fixing / improving printFloat() from Mikal Hart

This commit is contained in:
David A. Mellis 2009-01-25 15:44:17 +00:00
parent 473cf103b8
commit 854c69dfde
1 changed files with 32 additions and 15 deletions

View File

@ -81,7 +81,7 @@ void Print::print(long n, int base)
void Print::print(double n) void Print::print(double n)
{ {
printFloat(n*100, 2); printFloat(n, 2);
} }
void Print::println(void) void Print::println(void)
@ -167,20 +167,37 @@ void Print::printNumber(unsigned long n, uint8_t base)
'A' + buf[i - 1] - 10)); 'A' + buf[i - 1] - 10));
} }
void Print::printFloat(double number, uint8_t scale) void Print::printFloat(double number, uint8_t digits)
{ {
double mult = pow(10,scale); // Handle negative numbers
double rounded = floor(number /mult); if (number < 0.0)
double biground = rounded * mult; {
double remainder = (number - biground); print('-');
remainder = remainder / mult; number = -number;
print(long(rounded)); }
print(".");
while (scale--) { // Round correctly so that print(1.999, 2) prints as "2.00"
double toPrint = floor(remainder * 10); double rounding = 0.5;
print(int(toPrint)); for (uint8_t i=0; i<digits; ++i)
remainder -= (toPrint/10); rounding /= 10.0;
remainder *= 10;
number += rounding;
// Extract the integer part of the number and print it
unsigned long int_part = (unsigned long)number;
double remainder = number - (double)int_part;
print(int_part);
// Print the decimal point, but only if there are digits beyond
if (digits > 0)
print(".");
// Extract digits from the remainder one at a time
while (digits-- > 0)
{
remainder *= 10.0;
int toPrint = int(remainder);
print(toPrint);
remainder -= toPrint;
} }
} }