K&R c 编程书第 2 版，第 2 章中的 strcat 示例

Question

Initial I didn't intend to go deeper of this example, coz I think it's just several lines, not big deal, but when I use the hello world to test and verify my thought, I find something quite abnormal: here is the code part without too much comments:

#include<stdio.h>
void concatenate(char s[], char c[]);
int main(){
    char s[] = "hello";
    char c[] = "world";
    concatenate(s, c);
    printf("%s\n",s);
    return 0;
}
void concatenate(char s[], char c[]){
    int i, j;
    i = j = 0;
    while(s[i] != '\0')/* find the end of s*/
        i++;
    while((s[i++] = c[j++]) != '\0') /* copy c string */
        printf("%d,%c\n", i, s[i]);
    printf("0,%c\n", s[0]);
    printf("1,%c\n", s[1]);
    printf("2,%c\n", s[2]);
    printf("3,%c\n", s[3]);
    printf("4,%c\n", s[4]);
    printf("5,%c\n", s[5]);
    printf("6,%c\n", s[6]);
    printf("7,%c\n", s[7]);
    printf("8,%c\n", s[8]);
    printf("9,%c\n", s[9]);
    printf("10,%c\n", s[10]);
}

how to run it? I'm using ubuntu 64-bit 18.04 version===> cc concat.c the output:

From the screenshot, it is so hard to imagine why the output is like this, could anyone plz spend a bit of time explaining why it is this?? really appreciate it!

Answer 1

On one hand, there is something wrong in here:

Due to the post-increment of i, you should use:

  while((s[i++] = c[j++]) != '\0') /* copy c string */
        printf("%d,%c\n", i, s[i-1]);

instead of s[i] , otherwise you would print garbage values like this:

6,(
7, 
8,(
9,
10,4
0,h
1,e
2,l
3,l
4,o
5,w
6,o
7,r
8,l
9,d
10,

on the other hand, you are affecting c[j++] to s[i++] , thus copying the text 'hello' from c to s from index i which is 5 to the index 10 . This allows for copying the content of c[] to s[] from the end of hello , making it containing helloworld\0 .

A good reference, for a first thought can be to look at iso/iec doc section 6.7.8 example 8 (page 130).

It is stated that:

The contents of the arrays are modifiable