Question

My questions are marked in code comments in the snippet below.

int *foo(){
    int a = 10;
    int b = 20;
    return &b;
}
int *foo2(){
    int aa = 44;
    int bb = 40;
    return &bb;
}

int main(int argc, char* argv[])
{
    int *p = foo();
    int *p2 = foo2();

    int  a = 700;
    int *b = & a;

//  cout<<*b<<endl; /* what & why the difference from adding & removing this line?? */
    cout<<a<<endl;
    cout<<*p<<endl; /* what & why happened to "p" */

    return 0;
}

Answer 1

In foo() and foo2() you are returning pointers to local variables. These locals have automatic storage and it is undefined behavior to use a pointer to them once they go out of scope. Therefor p and p2 contain nothing useful. It might work, it might not.

I can’t understand your question very well, so I hope I got it all.