linux-mips
[Top] [All Lists]

Re: __MIPSEL__ in sys32_rt_sigtimedwait

To: "Maciej W. Rozycki" <macro@ds2.pg.gda.pl>
Subject: Re: __MIPSEL__ in sys32_rt_sigtimedwait
From: Daniel Jacobowitz <dan@debian.org>
Date: Wed, 21 Jan 2004 10:22:47 -0500
Cc: Ralf Baechle <ralf@linux-mips.org>, Pavel Kiryukhin <savl@dev.rtsoft.ru>, linux-mips@linux-mips.org
In-reply-to: <Pine.LNX.4.55.0401211414040.11137@jurand.ds.pg.gda.pl>
Original-recipient: rfc822;linux-mips@linux-mips.org
References: <400D6877.1000105@dev.rtsoft.ru> <20040120183157.GB5495@linux-mips.org> <20040120193918.GA2108@nevyn.them.org> <Pine.LNX.4.55.0401211414040.11137@jurand.ds.pg.gda.pl>
Sender: linux-mips-bounce@linux-mips.org
User-agent: Mutt/1.5.1i
On Wed, Jan 21, 2004 at 02:47:17PM +0100, Maciej W. Rozycki wrote:
> On Tue, 20 Jan 2004, Daniel Jacobowitz wrote:
> 
> > No, I'm pretty sure Pavel's right.
> > 
> > -#ifdef __MIPSEB__
> >     case 1: these.sig[0] = these32.sig[0] | (((long)these32.sig[1]) << 32);
> > -#endif
> > -#ifdef __MIPSEL__
> > -    case 1: these.sig[0] = these32.sig[1] | (((long)these32.sig[0]) << 32);
> > -#endif
> > 
> > Consider a 64-bit sigset.  32-bit userland, 64-bit kernel.  Here's a
> > userland sigset with signal 33 set, only, on a little endian target.
> > Word 1, least significant bit, right?
> 
>  Right, but...
> 
> > byte address in memory
> >     1       2       3       4       5       6       7       8
> > val 0       0       0       0       0       0       0       1
> 
> ... this is incorrect -- it would be right for big-endian; word #1, bit #1
> for little-endian is:
> 
> byte address in memory
>       1       2       3       4       5       6       7       8
> val   0       0       0       0       1       0       0       0
> 
> 
> > Obviously, as a 64-bit integer the sigset looks different.  There it's
> > supposed to be 1 << (33 - 1).
> > val 0       0       0       1       0       0       0       0
> 
>  Again, for little-endian it should actually be:
> 
> val   0       0       0       0       1       0       0       0
> 
> i.e. the whole operation is actually a no-op, except that the 64-bit
> vector is assured to be properly aligned for doubleword accesses.

Re-reading what I wrote, the above was actually supposed to be a
big-endian example.  D'oh!  If you pretend I wrote "big endian" up at
the top, then it makes sense.

>  As a side note -- that's the reason certain C code portability problems
> related to the width of the machine word only get actually discovered when
> problematic software is run on a big-endian processor.  I've been hit by
> this property once -- I was porting a 16-bit program and it appeared to
> run just fine on both a 32-bit (i386) and a 64-bit (Alpha) little-endian
> CPU, but when run on a 32-bit big-endian one (SPARC) I discovered a few
> more bits to be cleaned up.
> 
> > So the correct algorithm to convert a userspace sigset to a kernel
> > sigset is to shift the second word left 32 bits, and leave the first
> > word right aligned, and or them together.  Which is what using the
> > __MIPSEB__ case does.
> 
>  But this conclusion is of course right.
> 
>   Maciej
> 
> -- 
> +  Maciej W. Rozycki, Technical University of Gdansk, Poland   +
> +--------------------------------------------------------------+
> +        e-mail: macro@ds2.pg.gda.pl, PGP key available        +
> 

-- 
Daniel Jacobowitz
MontaVista Software                         Debian GNU/Linux Developer

<Prev in Thread] Current Thread [Next in Thread>