char** C 中的数组越界行为

Question

有人可以解释一下，对于二维数组，当您超出索引范围但在初始化内存内时，C 中会发生什么吗？

例如:

    char t[3][3] = {{1,2,3},{4,5,6},{7,8,9}};

    for (int i = 0; i < 6; i++)
    {   
        printf("%i\n", t[0][i]); //prints 1 thru 6 on separate lines
    }

按预期工作，但是

//gcc 5.4.0

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main(void)
{
    char* s[] = {".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."};

/*
    printf("%i\n", s); 
    printf("%i\n",(s+1));
    printf("%i\n",(s+2));
    printf("%i\n", (s+3));
    printf("%i\n", (s+4));
*/

    for (int i = 0; i < 20; i++)
    {   
        printf("%c\n", s[0][i]);
    }

    return 0;

/* Output:
.
-

.
.
.
-

-
.
.
-

-
.
-
-

-
-
*/

显示每个“字符串”分配了 8 个字节的连续内存，但字符在数组中出现时并不连续打印(摩尔斯电码) - 前 2 个字符来自 s[0]，但最后一个字符来自 s[0]是从数组的末尾开始的。

Answer 1

s 不是二维数组。您没有将字符串存储在s中 - 您将每个字符串的第一个字符的地址存储在s<中 (在您的计算机上，char * 的宽度为 8 个字节)。字符串本身在内存中不必相邻，这样，如果您走过一个字符串的末尾，就会进入另一个字符串的开头。

IOW，你想象的 s 看起来像这样:

         +---+
s[0][0]: |'.'|
         +---+
s[0][1]: |'-'|
         +---+
s[0][2]: | 0 |
         +---+
s[1][0]: |'-'|
         +---+
s[1][1]: |'.'|
         +---+
s[1][2]: |'.'|
         +---+
          ...

事实并非如此。你实际上拥有的是这样的:

      +---+         +---+---+---+
s[0]: |   | ------> |'.'|'-'| 0 | <-------------+
      +---+         +---+---+---+               |
s[1]: |   | ---+                                |
      +---+    |    +---+---+---+---+---+       |
s[2]: |   | -+ +--> |'-'|'.'|'.'|'.'| 0 | <-----+---- not guaranteed to be adjacent
      +---+  |      +---+---+---+---+---+       |
       ...   |                                  |
             |      +---+---+---+---+---+       |
             +----> |'-'|'.'|'-'|'.'| 0 | <-----+
                    +---+---+---+---+---+