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From: Guillaume Melquiond (guillaume.melquiond_at_[hidden])
Date: 20080225 16:49:19
Le lundi 25 février 2008 à 22:08 +0100, Johan Råde a écrit :
> >>> I am missing the point of the changesign function. What is its
> >>> rationale? The documentation talks about copysign(x, signbit(x) ? 1 :
> >>> 1) but not about x. So I am a bit lost.
>
> >> Is x guaranteed to change the sign of zero and NaN?
> >> If it is, then I agree that the function is superfluous.
>
> > Back to the topic at hand, I don't know of any architecture/compiler
> > where negating a number does not flip the sign bit of the number, so x
> > works both for zero and NaN.
>
> I just found out that for the ia32 architecture,
> unary minus flips the signbit of NaN if you use the x87 units,
> but not if you use the SSE units.
> So the changesign function is useful.
Sorry to disappoint you, but you probably have not found anything but a
compiler optimization.
As I explained in a previous mail, the sign bit is not part of the NaN
payload. So a standardcompliant compiler is allowed to replace NaN(17)
by NaN(17) in the code. If you put a "volatile" between the NaN and the
negation, you will probably get your sign back. It doesn't have anything
to do with the arithmetic units.
As a matter of fact, the SSE units do not provide any "negate" opcode,
so compilers generate a "xor" opcode instead. You will have a hard time
convincing anybody that the xor instruction is failing to flip the sign
bit.
So I suggest you take a look at the assembly output of your compiler and
check what the generated code actually look like.
Best regards,
Guillaume
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