c - 指针算术 : warning: assignment makes pointer from integer without a cast [enabled by default]

Question

我在学习指针算术时发现了以下代码:

#include <stdio.h>
int main()
{
    int *p, *q, *r, a, b;
    p = &a;
    q = &b;
    p = p-q;             
    r = &a;
    r = (int*)(r-q);   
    printf("p = %p\n",p);
    printf("r = %p\n",r);
}

当我编译代码时，我收到以下警告:

test.c:7:11: warning: assignment makes pointer from integer without a cast [enabled by   default]
         p = q-p;             
           ^

现在，当我运行代码时，我得到以下输出:

p = 0x1
r = 0x1

由于输出相同，任何人都可以解释一下警告的重要性。预先感谢您。

Answer 1

从地址中减去地址，不会返回地址。它返回整数，即这些地址之间的距离。

N1570 - 6.5.6p9:

When two pointers are subtracted, both shall point to elements of the same array object, or one past the last element of the array object; the result is the difference of the subscripts of the two array elements. The size of the result is implementation-defined, and its type (a signed integer type) is ptrdiff_t defined in the stddef.h header. ...