C++ 指针增量

Question

Please help me understand how the value of q is initially 1 and later changes to 0.

Code:

#include <stdio.h>
int main()
{    
    int i=10;
    int *p, *q;
    p = &i;
    printf("%u\n%u\n%u\n",p,q,*q);
    *q++=*p++;
    printf("%u\n%u\n%u\n%u",p,q,*p,*q);
    return 0;
}

Output:

1527896876
1527897120
1
1527896880
1527897124
1527897124
0

Answer 1

int *p, *q; p = &i; printf("%u\n%u\n%u\n",p,q,*q);

int* is a wrong type for the format specifier %u. If you pass an argument of wrong type, then the behaviour of your program is undefined.

---

 printf("%u\n%u\n%u\n",p,q,*q); *q++=*p++;

Here, q has an indeterminate value. It doesn't point to any object. You indirect through the pointer to access an object that it points to. Which is a contradiction because it doesn't point to any object.

Indirecting through an indeterminate pointer results in undefined behaviour.

---

The behaviour of your program is undefined. That explains everything about the behaviour of the program.

Answer 2

I think what happens is that here

*q++=*p++;

the post-increment operator of q is executed after the assignment. So, you basically set the value of q to p, and then you increment it. Also, i don't think the increment operator of p does anything.

For a better understanding look at the following example

#include <stdio.h>
int main() {
  int i = 10;
  int *p;
  int *q = new int[2];
  q[0] = 9;
  q[1] = 99;
  p = &i;
  printf("%u\n", *q);
  *q++ = *p++;
  printf("%u\n", *q);
  printf("%u", *--q);
  return 0;
}

9
99
10

Notice how, in the end q points to the next element, thanks to the post-increment operator.

Anyway, i have no idea what you are trying to do.