INTRODUCTION



        pc^.row := V;
	pw^.col := H;

     Problem 2.
        (includes the above common code)
        (* again, procedure execution code *)
        pc := rowcrs;
        pw := colcrs;
        V := Pc^.row;
        H := pw^.col;

     Did you get lost? Don't feel bad. I really felt that this could be
written in a simpler fashion, but I wanted to present a version which I
felt reasonably sure would work under most circumstances.
     The type declarations are necessary simply to establish record
formats which can be pointed to (and it was these record formats which
I felt to be redundant). Then the variables which indeed point to these
record formats are declared. Most importantly, the "absolute" type
allows us to inform the Pascal compiler that we have a constant which
really is (honest, really, please let it be) the address of one of those
record formats we wanted to point to. (And it is this "absolute" type
which is the extension of Pascal which is not in the standard.)
     Once we have made all our declarations, the code looks
surprisingly like the C code: assign the absolute address to the pointer
and then fetch or store via the pointer. The overhead of the record
element reference (the ".row" and ".col") is the only real difference
(and perhaps unneeded, as I stated).

PILOT
And here we are at last at the simplest of the Atari languages. Again,
standard PILOT has no defined way of accessing individual memory
cells. And, again, the reason for this is that PILOT was (and is) a
language designed for use in schools, where the last thing you want is
poking around in memory and crashing the 100 megabyte disk with
next year's budget on it.
     However, when using PILOT on an Atari computer, the worst
anyone can do is to crunch their own copy of their own disk or cassette.
So Atari has thoughtfully provided a way to access memory cells from
PILOT; and they have done it in a fashion that is remarkably
reminiscent of BASIC. Once more, the solution is given first:

     Problem 1.
        C:@B84 = #V
        C:@B86 = #H/256	
	C:@B85 = #H\256

     Problem 2.
        C:#V = @B84
        C:#H = @B85 + (256 * @B86)