Date: Fri, 10 Jun 2005 13:59:22 +0300 From: Ruslan Ermilov <ru@freebsd.org> To: Poul-Henning Kamp <phk@phk.freebsd.dk> Cc: Dag-Erling Sm?rgrav <des@des.no>, current@freebsd.org Subject: Re: [current tinderbox] failure on ...all... Message-ID: <20050610105922.GA79872@ip.net.ua> In-Reply-To: <72420.1118396985@critter.freebsd.dk> References: <20050610094615.GC79474@ip.net.ua> <72420.1118396985@critter.freebsd.dk>
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On Fri, Jun 10, 2005 at 11:49:45AM +0200, Poul-Henning Kamp wrote:
> In message <20050610094615.GC79474@ip.net.ua>, Ruslan Ermilov writes:
>
>
> >> struct t1 { int a; } x;
> >> struct t2 { int a; } y = { 42 };
> >> x = y;
> >>
> >> The types `struct t1' and `struct t2' are not compatible and thus not
> >> assignable. See 6.2.7 and 6.5.16.1.
> >>
> >If you're to byte-copy say t1 to t2, is it guaranteed to work? That is,
> >do both types are guaranteed to have the same size and alignment of their
> >structure members? I'm pretty sure this is guaranteed, as lot of code
> >assumes this, for example, the sockaddr* structures.
>
> I do not belive that is guaranteed. (If it were the structs might as
> well have been made assignable). You need to make sure that the two
> definitions are covered by the exact same compilation conditions,
> and you can't tell if a compiler has an option along the lines of
>
> -fstruct_is_magic=t2
>
> Which does weird things you don't know about.
>
> The fact that it mostly works (and that we rely on this) is a
> testament to the fact that compiler writers emply their destructive
> creativity elsewhere.
>
Well, the above example isn't quite different from when compiling two
modules that use the same structure type t1 but one of them is compiled
with -fstruct_is_magic=t1. :-)
Cheers,
--
Ruslan Ermilov
ru@FreeBSD.org
FreeBSD committer
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