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Re: bash --debugger on a script with no arguments


From: Rocky Bernstein
Subject: Re: bash --debugger on a script with no arguments
Date: Wed, 29 Apr 2015 13:51:07 -0400

Now I am really confused and I am not sure we are talking about the same thing. In bash-4.3.30 on ftp://ftp.gnu.org/gnu/bash at line 723 of shell.c there is: 

  if (debugging_mode && locally_skip_execution == 0 && running_setuid == 0 && dollar_vars[1])
    start_debugger ();

Shouldn't dollar_vars[1] be dollar_vars[0] (the name of the script to be debugged)? 

It might be a good idea to add an else branch to give a warning, but that warning would warn about the fact that suid was set, or no debugged script was given, or even -n (skip_execution) was given so debugger main file is getting read. But this is different from the problem encountered.




On Wed, Apr 29, 2015 at 10:57 AM, Chet Ramey <address@hidden> wrote:
On 4/28/15 10:11 PM, Rocky Bernstein wrote:
> It's come to my attention that running bash --debugger doesn't source
> DEBUGGER_START_FILE when the script to be debugged isn't followed by any
> arguments.

This was changed last November in the devel branch as part of a set of
changes to make the absence of the debugger start file a visible error.
The change will be in the next bash release.

--
``The lyf so short, the craft so long to lerne.'' - Chaucer
                 ``Ars longa, vita brevis'' - Hippocrates
Chet Ramey, ITS, CWRU    address@hidden    http://cnswww.cns.cwru.edu/~chet/


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