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RE: memset (0, 0, 0);
- From: "Thomas,Stephen" <stephen dot thomas at superh dot com>
- To: "Andrew Cagney" <ac131313 at redhat dot com>, "Rennecke,Joern" <joern dot rennecke at superh dot com>
- Cc: <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 10:22:04 +0100
- Subject: RE: memset (0, 0, 0);
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)
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?
SuperH (UK) Ltd.
From: Andrew Cagney [mailto:ac131313 at redhat dot com]
Sent: 04 April 2003 16:17
Cc: gdb at sources dot redhat dot com; newlib at sources dot redhat dot com; bug-glibc at gnu dot org; Thomas,Stephen; McGoogan,Sean
Subject: Re: memset (0, 0, 0);
> This conflicts with gdb usage of memset (0, 0, 0); in some places.
> There are three practical questions here:
> - should gdb use this idiom?
> - should all target-specific variants of newlib's memset implement it?
> - should all target-specific variants of glibc's memset implement it?
I'm not sure why you're refering to GDB here. GDB assumes ISO C and
hence should never use memset in ways that violate the ISO C spec. If
it is, then someone gets to fix it.