Why do I need 17 significant digits (and not 16) to represent a double?

Question

Can someone give me an example of a floating point number (double precision), that needs more than 16 significant decimal digits to represent it?

I have found in this thread that sometimes you need up to 17 digits, but I am not able to find an example of such a number (16 seems enough to me).

Can somebody clarify this?

Answer 1

My other answer was dead wrong.

#include <stdio.h>

int
main(int argc, char *argv[])
{
    unsigned long long n = 1ULL << 53;
    unsigned long long a = 2*(n-1);
    unsigned long long b = 2*(n-2);
    printf("%llu\n%llu\n%d\n", a, b, (double)a == (double)b);
    return 0;
}

Compile and run to see:

18014398509481982
18014398509481980
0

a and b are just 2*(2⁵³-1) and 2*(2⁵³-2).

Those are 17-digit base-10 numbers. When rounded to 16 digits, they are the same. Yet a and b clearly only need 53 bits of precision to represent in base-2. So if you take a and b and cast them to double, you get your counter-example.

Answer 2

The correct answer is the one by Nemo above. Here I am just pasting a simple Fortran program showing an example of the two numbers, that need 17 digits of precision to print, showing, that one does need (es23.16) format to print double precision numbers, if one doesn't want to loose any precision:

program test
implicit none
integer, parameter :: dp = kind(0.d0)
real(dp) :: a, b
a = 1.8014398509481982e+16_dp
b = 1.8014398509481980e+16_dp
print *, "First we show, that we have two different 'a' and 'b':"
print *, "a == b:", a == b, "a-b:", a-b
print *, "using (es22.15)"
print "(es22.15)", a
print "(es22.15)", b
print *, "using (es23.16)"
print "(es23.16)", a
print "(es23.16)", b
end program

it prints:

First we show, that we have two different 'a' and 'b':
a == b: F a-b:   2.0000000000000000     
using (es22.15)
1.801439850948198E+16
1.801439850948198E+16
using (es23.16)
1.8014398509481982E+16
1.8014398509481980E+16