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Re: __MIPSEL__ in sys32_rt_sigtimedwait

To: Daniel Jacobowitz <dan@debian.org>
Subject: Re: __MIPSEL__ in sys32_rt_sigtimedwait
From: "Maciej W. Rozycki" <macro@ds2.pg.gda.pl>
Date: Wed, 21 Jan 2004 14:47:17 +0100 (CET)
Cc: Ralf Baechle <ralf@linux-mips.org>, Pavel Kiryukhin <savl@dev.rtsoft.ru>, linux-mips@linux-mips.org
In-reply-to: <20040120193918.GA2108@nevyn.them.org>
Organization: Technical University of Gdansk
Original-recipient: rfc822;linux-mips@linux-mips.org
References: <400D6877.1000105@dev.rtsoft.ru> <20040120183157.GB5495@linux-mips.org> <20040120193918.GA2108@nevyn.them.org>
Sender: linux-mips-bounce@linux-mips.org
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.

 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        +

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