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Date: Mon, 07 Apr 2003 13:07:00 -0000
From: Daniel Jacobowitz <drow@mvista.com>
To: "Thomas,Stephen" <stephen.thomas@superh.com>
Cc: Andrew Cagney <ac131313@redhat.com>,
	"Rennecke,Joern" <joern.rennecke@superh.com>,
	gdb@sources.redhat.com, newlib@sources.redhat.com,
	bug-glibc@gnu.org, "McGoogan,Sean" <sean.mcgoogan@superh.com>
Subject: Re: memset (0, 0, 0);
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	Andrew Cagney <ac131313@redhat.com>,
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On Mon, Apr 07, 2003 at 10:22:04AM +0100, Thomas,Stephen wrote:
> Hi,
> 
> gdb appears to call memset(0,0,0) from build_regcache() in gdb/regcache.c. I can't really claim to understand how this works, but this function appears to get called 3 times during gdb initialization:
> 
>   static void build_regcache (void)
>   {
>     ...
>     int sizeof_register_valid;
>     ...
>     sizeof_register_valid = ((NUM_REGS + NUM_PSEUDO_REGS) * sizeof (*register_valid));
>     register_valid = xmalloc (sizeof_register_valid);
>     memset (register_valid, 0, sizeof_register_valid);
>   }
> 
> On the 1st time of calling, none of the gdbarch stuff is set up, so NUM_REGS = NUM_PSEUDO_REGS = 0. So xmalloc gets called with size=0. That returns 0 as the 'address', which gets passed to memset. I guess this just works OK on other architectures (it does on x86 anyway).
> 
> Easy enough to fix I suppose, but is that really the point?

Yes, I think that really is the point.  It's just a bug, IMO.

-- 
Daniel Jacobowitz
MontaVista Software                         Debian GNU/Linux Developer