memset (0, 0, 0);

[Geoff Keating]

"Thomas,Stephen" <stephen.thomas@superh.com> 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 <geoffk@geoffk.org>