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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