Re: CVS commit: src/sys/dev/usb

[Valery Ushakov <uwe%stderr.spb.ru@localhost>]

On Sat, Apr 21, 2018 at 16:27:05 +0200, Manuel Bouyer wrote:

> On Sat, Apr 21, 2018 at 05:09:27PM +0300, Valery Ushakov wrote:
> > 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 <bouyer%antioche.eu.org@localhost>
> > > Subject: Re: CVS commit: src/sys/dev/usb
> > > To: Martin Husemann <martin%duskware.de@localhost>
> > > Cc: Christos Zoulas <christos%astron.com@localhost>, source-changes-d%NetBSD.org@localhost
> > > 
> > > 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
> 
> Yes, what's confusing is that in this case hashtbl and &hashtbl are
> the same thing, while for
> 	char *hashtbl;
> they are different objects.

Well, as the example above shows, they are _not_ the same thing.  They
just happen to have the same numeric value (memory address).

This is also exactly why in that expression above hashtbl is better
than &hashtbl.  Well, rahter, why &hashtbl is arguably incorrect.

-uwe