c - 最后一次 strcpy() 调用后 char* s2 到底如何以及为何发生变化？

Question

这是我几天后在学校进行的测试中可能会出现的问题之一。我无法理解 char* s2 在最后一次 strcpy() 调用后如何从“消息服务”更改为 “ice”。我认为这是指针的某种结果，但这就是我所知道的。

char s1[] = "Short Message Service";
char *s2, *s3;

s2 = strchr(s1, 'M');

s3 = strrchr(s2, 'S');

strncpy(s1 + 1, s2, 1);

strcpy(s1 + 2, s3);  

我尝试调试程序并观察变量和指针的变化，但我仍然不明白最后一次调用如何影响 char *s2。

我期望 s2 的值是“Message Service”，但实际输出最终是“ice”。不过，我确实理解 s1 和 s3 如何以及为何变化。

Answer 1

如果我们显示数组，并添加相应的指针，它将看起来像这样:

+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+------+
| 'S' | 'h' | 'o' | 'r' | 't' | ' ' | 'M' | 'e' | 's' | 's' | 'a' | 'g' | 'e' | ' ' | 'S' | 'e' | 'r' | 'v' | 'i' | 'c' | 'e' | '\0' |
+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+------+
^                                   ^                                               ^
|                                   |                                               |
s1                                 s2                                               s3

When you modify the contents of the array, the pointers s2 and s3 themselves doesn't change, they still point to the same locations in the array.

So after

strncpy(s1 + 1, s2, 1);

数组(带有指针)看起来像

+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+------+
| 'S' | 'M' | 'o' | 'r' | 't' | ' ' | 'M' | 'e' | 's' | 's' | 'a' | 'g' | 'e' | ' ' | 'S' | 'e' | 'r' | 'v' | 'i' | 'c' | 'e' | '\0' |
+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+------+
^                                   ^                                               ^
|                                   |                                               |
s1                                 s2                                               s3

The single character that s2 is pointing to is copied to s1[1].

Then with

strcpy(s1 + 2, s3);  

数组看起来像

+-----+-----+-----+-----+-----+-----+-----+-----+-----+------+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+------+
| 'S' | 'M' | 'S' | 'e' | 'r' | 'v' | 'i' | 'c' | 'e' | '\0' | 'a' | 'g' | 'e' | ' ' | 'S' | 'e' | 'r' | 'v' | 'i' | 'c' | 'e' | '\0' |
+-----+-----+-----+-----+-----+-----+-----+-----+-----+------+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+------+
^                                   ^                                                ^
|                                   |                                                |
s1                                 s2                                                s3

子字符串“Service”被复制到从s1[2]开始的字符串部分。

指针s2仍然指向相同的位置，但是如果我们将s2视为指向字符串第一个字符的指针，那么它的内容已经改变，因为你已覆盖该内容。