi_dad;
    h2 = h3;

    std::cout << "*h2 = " << *h2 << ", *h3 = " << *h3 << std::endl;
    std::cout << "*h = " << *h << std::endl;
}        

      

Its conclusion

*h = hi mom, *h2 = hi mom
*h2 = hi dad, *h3 = hi dad
*h = hi mom

      

+2


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Here's what's going on:

HasPtr h("hi mom");
HasPtr h2 = h;

h.ps ---> "hi mom"  <--- h2.ps

      

Then with the following statement, you change h2

, but it doesn't affect h

:

HasPtr h3("hi dad");
h2 = h3;

h2.ps ----> "hi dad" <--- h3.ps
 |
XXX changed
 |
"hi mom" <--- h.ps

      



Pointers are different instances pointing to the same address. Example:

int a = 5, c = 1; 
int *p1 = &a, *p2 = &a;

      

Change p1

(i.e. p1 = &c

) will not affect p2

.

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