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Re: memset (0, 0, 0);
- From: Daniel Jacobowitz <drow at mvista dot com>
- To: "Thomas,Stephen" <stephen dot thomas at superh dot com>
- Cc: Andrew Cagney <ac131313 at redhat dot com>,"Rennecke,Joern" <joern dot rennecke at superh dot com>,gdb at sources dot redhat dot com, newlib at sources dot redhat dot com,bug-glibc at gnu dot org, "McGoogan,Sean" <sean dot mcgoogan at superh dot com>
- Date: Mon, 7 Apr 2003 09:07:31 -0400
- Subject: Re: memset (0, 0, 0);
- References: <9FF3133289A7A84E81E2ED8F5E56B379604384@sh-uk-ex01.uk.w2k.superh.com>
On Mon, Apr 07, 2003 at 10:22:04AM +0100, Thomas,Stephen wrote:
> 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.
MontaVista Software Debian GNU/Linux Developer