C++ 在 C 中打印指针

我试图用指针理解一些东西，所以我写了以下代码:

#include <stdio.h>

int main(void)
{
    char s[] = "asd";
    char **p = &s;

    printf("The value of s is: %p\n", s);
    printf("The direction of s is: %p\n", &s);

    printf("The value of p is: %p\n", p);
    printf("The direction of p is: %p\n", &p);

    printf("The direction of s[0] is: %p\n", &s[0]);
    printf("The direction of s[1] is: %p\n", &s[1]);
    printf("The direction of s[2] is: %p\n", &s[2]);

    return 0;
}

使用gcc编译时，我会收到以下警告:

$ gcc main.c -o main-bin -ansi -pedantic -Wall -lm
main.c: In function ‘main’:
main.c:6: warning: initialization from incompatible pointer type
main.c:9: warning: format ‘%p’ expects type ‘void *’, but argument 2 has type ‘char (*)[4]’
main.c:11: warning: format ‘%p’ expects type ‘void *’, but argument 2 has type ‘char **’
main.c:12: warning: format ‘%p’ expects type ‘void *’, but argument 2 has type ‘char ***’

(gcc的标志是因为我必须是C89)

为什么指针类型不兼容？数组的名称不是指向它的第一个元素的指针吗？所以如果s是指向a的指针，&s一定是char **，不是吗？

run 时我会得到这样的结果:

$ ./main-bin
The value of s is: 0xbfb7c860
The direction of s is: 0xbfb7c860
The value of p is: 0xbfb7c860
The direction of p is: 0xbfb7c85c
The direction of s[0] is: 0xbfb7c860
The direction of s[1] is: 0xbfb7c861
The direction of s[2] is: 0xbfb7c862

s的值和它的方向(当然还有p的值)怎么会相同呢？