the c programing language 练习1-21 将空格字符替换为最少数量的制表符和空格

/*
 * K&R2 1-21
 * Author: Donmmi
 * Email:[email protected]
 */

#include 

/* 4 spaces to a tab */
#define NTAB    4

int main(void) {
    /*
     * len is the nums of input charactor
     * if (c == ' ' && len % NTAB == 0) translate the space(s) to tab
     */
    int     c, len, nspace, flag;

    len = nspace = flag = 0;
    while ((c = getchar()) != EOF) {
        ++len;
        if (c == ' ') {
            if (len % NTAB == 0) { /* space(s) translate to tab */
                if (flag == 0)
                    putchar('\t');
                flag = 1;
                nspace = 0;
            } else
                ++nspace;
        } else {
            /*
             * If the next charactor is not space and can't translate to tab
             * then output saved space and the charactor
             */
            while (nspace) {
                putchar(' ');
                --nspace;
            }
            putchar(c);
            if (c == '\n')
                len = 0;
            flag = 0;
        }
    }

    return 0;
}

记录字符个数len，和连续空格个数nspace

如果下个字符不为空格则输出之前保存的空格和非空格字符

如果下个字符为空格，根据len%NTAB判断是否可以将空格转换为tab