On Sat, Aug 05, 2000 at 11:31:02AM -0500, Philip C Mendelsohn wrote:
> On Fri, 4 Aug 2000, Eric G . Miller wrote:
>
> > I suggest going to comp.lang.c and checking the C FAQ. Specifically look
> > for the inherent problem of putting the same variable on both sides of a
> > relational operator: (x+y) > x
>
> In this case, a little algebra shows that all you need to check is y > 0.
> That's legal and defined.
Yup. But I think the assignment of z = x+y makes z->x quicker then the
evaluation of x+y -> x. It's a floating point precision thing...
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