/********************************************************************************************* @Name	 : 浮点型转字符* @Brief	 : none* @Input	 : str:字符串指针 num:浮点数 n:精度* @Output	 : none* @Return	 : none******************************************************************************************
**/
int ftoa(char *str, float num, int n)        //n是转换的精度，即是字符串'.'后有几位小数
{int     sumI;float   sumF;int     sign = 0;int     temp;int     count = 0;char *p;char *pp;if(str == NULL) return -1;p = str;/*Is less than 0*/if(num < 0){sign = 1;num = 0 - num;}sumI = (int)num;    //sumI is the part of intsumF = num - sumI;  //sumF is the part of float/*Int ===> String*/do{temp = sumI % 10;*(str++) = temp + '0';}while((sumI = sumI /10) != 0);/*******End*******/if(sign == 1){*(str++) = '-';}pp = str;pp--;while(p < pp){*p = *p + *pp;*pp = *p - *pp;*p = *p -*pp;p++;pp--;}*(str++) = '.';     //point/*Float ===> String*/do{temp = (int)(sumF*10);*(str++) = temp + '0';if((++count) == n)break;sumF = sumF*10 - temp;}while(!(sumF > -0.000001 && sumF < 0.000001));*str ='\0';return 0;}/********************************************************************************************* @Name	 : 转置函数* @Brief	 : none* @Input	 : s:待转换字符串* @Output	 : none* @Return	 : none******************************************************************************************
**/
void reverse(char *s)  
{  char temp;  char *p = s;    //p指向s的头部  char *q = s;    //q指向s的尾部  while(*q)  ++q;  q--;  //交换移动指针，直到p和q交叉  while(q > p)  {  temp = *p;  *p++ = *q;  *q-- = temp;  }   
}  /********************************************************************************************* @Name	 : 整型转字符* @Brief	 : none* @Input	 : n:整型 s:字符串* @Output	 : none* @Return	 : none******************************************************************************************
**/void itoa(uint16_t n,char s[])
{uint16_t i = 0;   do      //从各位开始变为字符，直到最高位，最后应该反转  {  s[i++] = n%10 + '0';  n = n/10;  }while(n > 0);  s[i] = '\0';    //最后加上字符串结束符 reverse(s);
}