Re: 0.3*3 == 0.9?

[Jordi Gutiérrez Hermoso]

2010/8/25 Michael Creel <address@hidden>:
>> Anyways, yes, this is completely normal. 0.3 has an infinite binary
>> expansion because its denominator has factors other than 2, so you'll
>> get a rounding error.

> Right, I'm aware of that. My question is whether or not Octave's comparison
> of floats takes it into account.

I don't see how == by itself could. I was looking for a function like
fcomp(a,b,e) := abs(a-b) < e, but it's pretty trivial to just do the
rhs of that function.