Graham Davis a écrit :
> I think that we should consider positions with similar NaN ordinates to
> be equal. (...snip...)
>
> So, I think I should leave this last part of the DirectPosition equals
> method the same. What do you think?
It depends what we want to do:
The current code will consider 2 points as equals if at least one point have NaN
value, even if the other point has valid value. For example (2,5,8) and
(2,NaN,8) are considered equals.
With the proposed code change, (2,5,8) and (2,NaN,8) are considered different
but (2,NaN,8) and (2,NaN,8) would be considered different as well.
If the goal is "(2,5,8) and (2,NaN,8) are different" but (2,NaN,8) and (2,NaN,8)
are equals", then a possible approach (admitly harder to read) is:
double a = position.getOrdinate(i);
double b = this.getOrdinate(i);
if (!(abs(a - b) <= tol) && (!Double.isNaN(a) || !Double.isNaN(b))) {
return false;
}
Whatever policy is choosen regarding NaN, it is probably worth to put a javadoc
comment or a comment in the code...
Martin
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