Determing alignment of (char *) pointer

[David Eppstein]

In <7381 at utzoo.UUCP>, henry at utzoo.UUCP (Henry Spencer) writes:
> >    if ((long)(p - (char *) 0) & 3) ...
> If I had to pick one, this would be it.

I'm responding to this partly because in his criticism of a different
and worse construction Henry mentioned that it wouldn't work on a PDP-10.

It's been too long since I worked on the relevant compiler, but I
doubt this works on the PDP-10 either.  Normal char pointers like p
have word addresses in their low bits and bitfield stuff in their high
bits as Henry said earlier, but (char *)0 is exactly the zero word
(for efficiency in comparisons and for losers who pass it as an
argument without casting it).  I don't remember what subtracting one
from the other does but it's probably not what you expect.

Also, the &3 part bothers me.  This is ok for the usual char pointers
you see on the PDP-10 because they use 9-bit bytes which pack four to
a word.  But 7-bit bytes packed 5 to a word are very common, and I
seem to recall some other architectures having 3 bytes per word (which
can also be done on the PDP-10 but I've never seen it actually used).

The best way to check pointer alignment on the PDP-10 is
	(int *) p == (int *) (p - 1) /* unaligned if p-1 is in same word */
which works for all types of byte pointers, but I would go with
 	p != (char *) (double *) p
since it works for the 9-bit pointers used most often by C and could
be made to work on most other architectures.

Or better rewrite the code to not need this information in the first place.
-- 
David Eppstein, eppstein at cs.columbia.edu, seismo!columbia!cs!eppstein