c - 为什么我的 wc 实现差了一个字？

Question

[已解决] 编写解析代码是一个陷阱。一行有 15 个空格将有 15 个单词。空白行也算作一个单词。对我来说回到 flex 和 bison。

#include        <stdio.h>                         
#include        <stdlib.h>                        

int     main(int argc, char     *argv[]) {

        FILE    *fp = NULL;
        int             iChars =0, iWords =0, iLines =0;
        int             ch;                             

        /* if there is a command line arg, then try to open it as the file
                otherwise, use stdin */                                   

        fp = stdin;
        if (argc == 2) {
                fp = fopen(argv[1],"r");
                if (fp == NULL) {       
                        fprintf(stderr,"Unable to open file %s. Exiting.\n",argv[1]);
                        exit(1);
                }
        }

        /* read until the end of file, counting chars, words, lines */
        while ((ch = fgetc(fp)) != EOF) {
                if (ch == '\n') {
                        iWords++;
                        iLines++;
                }

                if (ch == '\t' || ch == ' ') {
                        iWords++;
                }

                iChars++;
        }

        /* all done. If the input file was not stdin, close it*/
        if (fp != stdin) {
                fclose(fp);
        }

        printf("chars: %d,\twords: %d,\tlines: %d.\n",iChars,iWords,iLines);
}

测试数据 foo.sh

#!/home/ojblass/source/bashcrypt/a.out
This is line 1
This is line 2
This is line 3

ojblass@linux-rjxl:~/source/bashcrypt> wc foo.sh

5 13 85 foo.sh

ojblass@linux-rjxl:~/source/bashcrypt> a.out foo.sh

chars: 85, words: 14, lines: 5.

Answer 1

你的算法是错误的。如果您在测试文件中有 2 个连续的空白字符，则单词计数器将递增两次，但它应该只递增一次。

一个解决方案是记住最后读取的字符。如果读取的字符是一个特殊字符(空白、换行、...)并且前一个字符是字母数字，那么您将增加单词的计数器。