Compare double to zero using epsilon

Question

Today, I was looking through some C++ code (written by somebody else) and found this section:

double someValue = ...
if (someValue <  std::numeric_limits<double>::epsilon() && 
    someValue > -std::numeric_limits<double>::epsilon()) {
  someValue = 0.0;
}

I'm trying to figure out whether this even makes sense.

The documentation for epsilon() says:

The function returns the difference between 1 and the smallest value greater than 1 that is representable [by a double].

Does this apply to 0 as well, i.e. epsilon() is the smallest value greater than 0? Or are there numbers between 0 and 0 + epsilon that can be represented by a double?

If not, then isn't the comparison equivalent to someValue == 0.0?

Answer 1

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...