INTRODUCTION



locations 85 and 86, a "two byte" value. What does that mean? How
can something occupy two locations? Actually, it all stems from the
fact that a single location (byte, memory cell, character, etc.) in an
Atari computer can store only 256 different values (usually numbered
0 to 255). If you need to store a bigger number, you have to use more
bytes. For example, two contiguous bytes can be used to store 65536
different values, three bytes can store 16,777,216 different values, etc.
     Since the Atari graphics mode can have as many as 320 columns,
we can't use a single one-byte location to store the column number.
Great! We'll simply use two bytes and tell BASIC that we want to talk to
a bigger memory cell. What's that? You can't tell BASIC to use a
bigger memory cell? Oops.
     Ah, but have no fear. We can still perform the task; it just takes a
little more work in BASIC. The first sub-problem is to break the
column number (variable "H") into two "pieces," one for the first byte
and one for the second. The clearest way to accomplish this is with the
following code:

     H1 = INT(H/256)
     H2 = H - 256 * H1

     Because of the nature of machine language "arithmetic," numbers
designed to be two-byte integers must usually be divided as shown: the
"high order byte" must be obtained by dividing the number by 256,
and any fractional part of the quotient must be discarded. The "low
order byte" is actually the remainder after all units of 256 have been
extracted (often designated as "the number modulo 256").
     So, if we have obtained "H1" and "H2" as above, we can change
the cursor row as follows:

     POKE 85,H2
     POKE 86,H1

     Notice the reversal of the order of the bytes! For the Atari (and
many other microcomputers), the low order (or least significant) byte
comes first in memory, followed by the high order (or most significant)
byte.
     Now, suppose we wish to avoid the use of the temporary variables
"H1" and "H2" and further suppose that we would now like to write the
entire solution to the first problem here. Voilą:

     POKE 84,V
     POKE 86,INT(H/256)
     POKE 85,H -256 * INT(H/256)

     And we wrote those last two lines in "reverse" order so that we
could offer a substitute last line, which will not be explained here but
which should become clear a few paragraphs hence:

     POKE 85,H - 256 * PEEK(86)