Compare double to zero using epsilon

Question

Assuming 64-bit IEEE double, there is a 52-bit mantissa and 11-bit exponent. Let's break it to bits:

1.0000 00000000 00000000 00000000 00000000 00000000 00000000 × 2^0 = 1

The smallest representable number greater than 1:

1.0000 00000000 00000000 00000000 00000000 00000000 00000001 × 2^0 = 1 + 2^-52

Therefore:

epsilon = (1 + 2^-52) - 1 = 2^-52

Are there any numbers between 0 and epsilon? Plenty... E.g. the minimal positive representable (normal) number is:

1.0000 00000000 00000000 00000000 00000000 00000000 00000000 × 2^-1022 = 2^-1022

In fact there are (1022 - 52 + 1)×2^52 = 4372995238176751616 numbers between 0 and epsilon, which is 47% of all the positive representable numbers...

The test certainly is not the same as someValue == 0. The whole idea of floating-point numbers is that they store an exponent and a significand. They therefore represent a value with a certain number of binary significant figures of precision (53 in the case of an IEEE double). The representable values are much more densely packed near 0 than they are near 1.

To use a more familiar decimal system, suppose you store a decimal value "to 4 significant figures" with exponent. Then the next representable value greater than 1 is 1.001 * 10^0, and epsilon is 1.000 * 10^-3. But 1.000 * 10^-4 is also representable, assuming that the exponent can store -4. You can take my word for it that an IEEE double can store exponents less than the exponent of epsilon.

You can't tell from this code alone whether it makes sense or not to use epsilon specifically as the bound, you need to look at the context. It may be that epsilon is a reasonable estimate of the error in the calculation that produced someValue, and it may be that it isn't.

There are numbers that exist between 0 and epsilon because epsilon is the difference between 1 and the next highest number that can be represented above 1 and not the difference between 0 and the next highest number that can be represented above 0 (if it were, that code would do very little):-

#include <limits>

int main ()
{
  struct Doubles
  {
      double one;
      double epsilon;
      double half_epsilon;
  } values;

  values.one = 1.0;
  values.epsilon = std::numeric_limits<double>::epsilon();
  values.half_epsilon = values.epsilon / 2.0;
}

Using a debugger, stop the program at the end of main and look at the results and you'll see that epsilon / 2 is distinct from epsilon, zero and one.

So this function takes values between +/- epsilon and makes them zero.

An aproximation of epsilon (smallest possible difference) around a number (1.0, 0.0, ...) can be printed with the following program. It prints the following output:
epsilon for 0.0 is 4.940656e-324
epsilon for 1.0 is 2.220446e-16
A little thinking makes it clear, that the epsilon gets smaller the more smaller the number is we use for looking at its epsilon-value, because the exponent can adjust to the size of that number.

#include <stdio.h>
#include <assert.h>
double getEps (double m) {
  double approx=1.0;
  double lastApprox=0.0;
  while (m+approx!=m) {
    lastApprox=approx;
    approx/=2.0;
  }
  assert (lastApprox!=0);
  return lastApprox;
}
int main () {
  printf ("epsilon for 0.0 is %e\n", getEps (0.0));
  printf ("epsilon for 1.0 is %e\n", getEps (1.0));
  return 0;
}