From: Francesco Potorti` <pot@gnu.org>
Cc: "Richard M. Stallman" <rms@gnu.org>, emacs-devel@gnu.org
Subject: Re: Problem report #1
Date: Tue, 02 May 2006 11:40:48 +0200 [thread overview]
Message-ID: <E1FarNA-0003c5-00@pot.isti.cnr.it> (raw)
In-Reply-To: "dann@ics.uci.edu"'s message of Sun, 30 Apr 2006 09:39:50 -0700
>CID: 1
>Checker: DEADCODE (help)
>File: emacs/lib-src/pop.c
>Function: pop_open
>Description: After this line (or expression), the value of "password" cannot be 0
I do not maintain pop.c, but I looked into this since rms asked for it.
>Event cannot_single: After this line (or expression), the value of "password" cannot be 0
>Also see events: [dead_error_line][dead_error_condition][cannot_single]
>
>255 if ((! password) && (! DONT_NEED_PASSWORD))
>256 {
>257 if (! (flags & POP_NO_GETPASS))
>258 {
>259 password = getpass ("Enter POP password:");
>260 }
>261 if (! password)
>262 {
>263 strcpy (pop_error, "Could not determine POP password");
>264 return (0);
>265 }
>266 }
>267 if (password)
>268 flags |= POP_NO_KERBEROS;
>269 else
>
>Event dead_error_line: Cannot reach this line of code
>Also see events: [dead_error_condition][cannot_single][cannot_single]
>
>270 password = username;
This is true. Given the above check and error message, we always have a
password at this stage, which means that the POP_NO_KERBEROS flag is
unconditionally set. As a consequence, KPOP is never used.
I have no idea why this is it, maybe we should ask the authors of the
kerberos code. If no one else can do it, I can look into it, but I
never programmed kerberos before.
Please put me in CC when answering this.
next reply other threads:[~2006-05-02 9:40 UTC|newest]
Thread overview: 4+ messages / expand[flat|nested] mbox.gz Atom feed top
2006-05-02 9:40 Francesco Potorti` [this message]
2006-05-02 21:37 ` Problem report #1 Richard Stallman
2006-05-03 9:54 ` Francesco Potorti`
-- strict thread matches above, loose matches on Subject: below --
2006-04-30 16:39 Dan Nicolaescu
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