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Date: | Wed, 22 Mar 2000 10:28:54 +0200 (IST) |
From: | Eli Zaretskii <eliz AT is DOT elta DOT co DOT il> |
X-Sender: | eliz AT is |
To: | ams AT ludd DOT luth DOT se |
cc: | djgpp-workers AT delorie DOT com |
Subject: | Re: Unnormals??? |
In-Reply-To: | <200003210709.IAA03083@father.ludd.luth.se> |
Message-ID: | <Pine.SUN.3.91.1000322102807.17945F-100000@is> |
MIME-Version: | 1.0 |
Reply-To: | djgpp-workers AT delorie DOT com |
Errors-To: | dj-admin AT delorie DOT com |
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On Tue, 21 Mar 2000 ams AT ludd DOT luth DOT se wrote: > The copysign functions produce a value with the magnitude of x and the > sign of y. They produce a NaN (with the sign of y) if x is a NaN." > > This means the standard do think that NaNs have a sign (however > misguided that is) No, they only think a NaN *might* have a sign. If there's no notion of a sign of a NaN, the copying a sign to it does nothing. > hence we really do need to print the "-" of a negative NaN. Period. I don't see how this argument changes anything. I really think that it's up to us to decide what we think is appropriate in this case. The standard clearly doesn't provide enough guidance on this issue. Even this last argument seems to be far-fetched (why should the standard hide such an important info in the description of an obscure function?).
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