From: Yuri Gribov <tetra2005@gmail.com>
To: Pedro Alves <palves@redhat.com>
Cc: Yury Gribov <y.gribov@samsung.com>,
gdb-patches@sourceware.org, Stan Shebs <stanshebs@google.com>,
Paul Pluzhnikov <ppluzhnikov@google.com>
Subject: Re: [PATCH][PING][PR gdb/19361] Fix invalid comparison functions
Date: Sat, 02 Jan 2016 02:18:00 -0000 [thread overview]
Message-ID: <CAJOtW+5gASN5cyWZzVxBsfyDg_6wA2WGqF0T-BmBGAw6rzjZaA@mail.gmail.com> (raw)
In-Reply-To: <56844E1B.20507@redhat.com>
On Thu, Dec 31, 2015 at 12:35 AM, Pedro Alves <palves@redhat.com> wrote:
> On 12/30/2015 09:25 PM, Pedro Alves wrote:
>> On 12/30/2015 08:18 PM, Yuri Gribov wrote:
>>
>>> Sorry, I should have been more wordy about the actual problem. With
>>> current approach i.e.
>>>
>>> if (pid1 == pgid1)
>>> return -1;
>>> else if (pid2 == pgid2)
>>> return 1;
>>>
>>> comparison of two group leaders is not going to be symmetric:
>>>
>>> cmp(lead_1, lead_2) == cmp(lead_2, lead_1) == -1
>>
>> Aaaaaaah, d'oh! Thanks, it's obvious now, yes, we fail to consider
>> the case of both elements being leaders. I couldn't see that
>> even after staring at the code for a while. That hunk is OK as
>> is then. (Please clarify this in the commit log.)
>
> Wait, no we don't... If both are leaders when you get there, then
> they must have different pgid's, and that case is handled before:
>
> /* Sort by PGID. */
> if (pgid1 < pgid2)
> return -1;
> else if (pgid1 > pgid2)
> return 1;
> else
>
> But let's assume not. Let's assume we see two leaders when you get to the
> code in question. That means they have pgid1==pgid2. Since by definition
> being leader means pgid==pid, it must be that pid1 == pid2 as well. That
> is, this is really about comparing equivalent elements. Which
> brings us back again to:
>
> /* Easier to check for equivalent element first. */
> if (pid1 == pid2)
> return 0;
>
> Or am I confused again?
Hm, this sounds reasonable. Let me see if I can repro this bug and get
the exact inputs which caused the problem (unfortunately I didn't
collect them during testing).
-Y
prev parent reply other threads:[~2016-01-02 2:18 UTC|newest]
Thread overview: 8+ messages / expand[flat|nested] mbox.gz Atom feed top
2015-12-15 11:51 [PATCH][PR " Yury Gribov
2015-12-29 7:31 ` [PATCH][PING][PR " Yury Gribov
2015-12-29 17:27 ` Pedro Alves
2015-12-29 18:09 ` Yury Gribov
2015-12-30 20:18 ` Yuri Gribov
2015-12-30 21:25 ` Pedro Alves
2015-12-30 21:35 ` Pedro Alves
2016-01-02 2:18 ` Yuri Gribov [this message]
Reply instructions:
You may reply publicly to this message via plain-text email
using any one of the following methods:
* Save the following mbox file, import it into your mail client,
and reply-to-all from there: mbox
Avoid top-posting and favor interleaved quoting:
https://en.wikipedia.org/wiki/Posting_style#Interleaved_style
* Reply using the --to, --cc, and --in-reply-to
switches of git-send-email(1):
git send-email \
--in-reply-to=CAJOtW+5gASN5cyWZzVxBsfyDg_6wA2WGqF0T-BmBGAw6rzjZaA@mail.gmail.com \
--to=tetra2005@gmail.com \
--cc=gdb-patches@sourceware.org \
--cc=palves@redhat.com \
--cc=ppluzhnikov@google.com \
--cc=stanshebs@google.com \
--cc=y.gribov@samsung.com \
/path/to/YOUR_REPLY
https://kernel.org/pub/software/scm/git/docs/git-send-email.html
* If your mail client supports setting the In-Reply-To header
via mailto: links, try the mailto: link
Be sure your reply has a Subject: header at the top and a blank line
before the message body.
This is a public inbox, see mirroring instructions
for how to clone and mirror all data and code used for this inbox