Xref: utzoo comp.lang.c:9350 comp.arch:4350
Path: utzoo!mnetor!uunet!lll-winken!lll-tis!mordor!sri-spam!sri-unix!quintus!ok
From: ok@quintus.UUCP (Richard A. O'Keefe)
Newsgroups: comp.lang.c,comp.arch
Subject: Re: Bit Addressable Architectures
Message-ID: <878@cresswell.quintus.UUCP>
Date: 15 Apr 88 03:56:47 GMT
References: <11702@brl-adm.ARPA> <243@eagle_snax.UUCP> <2245@geac.UUCP> <23396@bbn.COM>
Organization: Quintus Computer Systems, Mountain View, CA
Lines: 19

In article <23396@bbn.COM>, emiller@bbn.com (ethan miller) writes:
> In article <877@cresswell.quintus.UUCP> ok@quintus.UUCP (Richard A. O'Keefe) writes:
> =>In article <8646@eleazar.Dartmouth.EDU>, major@eleazar.Dartmouth.EDU (Lou Major) writes:
> =>> sizeof (foo) == sizeof (char *)
> =>Wrong.  The answer *is* 16.  This is one of the few cases where
> =>foo and &(foo[0]) are different.  I _tried_ this to make sure I was right.
> =>That's always a good idea.
> Sure is.  What did you try?  _I_ just tried printing foo and &(foo[0]), and
> they are the same.  BTW, I also tried sizeof (foo), and it is 16.

foo and &(foo[0]) as expressions have different types:
	foo	  : array-of-16-chars
	&(foo[0]) : pointer-to-char
sizeof notices this difference.  In almost any other context, there is an
implicit conversion to pointer-to-char form.  In particular, print() is not
going to reveal the difference.  Consider
	short x = 1; int y = 1;
	printf("%d %d\n", x, y);
Printing obscures the difference between x and y, and in just the same way
it obscures the difference between foo and &(foo[0]).