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<== Date ==> <== Thread ==>
Subject: Re: Bug on Pentium IOC
From: Eric Norum <[email protected]>
To: Dirk Zimoch <[email protected]>
Cc: Bernd Schoeneburg <[email protected]>, TECHTALK <[email protected]>
Date: Tue, 17 Dec 2002 08:38:54 -0600

On Tuesday, December 17, 2002, at 06:40 AM, Dirk Zimoch wrote:



Bernd Schoeneburg wrote: 
Hi all,
can anyone explain this:
~(~0UL << 32)        is 0xFFFFFFFF on a Motorola VME162 i.e., which is
correct
                                the same on a Pentium is 0, which is
false!
Ideas?
Bernd


The Pentium (and all Intel x86s, I think) does all shifts mod 32.
Thus (~0 << 32) is not shifted at all.
=> ~(~0 << 32) = ~~0 = 0

There may be other architectures that behave similar.

From the C standard description of the shift operators: ``The result is undefined if the right operand is negative, or greater than or equal to the number of bits in the left expression's type.
Since 0UL is 32 bits on both the M68k and I386 architectures, the above expression is undefined and the compiler is free to do anything it wants.
The only bug here is in the code -- not on the Pentium nor in the compiler.

--
Eric Norum <[email protected]>
Department of Electrical Engineering
University of Saskatchewan
Saskatoon, Canada.
Phone: (306) 966-5394   FAX:   (306) 966-5407

References:
Re: Bug on Pentium IOC Dirk Zimoch
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