Re: gdb command line option -e or -exec usage

[Eli Zaretskii <eliz@gnu.org>]

> Date: Fri, 12 Feb 2010 11:10:08 -0800
> From: Michael Snyder <msnyder@vmware.com>
> CC: "gdb@sourceware.org" <gdb@sourceware.org>,   "Paragi, Ajeykumar B" <ajeykumar.paragi@hp.com>,  "gdb@gnu.org" <gdb@gnu.org>
> 
> Pedro Alves wrote:
> > On Friday 12 February 2010 18:45:01, Michael Snyder wrote:
> >> Paragi, Ajeykumar B wrote:
> >>> Hi,
> >>>
> >>> I have invoked gdb with '-e' command line option on an executable, 
> >>> The gdb manual says '-e' is used to set the executable to run.
> >>> But I could not see the symbols read in by gdb when I use -e and could
> >>> not insert breakpoints.
> >>>
> >>> What is the intent of having the command line option "-e" or "-exec"?
> >> Hmmm, it does seem to be broken.
> > 
> > It's not.  `-e' is equivalent to "(gdb) exec-file".  It doesn't read
> > its argument for symbols, only for executable to run, or for pure
> > memory contents.
> > 
> >> You can get the same effect, though, by just leaving out the -e flag.
> >>
> >>      % gdb my-executable
> > 
> > That's equivalent to "(gdb) file", which is equivalent to
> > "gdb -e my-executable -s my-executable", or just "gdb -se my-executable".
> > 
> 
> Right!  Thanks, Pedro.

Are we missing something on the manual regarding this?