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NaN encodings
In his careful critique of 754-2008 edge cases, Vincent LefÛvre mentioned
the 80-bit x87 format as a possible exception to the encoding of the NaN
code into p-2 bits.
I checked; the rule still aplies. The exposed high-order bit is always 1
for a real NaN; the next bit is the Quiet/Signalling bit, and the remaining
p-2 (62) bits comprise the NaNcode.
When the 80-bit register is loaded with a bit pattern that looks like a NaN
except for a zero exposed high-order bit, it is called a pseudo-NaN -- but
I could not find a statement describing what happens when a pseudo-NaN is
stored into a Binary64 or a Binary32. I assume the exposed bit is simply
ignored -- can anybody confirm this?
Michel.
Sent: 2009-03-25 05:21:09 UTC