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Re: memset (0, 0, 0);
From: |
Geoff Keating |
Subject: |
Re: memset (0, 0, 0); |
Date: |
07 Apr 2003 10:17:38 -0700 |
User-agent: |
Gnus/5.0808 (Gnus v5.8.8) Emacs/20.7 |
"Thomas,Stephen" <address@hidden> writes:
> 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?
xmalloc is never supposed to return 0, and in fact, there's code to
prevent it:
if (size == 0)
size = 1;
newmem = malloc (size);
if (!newmem)
xmalloc_failed (size);
return (newmem);
xmalloc_failed finishes with
xexit (1);
so xmalloc should never return NULL.
--
- Geoffrey Keating <address@hidden>
- memset (0, 0, 0);, Joern Rennecke, 2003/04/04
- Re: memset (0, 0, 0);, Daniel Jacobowitz, 2003/04/04
- Re: memset (0, 0, 0);, Andrew Cagney, 2003/04/04
- Re: memset (0, 0, 0);, Andreas Schwab, 2003/04/04
- Re: memset (0, 0, 0);, Petr Vandrovec, 2003/04/04
- RE: memset (0, 0, 0);, Thomas,Stephen, 2003/04/07
- Re: memset (0, 0, 0);, Wolfram Gloger, 2003/04/07
- Re: memset (0, 0, 0);, Daniel Jacobowitz, 2003/04/07
- Re: memset (0, 0, 0);,
Geoff Keating <=
- RE: memset (0, 0, 0);, Thomas,Stephen, 2003/04/07
- RE: memset (0, 0, 0);, Thomas,Stephen, 2003/04/08