Re: Bug on Pentium IOC

[Eric Norum <eric.norum@usask.ca>]

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 <eric.norum@usask.ca>
Department of Electrical Engineering
University of Saskatchewan
Saskatoon, Canada.
Phone: (306) 966-5394   FAX:   (306) 966-5407