May June 1980 - Commodore Computers

1OO 100 COMPUTE. MAY/JUNE. I 198O. 1980. ISSUE 4.
in line 10. Remember that this number is the size
of your dimensioned variables.
Refer R to Table T 1. Locations 140 and 141 form
the program-end address. . Locations 142 and 143 form
the stack address and locations 144 and 145 form the
pointer to the end of memory used by the program.
The RUN command sets all of them equal to the
incorrect end-of-program pointer. . Lines 10 through 40
correct them.
Here H comes the only hard part. You are going
to have to peek around in memory. The RUN R U N
command sets the length of all strings to zero. You
must repair their lengths if you want to save string
data.
Table 1.
Table 1.
These two byte addresses point to (Q important a nt areas used by
Atari Acari Basic.
Use this
To get the location of this
PEEK(130) PEEK( 130) + PEEK(131)*256 PEEK(131)"256 Variable Va,;able name table ,able
PEEK(134) + PEEK(135)'256 PEEK( 135) "256 Variable Va,;able value table ,able
PEEK(136) +PEEK(137)*256+ PEEK(137)"2S6 Beginning Seg;n n; ng of program
Use these only when program running
Use these only when program running
PEEK(HO) PEEK(I40) + PEEK(141)'256
PEEK(141)"256
End of program and d beginning of dimensioned
variables
PEEKO42) PEEK( 142) + PEEK(143)*256
PEEK( 143)"256
End of dimensioned variables and beginning of
stack
PEEK(144) I44) + PEEK(145)*256
PEEK(145)"256
End of memory mory used by program
Look at Table 3. The third entry in the variable
value table is the string ALPHAS. ALPHA$. Its current length
is 5 + +0*256 0'256 or 5. These two bytes must be set to
the correct length of the string. Type the command:
PRINT PEEK(134) + PEEK(135),256. PEEK(135)*256.
Now you
know where the variable value table is. If you have
been writing the program in the listings, you should
get the answer 2056. Assume the string is the first
entry in the table. The locations of the length is
Table 3
Table 3
The variable value table; table: Each entry is eightt bytes.
V ariable
Variable
ABI AB1
Table Entry
Table T 2
The T variable name table: Entry lengths are different. differcllI.
Box symbolizes that 128 is added to ASCII value of last
character to show the name's end. cnd .
Variable
Variable name
Variable Variable name
AB1 ASI
AB [Tj
AR{3,4) AR(3,4)
AR[JJ
CD
CDOG(17)
I CDOGjTJ
IT]
ALPHAS (10) ALPHA[|j ALPHAI1I
E [El [])
FIG
FI [G]
AS ill
Note: Variable names can be up to 120 characters long
and are a completely unique. Variable ABCA i.s is different
from variable ABCD. A Variable names DO NOT appear
in the program in memory. Only a I1 byte pointer to
the variable name in the variable name table appears.
Table Entry
Contents
1234561 2 3 4 5 6 7 8 Meaning
, 01 0/ 0/64/,.0, 0/64/5,0. 0,0, 0 First byte is 0; 0: this is a number. Second byte is 0: this is thee first entry. 64
is the exponent. 5 is the binary coded decimal value.
FI IQ)
3 character number name
2 character array name
e
4 character array a name
6 character string name
1 character number name
3 character number name
Note; Variable names can be up to 120 characters long:
the variable name in the variable name table appears.
2060 anda 2061. Since the length of the string of data
being saved in the example is 10, I set location
2060 to a 10 lOin line 60 of the program.
Try it out
The program is complete. Save it. Now RunR it. You
will probably get garbage in the print out. Put a
10 character string in the string variable. Now put
numbers in various entries in the B array. Typing
a 9 for the I subscript will end the program with a
CSAVE. C Do this CSAVE onto a new tape. TurnT the
computer off and on. Now load this new copy of the
program and run it. Viola! TheT data is still there!
Now load this new copy of the program and run it.
Viola! V ! The data is still there! Now just imagine that
this was your budget information, , address book or
other files. . You have a no-hassle one-tape system.
.
Conclusion
I have provided more information about this internals
of the Atari than is really necessary to solve
thiss problem. If you are interested in this kind of
information, , study it. If not, skip it. If you have
any questions, I would be glad to answer them. . One
warning. Doo not press break while the program p'rogram is
running and then type RUN. Always use the CONT
AR(3,4)
AR(3,4)
doe,n',
doesn't
matter
matter
64 + lillO, 1/1/0, 0/4, 0/5,0
64 makes this an array. + I 1 means thatt it has been dimensioned. 0 ° +0*256 is the
displacement displacement into thee array area. 4 + 00*256 is the size of the first dimension and
5 +0-256 +0*256 is the size of the second dimension. .
CDOG( 17)
CDOG(17)
doe,n',
doesn't
matter
64 + 1/2/120,0/18,0/1, 120.0/ 18,011,00
This array is displaced 120 bytes into the aarray aarea, and it is ddimensioned
18+0*256+ 0 by 1 +0*256.
ALPHAS( IO) " 1234S "
ALPHA$(10)
FIG
"12345'
E .05
.05
FIG -S
128 + 1/31228,Of!), 1/3/228,0/5, OliO, 0/10, 0
01 4/63/5,0, 0,0, 0
0/ 4/63/5,0, 0,0. 0
01 S/64 + 12815. 0.0, 0
0/ 5/64+ 128/5, 0.0, 0
128 makes this a string. + I 1 means that it has been dimensioned. . It starts
228 + +0*256 bytes into the array area. . The current length of the string is
,) 5 +0*256. The maximum imulll size of the string is 10 +0*256.
This is a number. The exponent is now 63 so the number is only 11 1/100100 of its
integer value, or .05.
This is a minus number ( + 128 on exponent) nt)