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SubjectRe: if (a & X || b & ~Y) in dasd.c
Hi Pete, Hi Kevin,

it's just like Kevin explained, the singe &/|/~ are bit operations
while &&/|| are logical operations. The bit operations are handled
before the logical ones (am I correct?)- so compiler-wise this should
be fine.
Sure, one should use some paranthesises to make things clearer,
but why are you examining such ancient code? 2.2.14. seems a little
bit historical for me. The current dasd device driver is much more
readable and better designed (is redesigned in the 2.4 series).

mit freundlichem Gruß / with kind regards
Carsten Otte

IBM Deutschland Entwicklung GmbH
Linux for 390/zSeries Development - Device Driver Team
Phone: +49/07031/16-4076
IBM internal phone: *120-4076
We are Linux.
Resistance indicates that you're missing the point!

Kevin <> on 11/08/2001 10:10:51 PM

Please respond to Kevin <>

To: Pete Zaitcev <>
cc: Carsten Otte/Germany/IBM@IBMDE, <>,
Subject: Re: if (a & X || b & ~Y) in dasd.c

On Thu, 8 Nov 2001, Pete Zaitcev wrote:

[zaitce] Date: Thu, 8 Nov 2001 15:57:49 -0500
[zaitce] From: Pete Zaitcev <>
[zaitce] To:
[zaitce] Cc:,
[zaitce] Subject: if (a & X || b & ~Y) in dasd.c
[zaitce] Carsten and others:
[zaitce] this code in 2.2.14 looks suspicious to me:
[zaitce] ./drivers/s390/block/dasd.c:
[zaitce] /* first of all lets try to find out the appropriate
era_action */
[zaitce] if (stat->flag & DEVSTAT_FLAG_SENSE_AVAIL ||
[zaitce] stat->dstat & ~(DEV_STAT_CHN_END | DEV_STAT_DEV_END))
[zaitce] /* anything abnormal ? */
[zaitce] if (device->discipline->examine_error == NULL ||
[zaitce] stat->flag & DEVSTAT_HALT_FUNCTION) {
[zaitce] era = dasd_era_fatal;
[zaitce] } else {
[zaitce] era = device->discipline->examine_error
(cqr, stat);
[zaitce] }
[zaitce] DASD_DRIVER_DEBUG_EVENT (1, dasd_int_handler,"
era_code %d",
[zaitce] era);
[zaitce] }
[zaitce] Are you sure any parenthesises are not needed here?

I'm not going to pretend to know what the code is accomplishing, but as
a matter of C, those &'s aren't the same as &&'s and will get executed as
the || is tested. So the first one will get &'ded and if false then the
second will and if neither return true then we move on.

Or, perhaps I'm being an idiot and misunderstanding the question and the
code. In that case, I'll pretend I didn't say anything. :)

-[ ]-
-[ Rather be forgotten, than remembered for giving in. ]-
-[ ZZ = g ^ (xb * xa) mod p g = h^{(p-1)/q} mod p ]-

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