Date: Mon, 15 Jul 2002 14:57:35 +0200 From: Thomas Moestl <tmoestl@gmx.net> To: Terry Lambert <tlambert2@mindspring.com> Cc: Luigi Rizzo <luigi@FreeBSD.org>, Mike Barcroft <mike@FreeBSD.org>, current@FreeBSD.org Subject: Re: different packing of structs in kernel vs. userland ? Message-ID: <20020715125735.GC314@crow.dom2ip.de> In-Reply-To: <3D32B0F1.27EA45EE@mindspring.com> References: <20020714011810.A72236@iguana.icir.org> <20020714203642.GD314@crow.dom2ip.de> <20020714230821.C64412@espresso.q9media.com> <20020715105158.GA314@crow.dom2ip.de> <20020715040008.A85276@iguana.icir.org> <3D32B0F1.27EA45EE@mindspring.com>
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On Mon, 2002/07/15 at 04:24:33 -0700, Terry Lambert wrote:
> Luigi Rizzo wrote:
> > sorry but all this just does not make sense to me.
> >
> > sizeof(foo) should give the same result irrespective of
> > where you use it.
> >
> > Perhaps the best thing would be to put a
> >
> > printf("struct ip_fw has size %d\n", sizeof(struct ip_fw));
> >
> > both in ipfw2.c and somewhere in ip_fw2.c and see if there is
> > a mismatch between the two numbers.
>
> I have to assume that what didn't make sense was that his patch
> worked? 8-).
>
> He's making the valid point that for:
>
> struct foo *fee;
>
> It's possible that:
>
> sizeof(struct foo) != (((char *)&fee[1]) - ((char *)&fee[0]))
No, I do not. In fact, the opposite:
sizeof(struct foo) = (((char *)&fee[1]) - ((char *)&fee[0]))
_must_ always be true, since it is legal to compute the size of
storage needed for an n-element array of struct foo by using
(sizeof(struct foo) * n).
My point was that, because of the above, any padding that might be
required between the first and last member of two struct foo's
immediately following each other must be _included_ in struct foo,
after the last element.
- thomas
--
Thomas Moestl <tmoestl@gmx.net> http://www.tu-bs.de/~y0015675/
<tmm@FreeBSD.org> http://people.FreeBSD.org/~tmm/
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