【C语言】switch状态机计算字符串中有多少个单词（简易版）

switch实现状态机

实现条件：只考虑了多个空格，暂未考虑其他单词合法性等问题，注重思路即可

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

//枚举状态0开始，1在单词中，2在单词外，3结束
enum status
{
    
    
	BEGIN = 0, 
	INWORD = 1,
	OUTWORD = 2
};

int main()
{
    
    
	char str[] = {
    
    " hello world time   is life ohhhh"};
	
	//默认状态开始
	enum status status = BEGIN;
	int count = 0;		//单词个数
	for(int i = 0; i < strlen(str); i++)
	{
    
    
		switch(status)
		{
    
    
			case BEGIN:
				if(isalpha(str[i]))
				{
    
    
					status = INWORD;
				}
				else
				{
    
    
					status = OUTWORD;
				}
				break;
			case INWORD:
				if(!isalpha(str[i]))
				{
    
    
					status = OUTWORD;
					count++;
				}
				break;
			case OUTWORD:
				if(isalpha(str[i]))
				{
    
    
					status = INWORD;
				}
				break;
		}
	}
	if(status == INWORD) 
		count++;
	
	printf("str:\"%s\" of words = %d\n", str, count);
}

结果：