On Friday 04 February 2005 12:11, Timothy Miller wrote:
> > What if you're adding 0 (2^-127) and 2^-120? That won't be equal to
> > 2^-120 since (I guess) you have more than 7 bits of precision. What if
> > then, the result is multiplied?
> >
> > Well, maybe you should read this mail as: consider what the output is
> > used for. If it doesn't go into a multiplier, you should be fine.
>
> Yes, that is a good point.  The problem case is if you're multiplying
> a really big number by a really small number.  That could,
> theoretically, put you back into range of whatever clamping happens
> later and cause a problem.

It's a common case, just look at a polygon edge-on.

Regards,

Daniel
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