Re: memset (0, 0, 0);

[Daniel Jacobowitz]

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