C printf function specifier

Question

I need to print out a float value with 16 significant digits.

Right now when I print, it prints out

i.e 2.555556

when I want it to print out: 2.555555555555556 /// A total of 16 digits

My printf specifier currently is : "%20.18lf"

ideas?

You're aware that a float doesn't have 16 digits of precision? — Oliver Charlesworth, Feb 23 '12 at 19:12
I tried "%.16f" and it worked just fine for the number 1.0 / 3.0. — Tibi, Feb 23 '12 at 19:14

score 3 · Answer 1 · answered Feb 23 '12 at 19:20

If the value is of type float, you're probably not going to get more than about 7 significant digits after the decimal point. If it's of type double, you can get about 15. Depending on the system, long double may be no more precise than double; on mine, I get about 17 significant digits after the decimal point.

Most floating-point code uses type double; float is much less precise and often not significantly faster.

I'm a little surprised that you got 2.555556 rather than something like 2.555555582046508789 with a "%20.18lf" format.

Incidentally, the l (lowercase L) is not necessary. The format for double is "%f", and float arguments to printf are promoted to double, so "%f" is correct for both float and double. The format for long double is "%Lf".

Here's a test program I wrote:

#include <stdio.h>
int main(void) {
    const float f        = 23.0/9.0;
    const double d       = 23.0/9.0;
    const long double ld = 23.0L/9.0L;
    printf("%20.18f\n%20.18f\n%20.18Lf\n", f, d, ld);
    return 0;
}

and the output I got on my system:

2.555555582046508789
2.555555555555555358
2.555555555555555556

score 0 · Answer 2 · answered Feb 23 '12 at 19:20

The following code works well (you have to enable C99 as %lf specifier is not part of earlier standards):

#include <stdio.h>

int main(void)
{
        volatile float v = 2.555555555555556;
        printf("%20.18lf\n", v);
        return 0;
}

Prints the following:

$ gcc -std=c99 -Wall -pedantic -o test ./test.c 
$ ./test 
2.555555582046508789
$

If value is shorter, say 2.5, it right-pads it with zeros - 2.500000000000000000.

So you have an error elsewhere. Generally, you have to post a minimal self-consistent code example that reproduces the problem. Otherwise you are on your own.

I figured out the problem guys. When I was inputting, I was using the same specifier for the input also, so it would crash — user1050632, Feb 23 '12 at 19:45

score 0 · Answer 3 · answered Feb 23 '12 at 19:57

C float or double is not precise enough, basically you need a high precision division function. Here is what I wrote. It is for integer division. You may need to revise it for float division.

int dividend=121, divisor=9;
int quotient,remainder;

//first print out the number before decimal point
quotient=dividend/divisor;
printf("%d", quotient);
remainder=dividend-quotient*divisor;

//print out decimal point
printf(".");

int digit_num=20;//this can be any precision you want

//second we calculate the number after decimal point
for(int i=0;i<digit_num;i++)
{
    remainder=remainder*10;
    quotient=remainder/divisor;
    remainder=remainder-quotient*divisor;
    printf("%d", quotient );
    dividend=remainder;
}

printf("\n");

printf("%d digits after decimal point has been printed out\n", digit_num );

C printf function specifier

3 Answers3