一个统计代码行数的程序C语言

我这么说其实是有歧义的,因为这个东西根本没有那么通用.不过有很简单的办法让这个程序更加通用,我介绍下运行此程序前的前期准备.

首先, 将所有代码文件放置到同一个空文件夹中.选定所有代码,重命名为1.于是有了下面的效果:

所有文件都是排列好的了,这样方便我操作.下面贴出来效果图,还真是令我惊讶的一个结果呢.

从5月24号开始自己已经写了这么多代码了啊.

下面附上源码,很简单的实现方式.不过居然也费了我两个小时,真搞不懂这两个小时我都干嘛了...可能就是我没瞧得起这么简单的一个任务,计划不周详,才导致这个结果的吧.还可能什么啊..就是这个样子!

#include <stdio.h> #include <string.h> #include <stdlib.h> #define INITIAL_VALUE 1 #define NUMBER_LENTH 4 #define BIG_SPACE 10 #define FULL_NAME_LENTH 15 #define PF '/n' #define FINISH printf ("全部文件统计完毕,报告如下:/n") #define ERROR(i) printf ("第%d个文件关闭失败/n", i) struct item { int the_number_of_char ; int the_number_of_enter ; } ; int main (void) ; void get_full_name_base_on_number (const int integer, char * string) ; /* 根据数字值获得文件全名 */ struct item count_and_printf (FILE * const fp, const int i) ; /* 计算并返回当前文件字符数以及行数 */ int main (void) { FILE * fp ; char file_name[FULL_NAME_LENTH] ; int i, max_char_file, max_enter_file ; struct item temp, sum, max ; sum.the_number_of_char = sum.the_number_of_enter = 0 ; max.the_number_of_char = max.the_number_of_enter = 0 ; for (i = INITIAL_VALUE; ; i++) { get_full_name_base_on_number (i, file_name) ; if (NULL == (fp = fopen (file_name, "r"))) { FINISH ; i-- ; break ; } else { temp = count_and_printf (fp, i) ; sum.the_number_of_char += temp.the_number_of_char ; sum.the_number_of_enter += temp.the_number_of_enter ; if (temp.the_number_of_char > max.the_number_of_char) { max.the_number_of_char = temp.the_number_of_char ; max_char_file = i ; } if (temp.the_number_of_enter > max.the_number_of_enter) { max.the_number_of_enter = temp.the_number_of_enter ; max_enter_file = i ; } } if (fclose (fp) != 0) { ERROR (i) ; exit (EXIT_FAILURE) ; } } printf (" 文件总数:%-5d全部字符数:%-10d全部行数:%-10d/n" , i, sum.the_number_of_char, sum.the_number_of_enter) ; if (i) printf (" 最大字符数:%-10d出现在文件:1 (%d).c中/n 最大行数:%-10d出现在文件:1 (%d).c中/n", max.the_number_of_char, max_char_file, max.the_number_of_enter, max_enter_file) ; getchar () ; return 0 ; } void get_full_name_base_on_number (const int integer, char * string) { char number[NUMBER_LENTH] ; strcpy (string, "1 (") ; sprintf (number, "%d", integer) ; strcat (string, number) ; strcat (string, ").c") ; } struct item count_and_printf (FILE * const fp, const int i) { struct item sum ; char ch ; sum.the_number_of_char = 0, sum.the_number_of_enter = 0 ; while ((ch = getc (fp)) != EOF) { if (PF == ch) sum.the_number_of_enter++ ; else sum.the_number_of_char++ ; } sum.the_number_of_enter++ ; printf ("当前第:%-2d个文件: 字符数:%-5d 行数:%-10d/n", i, sum.the_number_of_char, sum.the_number_of_enter) ; return sum ; }