On Sat, Apr 21, 2018 at 09:44:59 +0200, Manuel Bouyer wrote:
> Date: Sat, 21 Apr 2018 09:44:59 +0200
> From: Manuel Bouyer <bou...@antioche.eu.org>
> Subject: Re: CVS commit: src/sys/dev/usb
> To: Martin Husemann <mar...@duskware.de>
> Cc: Christos Zoulas <chris...@astron.com>, source-changes-d@NetBSD.org
>
> On Sat, Apr 21, 2018 at 09:37:46AM +0200, Martin Husemann wrote:
> > On Sat, Apr 21, 2018 at 09:30:20AM +0200, Manuel Bouyer wrote:
> > > > >
> > > > >- axe_cmd(sc, AXE_CMD_WRITE_MCAST, 0, 0, (void *)&hashtbl);
> > > > >+ axe_cmd(sc, AXE_CMD_WRITE_MCAST, 0, 0, hashtbl);
> > > > >
> > > > >missing & ?
> > > >
> > > > uint8_t hashtbl[8] = { 0, 0, 0, 0, 0, 0, 0, 0 };
> > >
> > > So I guess the code was wrong before; not sure how multicast could have
> > > worked.
> >
> > No, the address is only needed as rhs of the cast. If passed directly,
> > the address will be used (due to arrays being passed as pointer to first
> > element in C).
> >
> > Try it:
> >
> > #include <stdio.h>
> > #include <inttypes.h>
> >
> > int main(void)
> > {
> > static uint8_t hashtbl[8] = { 0, 0, 0, 0, 0, 0, 0, 0 };
> >
> > printf("%p vs %p\n", (void *)&hashtbl, hashtbl);
> > return 0;
> > }
>
> I didn't know this. hashtbl and &hashtbl[0] are equivalent, and I would
> assume that &hashtbl is always a pointer to pointer. So the behavior depends
> on hashtbl being declared as pointer or as array ?
> this is confusing ...
&hashtbl is a pointer to an array of size 8. You can see this with
pointer arithmetic:
char hashtbl[8];
printf("%p %p\n%p %p\n", hashtbl, hashtbl + 1, &hashtbl, &hashtbl + 1);
prints
0x7fff54e4b720 0x7fff54e4b721
0x7fff54e4b720 0x7fff54e4b728
-uwe
