Experimental Physics and Industrial Control System
|
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
- Navigate by Date:
- Prev:
Re: Bug on Pentium IOC Dirk Zimoch
- Next:
Re: my base download account cann't use M.C.Shao
- Index:
1994
1995
1996
1997
1998
1999
2000
2001
<2002>
2003
2004
2005
2006
2007
2008
2009
2010
2011
2012
2013
2014
2015
2016
2017
2018
2019
2020
2021
2022
2023
2024
- Navigate by Thread:
- Prev:
Re: Bug on Pentium IOC Dirk Zimoch
- Next:
some JOImint for xmas Matthias Clausen
- Index:
1994
1995
1996
1997
1998
1999
2000
2001
<2002>
2003
2004
2005
2006
2007
2008
2009
2010
2011
2012
2013
2014
2015
2016
2017
2018
2019
2020
2021
2022
2023
2024
|
ANJ, 10 Aug 2010 |
·
Home
·
News
·
About
·
Base
·
Modules
·
Extensions
·
Distributions
·
Download
·
·
Search
·
EPICS V4
·
IRMIS
·
Talk
·
Bugs
·
Documents
·
Links
·
Licensing
·
|