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- Path: sparky!uunet!cs.utexas.edu!sun-barr!sh.wide!fgw!fdm!ace!melby
- From: melby@dove.yk.fujitsu.co.jp (John B. Melby)
- Newsgroups: comp.programming
- Subject: Re: scalb and rint
- Message-ID: <MELBY.93Jan25113448@dove.yk.fujitsu.co.jp>
- Date: 25 Jan 93 11:34:48 GMT
- References: <1993Jan21.112537.603@mnemosyne.cs.du.edu>
- Sender: news@ace.yk.fujitsu.co.jp
- Organization: Open Systems Group, Fujitsu Limited, Yokohama
- Lines: 18
- In-reply-to: mcasal@nyx.cs.du.edu's message of 21 Jan 93 11:25:37 GMT
-
- >moreover i'd like to know how 'NaN' works...
-
- 'NaN' stands for "not a number." In a floating point representation, it
- represents a result that is not compatible with the definition of '+Inf'
- and '-Inf.' This is dependent on how the representation defines infinity.
-
- For example, defining positive overflow to be +Inf is inconsistent with
- defining +Inf - a to be +Inf (where a is a positive real). If one is
- +Inf, the other must be +NaN or NaN.
-
- I think that the IEEE floating point representation treats the result of
- division by zero as infinity, which necessitates the use of an unsigned
- NaN.
-
- -----
- John B. Melby
- Fujitsu Limited, Yokohama
- melby@yk.fujitsu.co.jp
-