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Date: Sat, 20 Feb 2010 14:06:00 -0000
From: Stefano Sabatini <stefano.sabatini-lala@poste.it>
To: "Alfred M. Szmidt" <ams@gnu.org>
Cc: gdb@sources.redhat.com
Subject: Re: How to show the file being executed?
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On Saturday 2010-02-20 07:37:22 -0500, Alfred M. Szmidt wrote:
>    (gdb) info target
>    Symbols from "/home/stefano/src/PROGRAM".
>    Unix child process:
> 	   Using the running image of child Thread 0xb5f33b70 (LWP 8998).
> 	   While running this, GDB does not access memory from...
>    Local exec file:
> 	   `/home/stefano/src/PROGRAM', file type elf32-i386.
> 	   Entry point: 0x804bc70
> 	   0x08048134 - 0x08048147 is .interp
> 	   0x08048148 - 0x08048168 is .note.ABI-tag
> 	   0x08048168 - 0x0804818c is .note.gnu.build-id
> 	   [...]
> 
>    There is some way to show *only* this information?, otherwise I suggest
>    to implement a show file command.
> 
> You can use readelf to get most of that information.

I want to show *only* the local exec file as specified by the "file"
command, and within a gdb script, currently the only way I see is to
use info target.

Regards.