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 <chet.ra...@case.edu> 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 c...@case.edu
> http://cnswww.cns.cwru.edu/~chet/
>
