//Warning: if the float variable is larger than the 10000,the decimal part digit will be not precise(but the func logic is correct)!!!
int FtoStr(float f, char *str)
{
char buf[20] = {0};
char *pos = buf;
int tmp;
if(NULL == str) return -1;
if(f < 0)
{
f = -f;
*str++ = '-';
}
tmp = ((int)(f * 10000) % 10) >= 5 ? (f * 1000) + 1 : (f * 1000); //keep 3 decimal precision digit
do {
if(pos == buf + 3)//insert the decimal point position
*pos++ = '.';
*pos++ = tmp % 10 + '0';
}while(tmp /= 10);//check whether the copied char is the last char or not
while(pos < buf + 3)//less than 1.0
*pos++ = '0';
if(pos == buf + 3){
*pos++ = '.';
*pos++ = tmp +'0';
}
while(pos > buf)
*str++ = *--pos;
*str = '\0';
return 0;
}
本文介绍了一种将浮点数转换为字符串的方法,并特别关注了小数点后三位的有效精度处理。该方法通过一系列数学运算确保数值精度,并正确处理小于1.0的数值情况。
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