int array[8]; &array 的类型是 int (*)[8]

deference_pointer.c

#include <stdio.h>

int main(void)
{
    int array[8] = {10,11,12,13,14,15,16,17};
    int *p = array;
    int *q = &array; //提示不兼容的类型，这里其实是转换了指针类型
    int (*z)[8] = &array;

    printf("the size of array:     %ld, the size of &array:         %ld\n", sizeof(array), sizeof(&array));
    printf("the value of array[1]: %d, the value of (&array)[0][1]: %d\n", array[1], (&array)[0][1]);
    printf("the address of pointer array:  %p, the address of pointer array+1: %p\n", (void*)array, (void*)(array+1));
    printf("the address of pointer &array: %p, the address of pointer &array+1: %p\n", (void*)&array, (void*)(&array+1));

    printf("the size of p:     %ld, the size of q:         %ld\n", sizeof(p), sizeof(q));
    //printf("the value of p[1]: %d, the value of q[0][1]: %d\n", p[1], q[0][1]);
    printf("the address of pointer p:  %p, the address of pointer p+1: %p\n", (void*)p, (void*)(p+1));
    printf("the address of pointer q: %p, the address of pointer    q: %p\n", (void*)q, (void*)(q+1));

    return 0;
}

/*
示例输出:
the size of array:     32, the size of &array:         8
the value of array[1]: 11, the value of (&array)[0][1]: 11
the address of pointer array:  0x7ffdaff18c90, the address of pointer array+1: 0x7ffdaff18c94
the address of pointer &array: 0x7ffdaff18c90, the address of pointer &array+1: 0x7ffdaff18cb0
the size of p:     8, the size of q:         8
the address of pointer p:  0x7ffdaff18c90, the address of pointer p+1: 0x7ffdaff18c94
the address of pointer q: 0x7ffdaff18c90, the address of pointer    q: 0x7ffdaff18c94
*/