我想请教关于二维数组的指针地址的问题。

Question

我理解行指针和单个元素的指针是分层关系就是小区和住户的关系。a+1就是下一个小区的地址。*（a）+1就是这个小区下一个住户的地址。可是计算机内部是线性内存啊，（地址应该是简单的线性增加吧？）怎么实现这种区分的呢？像进制一样前面加符号？

Answer 1

比如a[3][3]={"a1","b2","c3"}

按顺序 a[0][0],a[0][1],a[0][2],a[1][0],a[1][1],a[1][2],........a[2][2]

其中 a ,a+1,a+2表示 3个连续地址的地址常量 补充！（数组名不=指针，数组名是常量指针，不能改变其值）

分别指向3个连续地址的首地址，分别&a[0][0]-&a[0][2]首地址；&a[1][0]-&a[1][2]首地址；&a[2][0]-&a[2][2]首地址

所以*(a)[0]='a'    *(a+1)[0]='b'    *(a+2)[0]='c'

补充！！！说得多不如上图

#include<stdio.h>
int main()
{
    char a[3][3]={"a1","b2","c3"};
    printf("       a=0x%x\n",(unsigned)a);
    printf("    *(a)=0x%x\n",(unsigned)*(a));
    printf("     a+1=0x%x\n",(unsigned)(a+1));
    printf("  *(a+1)=0x%x\n",(unsigned)*(a+1));
    printf("     a+2=0x%x\n",(unsigned)(a+2));
    printf("  *(a+2)=0x%x\n",(unsigned)*(a+2));

    printf("&a[0][0]=0x%x\n",(unsigned)(&a[0][0]));
    printf("&a[0][1]=0x%x\n",(unsigned)(&a[0][1]));
    printf("&a[0][2]=0x%x\n",(unsigned)(&a[0][2]));
    printf("&a[1][0]=0x%x\n",(unsigned)(&a[1][0]));
    printf("&a[1][1]=0x%x\n",(unsigned)(&a[1][1]));
    printf("&a[1][2]=0x%x\n",(unsigned)(&a[1][2]));
    printf("&a[2][0]=0x%x\n",(unsigned)(&a[2][0]));
    printf("&a[2][1]=0x%x\n",(unsigned)(&a[2][1]));
    printf("&a[2][2]=0x%x\n",(unsigned)(&a[2][2]));

    return 0;
}