Re: memset (0, 0, 0);

Daniel Jacobowitz
2003-04-07T13:07:47Z

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