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 +
