Re: memset (0, 0, 0);

[Wolfram Gloger]

Hi,

> 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).

Actually on x86-linux, malloc(0) (and therefore xmalloc(0)) returns a
unique address, not NULL.  Both behaviours are permitted by standard
C.

Regards,
Wolfram.