Re: bit ops on unsigned long?

[Richard Gooch <rgooch@ras.ucalgary.ca>]

Rusty Russell writes:
> In message <200203250245.g2P2jQa20821@vindaloo.ras.ucalgary.ca> you write:
> > Rusty Russell writes:
> > > Richard: 3 bugs in devfs.  Particularly note that the memset was
> > > bogus.  I can't convince myself that your memcpy & memset stuff is
> > > right anyway, given that you can ONLY treat them as unsigned longs
> > > (ie. bit 31 will be in byte 0 or byte 3, depending on endianness).
> > 
> > Yes, the memset is bogus because I didn't cast the pointer to a
> > char * or void *.
> 
> Yes.
> 
> > The memcpy should be fine, though. And so should
> > everything else, because the bitfield array is allocated in 16 byte
> > multiples.
> 
> No:
[...]
> These changed are required because otherwise you try to do set_bit on
> something not aligned as a long on all archs.

But of course. I'm not denying that. Naturally the type should be
changed. I thought that was obvious so I didn't bother agreeing. But
in fact, it already *is* aligned on a long boundary. Better, in
fact. It's aligned on a 16 byte boundary. Even though the type was
__u32.

> (Turning to the gallery) I assert: if you're going to do bitops on it,
> make it a "unsigned long".

Agreed.

> > So there should be no issues with big vs. little endian,
> > since memset/memcpy operations are done in blocks of sufficient
> > alignment.
> 
> I think you're right, as long as length is always a multiple of
> sizeof(long).  This is not obvious from this hunk of code alone, which
> is why I queried it...
> 
>     if (space->num_free < 1)
>     {
> 	if (space->length < 16) length = 16;
> 	else length = space->length << 1;

Assuming sizeof (long) <= 16 bytes, then length will always be a
multiple of sizeof (long). So, even for a 128 bit CPU, this code is
fine. It might break down on a 256 bit CPU...

Anyway, it looks like we're in agreement.

				Regards,

					Richard....
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