我在 Contiki 3.0 中使用 RPL,我需要在结果为浮动的情况下进行一些计算。但不是给我浮点数的结果,而是只计算整数,例如:5/2 = 2.0 而不是 2.5。我怎样才能得到正确的答案?我无法在 Contiki 3.0 中打印浮点数或双精度数,因此我使用此代码将浮点数转换为字符串:
// Reverses a string 'str' of length 'len'
void reverse(char* str, int len)
{
int i = 0, j = len - 1, temp;
while (i < j) {
temp = str[i];
str[i] = str[j];
str[j] = temp;
i++;
j--;
}
}
// Converts a given integer x to string str[].
// d is the number of digits required in the output.
// If d is more than the number of digits in x,
// then 0s are added at the beginning.
int intToStr(int x, char str[], int d)
{
int i = 0;
while (x) {
str[i++] = (x % 10) + '0';
x = x / 10;
}
// If number of digits required is more, then
// add 0s at the beginning
while (i < d)
str[i++] = '0';
reverse(str, i);
str[i] = '\0';
return i;
}
// Converts a floating-point/double number to a string.
void ftoa(float n, char* res, int afterpoint)
{
// Extract integer part
int ipart = (int)n;
// Extract floating part
float fpart = n - (float)ipart;
// convert integer part to string
int i = intToStr(ipart, res, 0);
// check for display option after point
if (afterpoint != 0) {
res[i] = '.'; // add dot
// Get the value of fraction part upto given no.
// of points after dot. The third parameter
// is needed to handle cases like 233.007
fpart = fpart * powf(10, afterpoint);
intToStr((int)fpart, res + i + 1, afterpoint);
}
}
感谢您的帮助谢谢 Hanin