Is it ok to compare floating points to 0.0 without epsilon?

Question

I am aware, that to compare two floating point values one needs to use some epsilon precision, as they are not exact. However, I wonder if there are edge cases, where I don\

Answer 1

It is a common misconception that floating point values are "not exact". In fact each of them is perfectly exact (except, may be, some special cases as -0.0 or Inf) and equal to s·2^{e – (p – 1)}, where s, e, and p are significand, exponent, and precision correspondingly, each of them integer. E.g. in IEEE 754-2008 binary32 format (aka float32) p = 24 and 1 is represented as ‭0x‭800000‬‬·2^{0 – 23}. There are two things that are really not exact when you deal with floating point values:

1. Representation of a real value using a FP one. Obviously, not all real numbers can be represented using a given FP format, so they have to be somehow rounded. There are several rounding modes, but the most commonly used is the "Round to nearest, ties to even". If you always use the same rounding mode, which is almost certainly the case, the same real value is always represented with the same FP one. So you can be sure that if two real values are equal, their FP counterparts are exactly equal too (but not the reverse, obviously).
2. Operations with FP numbers are (mostly) inexact. So if you have some real-value function φ(ξ) implemented in the computer as a function of a FP argument f(x), and you want to compare its result with some "true" value y, you need to use some ε in comparison, because it is very hard (sometimes even impossible) to white a function giving exactly y. And the value of ε strongly depends on the nature of the FP operations involved, so in each particular case there may be different optimal value.

For more details see D. Goldberg. What Every Computer Scientist Should Know About Floating-Point Arithmetic, and J.-M. Muller et al. Handbook of Floating-Point Arithmetic. Both texts you can find in the Internet.