The C Programming Language 练习题2-4

题目
squeeze(s1, s2)，将字符串s1 中任何与字符串s2 中字符匹配的字符都删除。

题目分析
先将s2中每个字符串拿出来在s1中寻找，然后生成不包含此字符的新的s1字符串，然后再找下一个。

编程实现
为了方便定位问题，加的打印信息较多。。。。（其实已经删掉许多打印了）

#include <stdio.h>#define MAXLINE 1000void squeeze(char s1[], char s2[]);int main(){    int i;    char c, sfirst[MAXLINE], ssecond[MAXLINE];    i = 0;    printf("Please input string1:");    while ((c = getchar()) != '\n')        sfirst[i++] = c;    sfirst[i] = '\0';    printf("The first string is:");    i = 0;    while (sfirst[i] != '\0')        printf("%c", sfirst[i++]);        printf("\n");    i = 0;    printf("Please input string2:");    while ((c = getchar()) != '\n')        ssecond[i++] = c;    ssecond[i] = '\0';    printf("The second string is:");    i = 0;    while (ssecond[i] != '\0')        printf("%c", ssecond[i++]);        printf("\n");    squeeze(sfirst, ssecond);    i = 0;    while (sfirst[i] != '\0')    {        printf("%c",sfirst[i]);        i++;    }}void squeeze(char s1[], char s2[]){    int m, n, l;    m = n = l = 0;    for (n = 0; s2[n] != '\0'; n++)        {            for (m = l = 0; s1[m] != '\0'; m++)                if (s1[m] != s2[n])                    s1[l++] = s1[m];            s1[l] = '\0';        }}