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 thi

Answer 1

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