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In the first of a new
series, Eric Doyle
introduces you to the
secret of your
Commodore's
memory.
WE ALL TAKE THE CHIPS INside
our computers for granted
but a greater understanding of
them can lead to better
programs. Over the next few
months I hope to clarify the Another difference is that
function of each chip and to ROM is a permanent, nonreveal
the inner workings and volatile store which cannot be
hidden secrets of the Comma- erased by turning the power on
dare range of computers: the and off but RAM is volatile and
Vic 20, C64, C-16, Plus/4 and its contents disappear when
C126. t h e computer is switched off.
The heart of any computer ROM is where the Basic
is the central processor and the operating system is stored and
most common application of at power up it reserves certain
the computer is to run Basic parts of RAM for storing the
programs so this is where our transient values generated as its
Odyssey will begin, r o u t i n e s are executed.
The microprocessor in
Commodore micros is a one of
several derivatives of the
Motorola 6500 series microchip.
This processor is the
number cruncher inside which
are the registers that assist in all
of the computer's mathematical
operations. Diagram 1
shows the basic architecture of
all 6500 range microprocessors.
As far as the processor is
concerned the whole of the
computer memory is an extension
of itself from which
numerical values can be loaded
(read) or sent (written).
A good way of imagining
memory is as a large, pigeon
hole internal mailing system.
Each box represents a byte of
memory and can contain a
value from zero to 255. Don't
worry if you don't know how
computers deal with numbers
larger than 255, all will become
clear later.
Memory comes in two
varieties Read Only Memory
(ROM) and Random Access
Memory (RAM). As the name
implies ROM can only be read
from but RAM can either be
read from or written to if it is
necessary to change its value.
For example variables defined
by a Basic program must be
stored in RAM for two reasons.
Firstly, because it must have a
value written to it to start with
and, secondly, because that
value may change later on in
the program.
Processing
Load and run Listing 1 to see
how the processor operates.
The program counter tells
the processor where the
current instruction is located in
RAM/ROM memory. When
commanded to execute a piece
of machine code, the current
value of the program counter is
stored in a special reserved
PRO6RAM: LISTING 1
0
180 L=0-0/16;60SUB200
10 0013:60SU8360
190 RETURN
20 LLsPEEX(S+1)4PEEK(S+2)125 200 IF 09 THEN VS=CHRS(550.
6
)+VS;6010220
30 LC20
210 VS.RIGHTS(STRICL1pLEN(ST
40 LN2PEEK(S+3)+PEEK(S+4)125 RS(L))-1)0$
61PRINT"(CLR)(WHT)LIMEN;LN:" 220 RETURN
(VW'
230 LL=PEEK(S+LC+1)+PEEK(S+L
50 LO.LNIGOSUB280:1FLCO0T Cf2)$256
HEN50
240 FLINO:IFLL.OTHEN LC=0:PRI
60 A:S400
NTNICYN)00 1
70 PRINT
1:0010 250 PRINT:00110:001PEEK(S+LCi.
6EASE
WAIT'
270 3)+PEEK(S+LC+4)$256:IFLN=40T
(YEL) 80 PIA+1:1FPEEK(A)21ANDPEEK( HENLC80:60T0270
(DOWN A4.1)1.143THEN100
260 PRINTI(DOWN)(WHT)LINPO.
1(DOW 90 SOT080
N)PL 100 PRINT"(UP) PRESS °
N;a(CYN)°
270 RETURN
280 WEEK(PLC):GOSUB170
110 6ET AS:IF A$O" THEN11 290 IF 00.2 AND FL.0 THEN PR
0
INTs(YEL)';
120 SA-4
300 IF CO.4 AND FL.0 THEN PR
130 LL.PEEK(S4.1).0PEEK(S4.2)12 INTICLDLUPpFL.1
56
310 IF CO.CN THEN PRINT:C0=0
140 LN=PEEK(64.3)+PEEK(S+4)12 320 IF FL.1 AND D>127 AND D<
56:PRINT*(CLR)(WHT)LINE";LN; 203 THEN PRINP(GRN)
':00.0:FL:0
1LU)';:60T0340
150 LO:LC+1:60SUB280:IFL000 VSe 330 PRINT (LB Vs";
THEN150
340 CORC04.1:IF S.H.C2LL-1 THE
160 PRINT
N 60SUB230
:END 1
350 RETURN
170 00(LBL YS.":L=D AND 15:60SUD20 360 PRINT
U)':LI6T
1
440—
(OLR)
•IO (DOWN)
(DOWN)
(
area of RAM memory (the
processor stack area) and the
internal stack pointer is adjusted
to point to the next free
location in the stack. The start
address of the new machine
code routine is then placed
into the counter. This value is
then loaded into the address
buffer which directs the data
bus to the correct location.
The data bus copies the
information found in the given
location and carries this back to
the processor which is expecting
a machine code operator.
When this is evaluated in the
instruction decoder it determines
whether an operand or
two will follow. Depending on
the type of operator, any
operands are evaluated and
stored in the X or Y register or
in the accumulator,
If the operator is a finite
address the address buffer is
given this value and the data
bus reacts accordingly. If the
value is an offset address, the
finite address has the value of
the X or Y register added to it
and this value is passed to the
address buffer.
All mathematical work is
performed in the ALL,' which
can access all of the registers
which merely act as passive
stores for transient values.
The instruction decoder
determines whether the data
bus is reading or writing its
encoded information and
which internal register
provides or accepts the
information.
When an RTS command is
detected the last value to be
stored on the stack is read into
the program counter and the
processor continues from
where it was before it was
asked to execute the routine.
I have said that the processor
only responds to machine
code routines so how does it
respond to Basic?