计算输入的一句英文语句中单词数

 1 //
 2 //  main.c
 3 //  统计输入单词数
 4 //
 5 //  Created by LongMa on 2019/6/27.
 6 //  Copyright © 2019 . All rights reserved.
 7 //
 8 
 9 #include <stdio.h>
10 #include <string.h>
11 
12 int main(int argc, const char * argv[]) {
13     
14     //输入一个英文句子,英文句以空格分隔。统计句子中有多少个单词。
15     //eg:how   are you,有3个单词
16     char string[80];
17     printf("请输入英文句子:");
18     gets(string);
19     
20     char c;
21     char kongGe[2] = " ";
22     int word = 0;
23     int num = 0;
24     
25     for (int i = 0; (c = string[i]) != \0; i++) {
26         if (c == 32) {//当前位为空格(ASCII码值为32)时,word记录为0.注意:xcode中strcmp(&c, " ")为-120,不为0,strcmp比较的是字符串,不能比较字符和字符串!
27             printf("%d,%d是否相等:%d\n",c,kongGe[0], strcmp(&kongGe[0], &c));
28             word = 0;
29         }else{//当前位不为空格时
30             if(0 == word){//上一位为空格/第一位是字母时
31                 num += 1;
32                 word = 1;
33             }else{//上一位不为空格时,不能算一个单词
34                 //do nothing
35             }
36         }
37     }
38     
39     printf("语句:%s\n", string);
40     printf("单词数:%d个",num);
41     return 0;
42 }

log:

请输入英文句子:warning: this program uses gets(), which is unsafe.
how are you
32,32是否相等:-120
32,32是否相等:-120
语句:how are you
单词数:3个Program ended with exit code: 0
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