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From: ark@alice.UucP (Andrew Koenig)
Newsgroups: net.lang.c,net.micro.pc,net.unix
Subject: Re: Fun with * and &
Message-ID: <5694@alice.uUCp>
Date: Mon, 23-Jun-86 15:20:16 EDT
Article-I.D.: alice.5694
Posted: Mon Jun 23 15:20:16 1986
Date-Received: Wed, 25-Jun-86 03:32:43 EDT
References: <1250@ncoast.UUCP>
Organization: Bell Labs, Murray Hill
Lines: 41
Xref: watmath net.lang.c:9561 net.micro.pc:8811 net.unix:8361

> Easy:  Use the Algol 68 method.  (Snide remarks about sh and adb source to
> /dev/null or /dev/bourne, please.)  Translate * as ``REF''.  Then they look
> like:
>
>		int *pi;	->	int REF pi;
>		x = *pi;	->	x = REF pi;
>
> REF is, of course, short for ``reference'', which is just another word for
> ``pointer''.  (Note that ``x = *pi'' is really a DEREFERENCE, since you're not
> merely using the ``refrence'' to the interger, but the integer itself.  That's
> the confusion of C.  ``*'' -> ``REF'' is a way to remember it.)

But that's not how Algol 68 works.  Instead, you write:

		REF INT pi;

which says that pi is bound to a reference to a reference to an
integer (less formally, a variable of type REF INT).  You can
also say:

		INT x;

which defines x as an integer variable (formally a REF INT), and then:

		x := pi;

which implicitly dereferences pi, or

		x := INT (pi);

which explicitly dereferences it.

The easy way to remember how C pointer declarations work is that

		int x;

says that x is an int, and

		int *x;

says that *x is an int, so x is a pointer to int.