I'm learning C++ and I have a question about the tutorial on pointers

[Aruisa3725]

// more pointers
#include <iostream>
using namespace std;

int main ()
{
int firstvalue = 5, secondvalue = 15;
int * p1, * p2;

p1 = &firstvalue; // p1 = address of firstvalue
p2 = &secondvalue; // p2 = address of secondvalue
*p1 = 10; // value pointed by p1 = 10
*p2 = *p1; // value pointed by p2 = value pointed by p1
p1 = p2; // p1 = p2 (value of pointer is copied)
*p1 = 20; // value pointed by p1 = 20

cout << "firstvalue is " << firstvalue << endl;
cout << "secondvalue is " << secondvalue << endl;
return 0;
}

I was wondering if since p1 = p2 and then *p1 = 20 if the code is retroactive and the *p1 = 20 makes p2 = 20. Thank you for your help

[oghmaosiris]

Try it and see.

put a cout in there and see if it changes.

[magnificence7]

I want to correct a few comments:

*p2 = *p1; // value pointed by p2 = value pointed by p1
to
*p2 = *p1; // value pointed by p2 now contains the value pointed by p1

p1 = p2; // p1 = p2 (value of pointer is copied)
to
p1 = p2; // p1 points to the same lcoation as p2.

I was wondering if since p1 = p2 and then *p1 = 20 if the code is retroactive and the *p1 = 20 makes p2 = 20.

Yes, after those two lines, they both point to the same value, which after that is changed to 20, so they both point to the same value, 20.

[Aruisa3725]

Okay, that solves my question! Thank you very much for your help :D