c程序的意外行为,试图从K&R解决E6-2

练习6-2。 编写一个程序,读取C程序并按字母顺序打印每组 变量名在前6个字符中相同,但在后面的某个地方不同。不算数 字符串和注释中的单词。使6成为可以从命令行设置的参数。(来自K&R)

我开始练习了,但我一开始就陷入了困境。我试图获取整个输入并将其逐行保存在数组中,然后将数组中的指针指向保存的行。后来我想到为每个单词分配空间,并将它们复制到一棵树上。然而,在以下代码中,当我输入时:

abc定义;'Ctrl-Z'

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

#define MAXLINE 100
#define MAXWORDS 10

int read_input(char *[]);

int main(int argc, char *argv[])
{
    char *variables[MAXWORDS] = {0};
    int i;
    int tmp;


    while((tmp = read_input(variables)) != EOF){
        for(i=0; i < tmp; i++){
            printf("%s ",*(variables+i));
        }
    }


    return 0;
}


/* read_input: mapping all the input words into *variables[], and returning the number of the words + 1 */
int read_input(char *variables[]){
    static char s[MAXLINE];
    int tmp;
    int i, j;

    i = 0;
    if((tmp = getline(s, MAXLINE)) == EOF)
        return EOF;

    for(j = 0; j < MAXWORDS && s[i] != '\0'; j++){ /* it does'nt work when j = MAXWORDS because the rest of the line is not saved */
        while(!isalpha(s[i]))
            i++;
        *(variables+j) = &s[i];
        while(isalnum(s[i]) || s[i] == '_')
            i++;
        if(s[i] == '\0')
            continue;
        s[i++] = '\0';
    }
    return j;
}

/* getline: read a line into s, return length */
int getline(char *s,int lim)
{
    static char end; /* static typed char that saves if the last not saved 'c' was an EOF */
    char c;
    int i;
    if(end == EOF){ /* the last input character was an EOF */
        end = 0;
        return EOF;
    }
    for (i=0; i < lim-1 && (c=getchar())!=EOF && c!=';'; ++i)
        s[i] = c;
    s[i] = '\0';
    if(i > 0 && c == EOF)
        end = EOF;
    else if(i == 0 && c == EOF){
        printf("!");
        return EOF;
    }
    else
        return i;

}

程序不会立即停止,main的while循环会要求另一个输入……我做错了什么?