format specifier for long double (I want to truncate the 0's after decimal) - iphone

Question

I have a 15-digit floating-point number and I need to truncate the trailing zeros after the decimal point. Is there a format specifier for that?

Answer 1

%Lg is probably what you want: see http://developer.apple.com/library/ios/#DOCUMENTATION/System/Conceptual/ManPages_iPhoneOS/man3/printf.3.html.

Answer 2

Unfortunately in C there is no format specifier that seems to meet all the requirements you have. %Lg is the closest but as you noted it switched to scientific notation at its discretion. %Lf won't work by itself because it won't remove the trailing zeroes.
What you're going to have to do is print the fixed format number to a buffer and then manually remove the zeroes with string editing (which can STILL be tricky if you have rounding errors and numbers like 123.100000009781).

Answer 3

Is this what you want:
#include <iostream>
#include <iomanip>
int main()
{
double doubleValue = 78998.9878000000000;
std::cout << std::setprecision(15) << doubleValue << std::endl;
}
Output:
78998.9878
Note that trailing zeros after the decimal point are truncated!
Online Demo : http://www.ideone.com/vRFlQ

Answer 4

You could print the format specifier as a string, filling in the appropriate amount of digits if you can determine how many:
sprintf(fmt, "%%.%dlf", digits);
printf(fmt, number);
or, just checking trailing 0 characters:
sprintf(fmt, "%.15lf", 2.123);
truncate(fmt);
printf("%s", fmt);
truncate(char * fmt) {
int i = strlen(fmt);
while (fmt[--i] == '0' && i != 0);
fmt[i+1] = '\0';
}

Answer 5

%.15g — the 15 being the maximum number of significant digits required in the string (not the number of decimal places)
1.012345678900000 => 1.0123456789
12.012345678900000 => 12.0123456789
123.012345678900000 => 123.0123456789
1234.012345678900000 => 1234.0123456789
12345.012345678900000 => 12345.0123456789
123456.012345678900000 => 123456.012345679