__MIPSEL__ in sys32_rt_sigtimedwait
Hi all,
my question - does endiannes matters in sigset translation in sys32_rt_sigtimedwait (arch/mips/signal32.c)?
===================
@@ -827,18 +827,10 @@
return -EFAULT;
switch (_NSIG_WORDS) {
-#ifdef __MIPSEB__
case 4: these.sig[3] = these32.sig[6] | (((long)these32.sig[7]) << 32);
case 3: these.sig[2] = these32.sig[4] | (((long)these32.sig[5]) << 32);
case 2: these.sig[1] = these32.sig[2] | (((long)these32.sig[3]) << 32);
case 1: these.sig[0] = these32.sig[0] | (((long)these32.sig[1]) << 32);
-#endif
-#ifdef __MIPSEL__
- case 4: these.sig[3] = these32.sig[7] | (((long)these32.sig[6]) << 32);
- case 3: these.sig[2] = these32.sig[5] | (((long)these32.sig[4]) << 32);
- case 2: these.sig[1] = these32.sig[3] | (((long)these32.sig[2]) << 32);
- case 1: these.sig[0] = these32.sig[1] | (((long)these32.sig[0]) << 32);
-#endif
}
/*===================
Regards,
Pavel Kiryukhin
Quoted 3 lines · Pavel Kiryukhin
Hi all,
my question - does endiannes matters in sigset translation insys32_rt_sigtimedwait(arch/mips/signal32.c)?
Think about where bit 33 ends for a big endian machine with an without the conversion.
Ralf
Quoted 8 lines · Ralf Baechle
Quoted 3 lines · Pavel Kiryukhin
Hi all,
my question - does endiannes matters in sigset translation insys32_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.
Signature · 2 lines
Daniel Jacobowitz MontaVista Software Debian GNU/Linux Developer
Quoted 9 lines · Daniel Jacobowitz
No, I'm pretty sure Pavel's right.
(quoted diff) +0
-5-#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
Quoted 3 lines
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...
Quoted 3 lines
byte address in memory
1 2 3 4 5 6 7 8val 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
Quoted 3 lines
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.
Quoted 4 lines
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
Signature · 3 lines
+ Maciej W. Rozycki, Technical University of Gdansk, Poland + +--------------------------------------------------------------+ + e-mail: [email protected], PGP key available +
Quoted 33 lines · Maciej W. Rozycki
Quoted 9 lines · Daniel Jacobowitz
No, I'm pretty sure Pavel's right.
(quoted diff) +0
-5-#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); -#endifQuoted 3 lines
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...
Quoted 3 lines
byte address in memory
1 2 3 4 5 6 7 8val 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 8val 0 0 0 0 1 0 0 0
Quoted 3 lines
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
Quoted 6 lines
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.
Quoted 22 lines
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.Quoted 4 lines
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
(quoted diff) +3
-0+ Maciej W. Rozycki, Technical University of Gdansk, Poland + +--------------------------------------------------------------+ + e-mail: [email protected], PGP key available +
Signature · 2 lines
Daniel Jacobowitz MontaVista Software Debian GNU/Linux Developer