Compare double to zero using epsilon
c++, double, floating-point
Solution
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...
Problem
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`?