Pointer arithmetic calculation

Question

#include <iostream>

using namespace std;

class temp {
    int a = 10;
};

class derived : temp {
    int b = 20;
};

int main()
{
    derived der;
    void * p = &der;
    cout<<"First element in der is : "<<*(int*)p<<endl;
    cout<<"(int*)p :"<<(int*)p<<endl;
    p += sizeof(int); //Way -1
    //p = (int*)p + sizeof(int); //Way-2
    cout<<"Second element in der is : "<<*(int*)p<<endl;
    cout<<"(int*)p :"<<(int*)p<<endl;

    return 0;
}

If I execute the above code, I get below output :

First element in der is : 10
(int*)p :0x7ffe9f05a928
Second element in der is : 20
(int*)p :0x7ffe9f05a92c

Where as if I comment the line of code in (Way-1) and un-comment line for (Way-2) I get below output :

First element in der is : 10
(int*)p :0x7ffc10e6de18
Second element in der is : 0
(int*)p :0x7ffc10e6de28

What is going wrong in second attempt ? Why pointer arithmetic results are different here ?

That's undefined behavior. The C++ standard doesn't specify what happens when you cast a pointer to an unrelated type and dereference it. The position of the elements in a class is also not specified. Your compiler can add padding. — Thomas Sablik
– Thomas Sablik, Commented May 8, 2020 at 14:13

cigien · Accepted Answer · 2020-05-08 18:41:29Z

1

When you do this:

void * p = &der;
int i = *(int*)p;

you are (indirectly) casting a derived pointer to an int pointer. This invokes undefined behaviour, and the program could print anything. It can even print different results when you run it a second time.

answered May 8, 2020 at 18:41

cigien

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