Re: [Q] Portability of <= and >= with real operands

[dewar@merv.cs.nyu.edu (Robert Dewar)]

Keith says

"Suppose you've computed two quantities, X and Y, such that X is
mathematically known to be less than Y, but it may be arbitrarily close.
Since floating-point is of finite precision, the representations of X
and Y may be equal.  (For example, X = 0.0, Y = some tiny value which
underflows to 0.0).  Then X <= Y may reflect the relationship more
accurately than X < Y.
"

Well first, surely you are not meaning to say that X<=Y is different
from not (X>Y), which you are arguing is somehow less portable than
X>=Y, so this argument is not self consistent!

In fact if your analysis is based on a view of the mathemaytical
proper values of variables, then this argument is almost certainly
bogus, since approximations may lead X > Y to be true, even if
X < Y mathematically.

You cannot substitute rules like this for proper analysis. Both > and
>= have perfectly well defined portable meanings for a given model
of floating-point arithmetic, and you have to analyze your code with
respect to a particular model.