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Re: __MIPSEL__ in sys32_rt_sigtimedwait

To: Ralf Baechle <ralf@linux-mips.org>
Subject: Re: __MIPSEL__ in sys32_rt_sigtimedwait
From: Daniel Jacobowitz <dan@debian.org>
Date: Tue, 20 Jan 2004 14:39:18 -0500
Cc: Pavel Kiryukhin <savl@dev.rtsoft.ru>, linux-mips@linux-mips.org
In-reply-to: <20040120183157.GB5495@linux-mips.org>
Original-recipient: rfc822;linux-mips@linux-mips.org
References: <400D6877.1000105@dev.rtsoft.ru> <20040120183157.GB5495@linux-mips.org>
Sender: linux-mips-bounce@linux-mips.org
User-agent: Mutt/1.5.1i
On Tue, Jan 20, 2004 at 07:31:57PM +0100, Ralf Baechle wrote:
> On Tue, Jan 20, 2004 at 08:42:15PM +0300, Pavel Kiryukhin wrote:
> 
> > Hi all,
> > my question - does endiannes matters in sigset translation in 
> > sys32_rt_sigtimedwait (arch/mips/signal32.c)?
> 
> Think about where bit 33 ends for a big endian machine with an without
> the conversion.

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?

byte address in memory
        1       2       3       4       5       6       7       8
val     0       0       0       0       0       0       0       1

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

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.

-- 
Daniel Jacobowitz
MontaVista Software                         Debian GNU/Linux Developer

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