From Newsgroup: alt.lang.asm

I'm working through the "Introduction To DECSYSTEM-20 Assembly
Programming" book by Stephen Longo. I have the Gorin one as well but the
Longo text jumps into writing actual programs faster. There's examples
that the student is supposed to complete at the end of the
chapters. There's no given sample output and no answers so the student
is on his own. I thought I'd post some of my solutions if people want to
follow along on their PDP-10 with the MACRO-20 assembler.

@exec ch2u5
LINK: Loading
[LNKXCT CH2U5 execution]
Type in an integer: 30
Sum: 50
Dif: 10
@cont
Type in an integer: 10
Sum: 30
Dif: -10


; Chapter 2 user exercise 5
; Accept an integer from the terminal. Add the integer to
; an integer stored in the program and then subtract it
; from the same integer. Assume the integer given is 20.
title ch2u5
search monsym ; Use monitor's symbols

ac1=1 ; Label accumulators
ac2=2
ac3=3

twent: 24 ; The given number 0o24=20 dec
usrnum: 0 ; Store user's type-in number
base10: 12 ; 0o12 = 10 decimal base
ttyin: .priin ; Primary input keyboard
ttyout: .priou ; Primary output TTY

msg1: asciz /Type in an integer: / ; Prompt for user data message
sum: asciz /Sum: / ; Label sum message
dif: asciz /Dif: / ; Label difference message
errmsg: asciz /IO Error/ ; Generic error message!

start: hrroi ac1,msg1 ; Point to prompt message
psout% ; Print it to TTY

; NIN% wants device, radix to 1, 3 and puts results in ac2
move ac1,ttyin ; Point to keyboard
move ac3,base10 ; Indicate base 10
nin% ; Get user integer
erjmp error ; Handle error
movem ac2,usrnum ; Save user's num to memory

; Do the sum part of the problem now
hrroi ac1,sum ; Print sum label
psout% ; To TTY

; NOUT% wants device, number, and base in accum 1, 2, 3
move ac1,ttyout ; Indicate TTY
move ac2,usrnum ; Get user's number
add ac2,twent ; Add 20 decimal
move ac3,base10 ; Indicate base 10
nout% ; Print sum to TTY
erjmp error ; Handle error

movei ac1,15 ; Print a \r
pbout%
movei ac1,12 ; Print a \n
pbout% ; TOPS-20 wants both

; Do the difference part of the problem now
hrroi ac1,dif ; Print diff label
psout% ; To TTY

move ac1,ttyout ; Indicate TTY
move ac2,usrnum ; Retrieve user's num
sub ac2,twent ; Sub 20 decimal
move ac3,base10 ; Indicate base 10 dec
nout% ; Print diff to TTY
erjmp error ; Handle error
haltf% ; Done
jrst start ; Restart prog if desired

error: hrroi ac1,errmsg ; Indicate error
psout% ; Print to screen
haltf% ; Exit
end start ; That's it
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From Newsgroup: alt.lang.asm

Chapter 3, ch3u1 and ch3u2. I find comparisons to be confusing on this
CPU. On x86, you use "cmp", the flags are set, and you jump based on the outcome. In MACRO-20, you seem to need to test for the thing you don't
want, and then continue the execution path from there. If the condition
fires off, it "eats" next instruction, else it goes on. Very odd. See
the "check:" label below for an example of what I mean.

@exec ch3u1
LINK: Loading
[LNKXCT CH3U1 execution]
Enter an integer: 6
6
@cont
Enter an integer: -4
NEG
@cont
Enter an integer: 0
@

; Chapter 3 user exercise 1
; Accept an integer via NIN% (the text has a mistake here).
; If the int is positive, display it
; If the int is negative, print "NEG"
; Assumed to do nothing if zero.
; CAM = compare accumulator to memory location
; CAI = compare accum to immediate value
; l, g, e, n = less than, greater, equal, not equal suffixes
title ch3u1
search monsym ; Use monitor's symbols

ac1=1 ; Label accumulators
ac2=2
ac3=3

base: 12 ; 0o12 = 10 decimal
ttyin: .priin ; Primary input keyboard
ttyout: .priou ; Primary output to TTY

prompt: asciz /Enter an integer: / ; User prompt
negmsg: asciz /NEG / ; Tell user it's negative

start: hrroi ac1,prompt ; Load user prompt
psout% ; Print it to TTY

; NIN% wants device, radix to 1, 3 and puts results in ac2
; Both NIN% and NOUT% require ERJMP, which is like an "else"
move ac1,ttyin ; Get input from keyboard
move ac3,base ; Signal what base to get
nin%
erjmp done ; Don't worry about error

; NIN% will give us an integer, not the ASCII octal for it
; Compare against the integer, not the ASCII value
check: cail ac2,0 ; Is ac2's num less than 0?
jrst pos ; No/else clause, check +/0

neg: hrroi ac1,negmsg ; It's negative, load message
psout% ; and print to TTY
jumpa done ; This condition is handled

pos: caig ac2,0 ; Is ac2's num greater than 0?
jrst done ; Else, it must be 0, goto done

; nout% wants device, number, base in accum 1,2,3 and erjmp
move ac1,ttyout ; Prepare to print positive num
move ac3,base ; Load desired number base
nout% ; Output number to TTY
erjmp done ; Handle error (or not)

done: haltf% ; Either way we're done
jrst start ; User can re-run program
end start ; Tell assember we're done and
; also where the prog starts at


User exercise 2, little caculator (add/sub).

@exec ch3u2
LINK: Loading
[LNKXCT CH3U2 execution]
Enter an integer: 24
Enter another integer: 6
Enter an operation (+ or -): +
30
@cont
Enter an integer: 10
Enter another integer: 30
Enter an operation (+ or -): -
-20
@cont
Enter an integer: *
Error: invalid input or operation
@

; Chapter 3 user exercise 2
; Get two integers via NIN%, PBIN% an operation as earlier
; in the text (only use + or -). Perform the operation and
; output the answer. Give error message if operation is
; not + or -.
title ch3u2
search monsym ; Use monitor's symbols

ac1=1 ; Label accumulators
ac2=2
ac3=3
op=7 ; Save user's operation

int1: 0 ; User's first num
int2: 0 ; User's second num

base: 12 ; 0o12 = 10 decimal
ttyin: .priin ; Keyboard
ttyout: .priou ; Primary output to TTY
add: "+" ; Mathmatical operations
sub: "-"

; User prompts and error messages
getnum: asciz /Enter an integer: /
getnm2: asciz /Enter another integer: /
errmsg: asciz /Error: invalid input or operation/
opmsg: asciz /Enter an operation (+ or -): /

start: hrroi ac1,getnum ; Load prompt for int1
psout% ; And print it

; NIN% wants device, radix to 1, 3 and puts results in ac2
; Both NIN% and NOUT% require ERJMP, which is like an "else"
move ac1,ttyin ; Prepare to get from kbd
move ac3,base ; Requested number base
nin% ; Fetch number to ac2
erjmp error ; Handle error
movem ac2,int1 ; Save user's 1st num

hrroi ac1,getnm2 ; Load prompt for int2
psout% ; And print it to TTY

; NIN% wants device, radix to 1, 3 and puts results in ac2
move ac1,ttyin ; Prepare to get from kbd
move ac3,base ; Requested number base
nin% ; Fetch number to ac2
erjmp error ; If error
movem ac2,int2 ; Save second number

; Numbers 1 and 2 are fetched. Now get the operation
hrroi ac1,opmsg ; Prompt for operation
psout% ; Print to TTY
pbin% ; Get char, either + or -
move op,ac1 ; Save it
pbin% ; Eat \r else stray input
pbin% ; Eat \n too

; Check the operation is a valid one (+ or - only)
camn op,add ; Is it '+'?
jrst plus ; It is equal, do plus
camn op,sub ; Is it '-'?
jrst minus ; It is equal, do minus
jumpa error ; Don't care if otherwise

plus: move ac2,int1 ; Fetch int1
add ac2,int2 ; Add int2 to it
jumpa disp ; Ready to display

minus: move ac2,int1 ; Fetch int1
sub ac2,int2 ; Subtract it from int1

; Display output section. Print the result that we calculated
; above; it is in accumulator 2. We're here by jump or fall-thru.
disp: move ac1,ttyout ; Output device is TTY
move ac3,base ; Load base
; nout% wants device, number, base in accum 1,2,3 and erjmp
nout% ; Print result, in ac2
erjmp error ; Handle error
haltf% ; Else we're finished here
jrst start ; Re-run/cont the program

error: hrroi ac1,errmsg ; Load error message
psout% ; And print to TTY
haltf% ; Exit to monitor
end start ; That's it, indicate start
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From Newsgroup: alt.lang.asm

jayjwa,

In MACRO-20, you seem to need to test for the thing you don't
want, .... If the condition fires off, it "eats" next instruction,
else it goes on. Very odd.

Its not as odd as you might think : imagine the possibly eaten instruction
is a jump. In that case you jump on "the thing you want".

Yes, its a bit of a mind-trick to think about it that way. :-)

You might also think of it as "skip next instruction if true". iow, if the "eaten" instruction is again a jump, the program will take that jump if the comparision is not true.

Ofcourse, nothing stops you from replacing that jump with the (re)setting of
a flag in a register. :-)

Regards,
Rudy Wieser


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From Newsgroup: alt.lang.asm

ch3u3
Lots of comparisons in this one. It still seems backwards.
@exec ch3u3
LINK: Loading
[LNKXCT CH3U3 execution]
Enter a character (A, B to echo, C to quit): A
A
Enter a character (A, B to echo, C to quit): B
B
Enter a character (A, B to echo, C to quit): D
Enter a character (A, B to echo, C to quit): F
Enter a character (A, B to echo, C to quit): C
Program complete
@cont
Enter a character (A, B to echo, C to quit): *
Enter a character (A, B to echo, C to quit): C
Program complete
@

Last night I found that DDT can display with your symbols already
inserted, like this:
@debug ch3u3
LINK: Loading
[LNKDEB DDT execution]
DDT
ch3u3$:
start$b $g

START/ MOVEI ITER,5

gdb seems so friendly and luxurious after using DDT. The text doesn't
warn about dangling newlines (actually \r\n in TOPS-10/20) but your
program will run incorrectly under EXEC but fine under DDT if you don't
handle them. The text for TOPS-10 mentions them and of course C
programmers have been bit by them at some point.

comment $
Chapter 3 user exercise 3
PBIN% a character. PBOUT% the character only if it is A or B.
Place the above in a loop that terminates after the 5th PBIN%
or if the letter C is entered.
$
title ch3u3
search monsym ; Use monitor's symbols

ac1=1 ; Label accumulators
iter=7 ; Iterator for loopage

char: "X" ; User's character

; User prompts and messages
getchr: asciz /Enter a character (A, B to echo, C to quit): /
compl: asciz /Program complete/

start: movei iter,5 ; Init iterator for loops
loop: hrroi ac1,getchr ; Load prompt
psout% ; And print to TTY

pbin% ; Get char to ac1
movem ac1,char ; Save user's char

; Input will be char\r\n, deal with \r\n on input stream. Oddly,
; the text does not tell you this and your program will run fine
; under DDT while debugging but will fail under EXEC.
pbin%
pbin%

; Now look at what character we have
move ac1,char
cain ac1,101 ; Compare to A (0o101)
jrst disp ; It's A
cain ac1,102 ; Compare to B
jrst disp ; It's B
cain ac1,103 ; Compare to C
jrst done ; C - exit loop to done
jumpa pass ; None of these, no output

disp: pbout% ; Display char in ac1
movei ac1,15 ; Load \r
pbout%
movei ac1,12 ; Load \n
pbout% ; End in \r\n

; Check loop condition. Do we restart the process?
pass: subi iter,1 ; Cut 1 from our looper
jumpn iter,loop ; Not zero? repeat loop

done: hrroi ac1,compl ; Signal program complete
psout%
haltf% ; Exit to monitor
jrst start ; Restartable program
end start ; That's it, halt assembler
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From Newsgroup: alt.lang.asm

Two this time, ch3u4 and ch3u5.

Enter two integers and display the greater.
@exec ch3u4
LINK: Loading
[LNKXCT CH3U4 execution]
Enter integer number one: -5
Enter number two: 0
0 is the greater number.
@cont
Enter integer number one: 10
Enter number two: 20
20 is the greater number.
@

comment $
Chapter 3 user exercise 4
Input two integers via NIN%. Output the greater via NOUT%. The
program should be restartable. The text does not say what to
do if they are the same so we'll assume they must be different.
$
title ch3u4
search monsym ; Use monitor symbols

ac1=1 ; Label accumulators
ac2=2
ac3=3
num1=6 ; Use higher accum for nums
num2=7 ; instead of mem locations

base: 12 ; 0o12 = 10 decimal
ttyin: .priin ; Keyboard
ttyout: .priou ; Primary output to TTY

; User prompts and error messages
entnm1: asciz /Enter integer number one: /
entnm2: asciz /Enter number two: /
errmsg: asciz /Error./ ; Very generic
isgtr: asciz / is the greater number./

start: hrroi ac1,entnm1 ; Load user prompt
psout% ; Output to TTY

; Get first integer from user
move ac1,ttyin ; Point to kbd
move ac3,base ; Load base
; NIN% wants device, radix to 1, 3 and puts results in ac2
nin% ; Get integer
erjmp error ; Handle error
move num1,ac2 ; Save integer 1

; Get second integer from user
hrroi ac1,entnm2 ; Load num #2 prompt
psout% ; Output to TTY

move ac1,ttyin ; Point to keyboard
move ac3,base ; Load base
nin% ; Get integer #2
erjmp error ; Handle error
move num2,ac2 ; Save integer 2

; Compare integer 1 to integer 2. Save num1 to ac3 before it
; gets over written and then put it back
move ac3,num1 ; Remember num1 because...
sub num1,num2 ; Cut num2 out of num1
jumpl num1,gtr2 ; Negative? num2 > num1
jumpg num1,gtr1 ; Positive? num1 > num2
jumpa done ; Neither (num1=num2)

gtr2: move ac2,num2 ; Load num2 for NOUT%
jumpa disp ; Display it

gtr1: move ac2,ac3 ; Fetch num1 saved from above

disp: move ac1,ttyout ; Point to TTY
move ac3,base ; Indicate base
; nout% wants device, number, base in accum 1,2,3 and erjmp
nout% ; ac2 already loaded
erjmp error ; Handle error else print
hrroi ac1,isgtr ; Load "greater" message
psout% ; Print to TTY

done: haltf% ; Exit to monitor
jrst start ; Restartable program

error: hrroi ac1,errmsg ; Load error message
psout% ; Print
haltf% ; Exit to monitor
end start ; That's it


Enter 5 integers and note the position of the greatest.

These comparisons still don't make any sense but I can just run them
with DDT and flip them to the condition I don't want so I get the
condition that I don't didn't want.

@exec ch3u5
LINK: Loading
[LNKXCT CH3U5 execution]
Enter number 1:10
Enter number 2:20
Enter number 3:2
Enter number 4:15
Enter number 5:8
The largest integer is at position 2
@cont
Enter number 1:10
Enter number 2:20
Enter number 3:50
Enter number 4:34
Enter number 5:60
The largest integer is at position 5
@

comment $
Chapter 3 user exercise 5
Set up a loop, accept five integers. Use NOUT% to display the
position of the largest integer after they have been entered.
$

title ch3u5
search monsym ; Use monitor symbols

ac1=1 ; Label used accumulators
ac2=2
ac3=3
gtrpos=4 ; Position of greatest int
num=5 ; Number being looked at
pos=6 ; Position of entry
iter=7 ; Iterator for loopage

base: 12 ; 0o12 = 10 decimal
ttyin: .priin ; Keyboard
ttyout: .priou ; Primary output to TTY

; User prompts and messages
etrmsg: asciz /Enter number /
errmsg: asciz /IO error/ ; Generic error message
lrgmsg: asciz /The largest integer is at position /

start: xor num,num ; Zero out number
movei pos,1 ; and init position to first
movei iter,5 ; Do 5 loops, init looper

getnum: hrroi ac1,etrmsg ; Load get number prompt
psout% ; And print

; nout% wants device, number, base in accum 1,2,3 and erjmp
move ac1,ttyout ; Print to TTY position num
move ac3,base ; Load base
move ac2,pos ; The numerical position
nout% ; And print it
erjmp error ; Handle error
movei ac1,72 ; To print ":"
pbout% ; To TTY

; nin% wants device, radix to 1, 3 and puts results in ac2
move ac1,ttyin ; Point to keyboard
move ac3,base ; Load base
nin% ; Get integer to ac2
erjmp error ; Handle error

; Examine number we just got that's in ac2. We have to test for the
; case we *do not* want, because the opcodes are backward (???)
caml num,ac2 ; Is it > what we have?
jrst lesser ; No, jump

move num,ac2 ; This is greater, note it
move gtrpos,pos ; And its position

lesser: addi pos,1 ; Go to next position
subi iter,1 ; Decrement loop counter
jumpn iter,getnum ; Not 0? Get next number

; By now we've the position of the greatest integer. Display it
disp: hrroi ac1,lrgmsg ; Load "largest" message
psout% ; Print to TTY
move ac1,ttyout ; Print to TTY position num
move ac3,base ; Load base
move ac2,gtrpos ; Pos of greatest integer
nout% ; And print it
erjmp error ; Handle error

haltf% ; Exit to monitor
jrst start ; Restartable program

error: hrroi ac1,errmsg ; Print error message
psout% ; To TTY
haltf% ; Exit to monitor level
end start ; That's it


One more user exercise in chapter 3 and I'm on to chapter 4.
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ch3u6

Check positive integers for even/odd without using division.
@exec ch3u6
MACRO: ch3u6
LINK: Loading
[LNKXCT CH3U6 execution]
Enter a positive integer to check for even or odd: 0
@cont
Enter a positive integer to check for even or odd: 3
The number is odd.
@cont
Enter a positive integer to check for even or odd: 120
The number is even.
@cont
Enter a positive integer to check for even or odd: 121
The number is odd.
@

comment $
Chapter 3 user exercise 6
Accept an integer via NIN%. Considering only positive integers,
PSOUT% a phrase stating whether the integer is EVEN or ODD.
Use repeated subtractions of 2 to achieve this.
$
title ch3u6
search monsym ; Use monitor's symbols

ac1=1 ; Label used accumulators
ac2=2
ac3=3
num=4 ; The user's integer

base: 12 ; 10 decimal
ttyin: .priin ; Keyboard is primary input
ttyout: .priout ; Output to TTY

; User promps and messages
etrmsg: asciz /Enter a positive integer to check for even or odd: /
numeve: asciz /The number is even./
numodd: asciz /The number is odd./
errmsg: asciz /Error invalid input./

start: hrroi ac1,etrmsg ; Load get num prompt
psout% ; Print to TTY

; nin% wants device, radix to 1, 3 and puts results in ac2
move ac1,ttyin ; Point to keyboard
move ac3,base ; Indicate base
nin% ; Get integer to ac2
erjmp error ; Handle error
move num,ac2 ; Save user's number

; The text says "consider only positive integers" but having
; the program return "odd" for zero just seems wrong so error
; check that the number is actually positive before we act
jumple num,done ; Exit early if <= 0

; Keep subtracting 2 from the user's number. If we hit zero, it
; is even, if we go below zero, it is odd. Keep looping if it's
; greater than zero until one of the above conditions occurs.
loop: subi num,2 ; Cut 2 from number
jumpg num,loop ; It's still greater than 0
jumpl num,odd ; It's negative, thus odd

even: hrroi ac1,numeve ; Fall-thru, num even
psout% ; Print to TTY
jumpa done ; Already indicated "even"

odd: hrroi ac1,numodd ; Load "odd" message
psout% ; Print to TTY

done: haltf% ; Exit to monitor level
jrst start ; Program is restartable

error: hrroi ac1,errmsg ; Load error prompt
psout% ; And print to TTY
haltf% ; Exit program to monitor
end start ; Tell assembler it's done
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ch4u3 and ch4u4. Exercises 1 and 2 didn't really require
programming. This chapter deals with "arrays", indexing, and address
modes. 1 of the modes seems pretty exotic, two simple, and 1 I can use
if I think about it.

@exec ch4u3
LINK: Loading
[LNKXCT CH4U3 execution]
10 plus 20 using direct addressing: 30
10+20 using immediate addressing: 30
10+20 using indirect addressing: 30
The sum of 10 and 20 using index addressing: 30
@

comment $
Chapter 4 user exercise 3
Write a program that will add the same two integers four times
using a different addressing move for each time.
Exercise 1 does not require a program, but is actually answered
by doing this exercise and exercise #2 is a flow char that's
already drawn in the text.
$
title ch4u3
search monsym ; Use monitor's symbols

ac1=1 ; Label accums
ac2=2
ac3=3

list: num1 ; Label location of first num
num1: 12 ; 10 decimal
num2: 24 ; 20 decimal

base: 12 ; 10 decimal
ttyin: .priin ; Keyboard is primary input
ttyout: .priout ; Output to TTY

; Prompts and user messages. Most of these have embedded \r\n
dirmsg: asciz /10 plus 20 using direct addressing: /
immmsg: asciz /
10+20 using immediate addressing: /
indmsg: asciz /
10+20 using indirect addressing: /
idxmsg: asciz /
The sum of 10 and 20 using index addressing: /

start: hrroi ac1,dirmsg ; Load direct prompt
psout% ; Print to TTY
move ac2,num1 ; Direct to ac2
add ac2,num2 ; Add num2 to ac2
; nout% wants device, number, base in accum 1,2,3 and erjmp
move ac1,ttyout ; Indicate dest is TTY
move ac3,base ; Indicate base
nout% ; Print to TTY
erjmp error ; Handle error

hrroi ac1,immmsg ; Load immediate message
psout%
movei ac2,12 ; Immediate 10 to ac2
addi ac2,24 ; Add 20 immed to ac2
move ac1,ttyout ; Indicate dest is TTY
move ac3,base ; Indicate base
nout% ; Print sum to TTY
erjmp error ; Handle error

hrroi ac1,indmsg ; Load "indirect" message
psout% ; Print to TTY
movei ac2,@list+1 ; Load 10 decimal
addi ac2,@list+2 ; Add next decimal number
move ac1,ttyout ; Point to TTY for output
move ac3,base ; Indicate base 10
nout% ; Print sum TTY
erjmp error ; Just exit on error

hrroi ac1,idxmsg ; Load "index" message
psout% ; And print it
movei ac3,1 ; Set up index (1) to ac3
move ac2,list(ac3) ; Move list+1
addi ac3,1 ; Set index (2) now
add ac2,list(ac3) ; Add to sum list+2
move ac1,ttyout ; Ready to print to TTY
move ac3,base ; Put back base (overwrote it)
nout% ; Number was alread in ac2
erjmp error ; Handle error (or not)

error: haltf% ; Error or not, done
jrst start ; Restartable program
end start ; Assembler's work is done



This next one is basically a FOR loop.

@exec ch4u4
LINK: Loading
[LNKXCT CH4U4 execution]
The sum is 15
@

comment $
Chapter 4 user exercise 4
Code the following high-level language notation into assembly
language:

FOR I = I TO 3
SUM = SUM + A(I)
NEXT I
PRINT SUM

Assume the array A contains 4, 2, 9.
$
title ch4u4
search monsym ; Use monitor's symbols

ac1=1 ; Label accumulators
sum=2 ; Tally array sum here
ac3=3
iter=4 ; Iterator for loopage

base: 12 ; 10 decimal
ttyin: .priin ; Keyboard is primary input
ttyout: .priout ; Output to TTY
A: exp 4, 2, 9 ; Our array/list, idx 0-2

start: xor sum,sum ; Clear sum
xor iter,iter ; Init iter, loop 0-2
loop: add sum,A(iter) ; Add element at A(iter)
addi iter,1 ; Increment loop counter
caie iter,3 ; Did we do ALL list yet?
jrst loop ; No, re-loop

; If we're here it's time to print the total sum. hrroi can
; take a message string directly like so:
hrroi ac1,[ asciz /The sum is /]
psout% ; And print

; nout% wants device, number, base in accum 1,2,3 and erjmp
move ac1,ttyout ; Indicate TTY is dest
move ac3,base ; Indicate base 10 dec
nout% ; sum should still be in ac2
erjmp done ; Handle error (or not)

done: haltf% ; Exit to monitor
jrst start ; Restartable program
end start ; Tell assembler it's done


Something odd happens with DDT: I label the accumulators, but there's no distinction between the accumulator's label that I gave it and a
plain-jane value that occurs. DDT does not differentiate. For example,
above in the "loop:" area, I compare to *values* but the debugger labels
those values with what I assigned to accumulators, like this:

LOOP+1/ ADDI ITER,AC1 $x
ITER/ SUM AC1
LOOP+2/ CAIE ITER,AC3

This is the code that made that:
addi iter,1
caie iter,3

In other words, in the above example, DDT should have left the "1" and
the "3" as is, as the values they are. I'm not sure if this can be
fixed, or how. Had I an instructor I'd certainly ask him because, in
that context, I'm refering to numerical values and not accumulators.
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ch4u5

I must remember to use an accumulator for indexing, else I would
scribble into some other memory. I'm guessing there's no equal of a
"segfault" on PDP-10? Once I overwrote part of premsg but the program
still ran. On MS-DOS you could scribble all over, too.

Reverse parts of an array, and I'm onto chapter 5. In the previous
example, I wrote array A: as 4, 2, 9. Of course there's no "9" in octal
but the assembler assumed it was decimal and the program ran anyway. It
should be as below in this example.

@exec ch4u5
LINK: Loading
[LNKXCT CH4U5 execution]
The array before swapping: 429
The array after swapping: 492

comment $
Chapter 4 user exercise 5
Write a program that will interchange the second and
third elements in the array 4, 2, 9 (decimal) and print
the new array. The program should operate as this high-level
example:

DUMMY = A(I)
A(I) = A(I + 1)
A(I + 1) = DUMMY
$
title ch4u5
search monsym ; Use monitor's symbols

ac1=1 ; Label accumulators
ac2=2
ac3=3
iter=4 ; Iterator for looping loopage
tmp=5 ; Temp swap space
idx=6 ; Indexer

base: 12 ; 10 decimal
ttyout: .priout ; Primary output to TTY
A: exp 4, 2, 11 ; Array in octal (4,2,9)

; User messages and carriage-return
premsg: asciz /The array before swapping: /
posmsg: asciz /The array after swapping: /
cr: asciz /
/

start: hrroi ac1,premsg ; Load pre-swapping message
psout% ; And print it

; Print the array as-is. This will be without commas or spaces
; but that's fine for an example execise. Note indexers MUST
; be IN accumulators.

xor iter,iter ; Clear iterator
; nout% wants device, number, base in accum 1,2,3 and erjmp
loop1: move ac1,ttyout ; Indicate dest is TTY
move ac2,A(iter) ; Load number from array A
move ac3,base ; Indicate base
nout% ; Print it
erjmp error ; Handle error

addi iter,1 ; Increment loop counter
caie iter,3 ; Time to stop or not?
jrst loop1 ; No it is not

hrroi ac1,cr ; Print a \r\n
psout%

; Swap the array elements by moving the contents of the memory to
; a temp accumulator and then moving that into its new spot in A()
; The second element is index 1 and the third is index 2. No other
; elements need swapping according to the directions. Indexers for
; MOVEM MUST be IN accumulators. Using 1, 2, etc directly does not work.
swap: movei tmp,@A+1 ; Load 2nd element to tmp location
movei ac1,@A+2 ; Load 3rd element to temp location
movei idx,1 ; Indicate we want index pos 1
movem ac1,A(idx) ; Overwrite 2nd element with 3rd
movei idx,2 ; Indeicate we want index pos 2
movem tmp,A(idx) ; Overwrite 3rd element with 2nd

; Print out revised array
hrroi ac1,posmsg ; Load "post" message
psout% ; Print to TTY

xor iter,iter ; Clear iterator
; nout% wants device, number, base in accum 1,2,3 and erjmp
loop2: move ac1,ttyout ; Indicate dest is TTY
move ac2,A(iter) ; Load number from array A
move ac3,base ; Indicate base
nout% ; Print it
erjmp error ; Handle error

addi iter,1 ; Increment loop counter
caie iter,3 ; Time to stop or not?
jrst loop2 ; No it is not

hrroi ac1,cr ; Print a \r\n
psout% ; To TTY because we're done
haltf% ; Done
jrst start ; Restartable program

error: haltf% ; Some error occured
end start ; Tell assembler we're done
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ch5u6

Make a table with A..Z and hex equal.

There's a small error in the text, explained below. This is the end of
chapter 5.

@exec ch5u6
LINK: Loading
[LNKXCT CH5U6 execution]
A 41
B 42
C 43
...
X 58
Y 59
Z 5A
@

comment $
Write a table to that will make an ASCII table of capital letters
and their hexadecimal values, A..Z. The text has an error in that
it labels A..Z as 0x41..0x7A when really it is 0x41..0x5A. 'z'
(lower case) is 0x7A, not 'Z' (capital letter).
$
title ch5u6
search monsym ; Use monitor's symbols

ac1=1 ; Label accumulators
ac2=2
ac3=3
iter=4 ; Loop iterator for loopage

base: 20 ; Base 16 in octal
ttyout: .priout ; Output to TTY
tab: 11 ; ASCII tab char in octal
cr: asciz /
/ ; \r and \n together

start: movei iter,"A" ; Init to "A". Need double quotes. prtchr: move ac1,iter ; Read to print via PBOUT%
pbout% ; Send character
move ac1,tab ; Read to send \t
pbout% ; Send character

; nout% wants device, number, base in accum 1,2,3 and erjmp
move ac1,ttyout ; Point to TTY
move ac2,iter ; Load num for printing hex
move ac3,base ; Base hexadecimal
nout% ; Print character in hex
erjmp done ; Error? Just exit.

hrroi ac1,cr ; Print \r\n after one complete line
psout% ; Ready for next line

cail iter,"Z" ; Is it still less than "Z"?
jrst done ; Yes, "eat" done. Did I mention
; how asinine PDP-10 compares are?
addi iter,1 ; Increment character
jumpa prtchr ; Do printing loop again

done: haltf% ; End program
jrst start ; Restartable program
end start ; Tell assembler we're done
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Disregarding appendices, Chapter 6 is almost 40% of the book. Why they
packed so much into this chapter I don't know; going will be slow. On
top of that there's a dozen or so new opcodes and it's all those logic/shift/rotate/bit commands that make you wonder why you started
into assembly in the first place.

ch6u1 - make alphabet chars lowercase
ch6u2 - make alphabet chars uppercase (as modified)

There's an issue with ch6u2 because, while it looks like the mirror of
ch6u1, it specifies bit 29 while likely they meant bit 30. My solution
has both. This wouldn't be the first time there was a discrepancy in the
text versus the example code.

@exec ch6u1
LINK: Loading
[LNKXCT CH6U1 execution]
Enter a single character: a
a
@cont
Enter a single character: B
b
@cont
Enter a single character: 6
6

@exec ch6u2
LINK: Loading
[LNKXCT CH6U2 execution]
Enter a single character: A
A
@cont
Enter a single character: b
B
@cont
Enter a single character: 8
ruA
@cont
Enter a single character: 3

comment $
PBIN% a character, set bit 30 and print it via PBOUT% What
happens when you PBIN% a capital letter? A digit?
$
title ch6u1
search monsym

ac1=1 ; Label used accumulators

bitmsk: ^B100000 ; Mask to set bit 30, equal 32 decimal
getmsg: asciz /Enter a single character: / ; User prompt message

start: hrroi ac1,getmsg ; Load user prompt message
psout% ; Print to TTY
pbin% ; Get the character
ior ac1,bitmsk ; Set bit in bitmsk
pbout% ; Print it
pbin% ; Eat \r
pbin% ; Eat \n
haltf% ; Exit to monitor
jrst start ; Restart program
end start ; Tell assembler that's all



comment $
Write a program that will PBIN%. After the character is accepted,
clear bit 29 and PBOUT% the result. What happens when you PBIN%
a capital letter? A digit?

Note the text says "bit 29" but, following the previous exercise,
it might actually mean "bit 30". Clearing bit 29 shifts on the
ASCII table, but clearing bit 30 does the opposite of the previous
exercise (but does not output ASCII numbers). Here I assume it
wants "clear bit 30" since it makes more sense to mirror exercise
ch6u1.
$
title ch6u2
search monsym

ac1=1 ; Label accumulators used

getmsg: asciz /Enter a single character: / ; User input prompt
mask1: ^B0100000 ; Mask for bit 30 set
mask2: ^B1000000 ; Mask for bit 29 set

start: hrroi ac1,getmsg ; Prompt user for character
psout% ; Print to TTY
pbin% ; Get char from keyboard
; Use 'mask1' for how I think the text meant the exercise, or 'mask2'
; for how the text is written verbatum.
andcm ac1,mask1 ; Clear bit according to mask
pbout% ; Print new value
pbin% ; Eat dangling \r
pbin% ; and also \n
haltf% ; Exit to monitor
jrst start ; Restartable program
end start ; End assembly
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ch6u3
This one took some thinking. Examine each bit in a user-entered integer
and print HIGH or LOW for each. My solution prints the original digit as
binary and then each number after (so that I could verify that it
worked). I examine the integer from the LSB side. The exercise does not
specify which direction to go in. Rigth-to-left was easiest.

@exec ch6u3
LINK: Loading
[LNKXCT CH6U3 execution]
Enter a decimal integer to examine: 54
110110
LOW
11011
HIGH
1101
HIGH
110
LOW
11
HIGH
1
HIGH
@cont
Enter a decimal integer to examine: 2
10
LOW
1
HIGH
@cont
Enter a decimal integer to examine: 0
0
LOW

comment $
Accept an integer in base 10. Examine each bit. If set, print the
word HIGH, followed by a CR. If not set, print LOW and a carriage
return (example of parallel-to-serial conversion).
$
title ch6u3
search monsym ; Use monitor's symbols

ac1=1 ; Label accumulators used
ac2=2
ac3=3
ac4=4

ttyin: .priin ; Primary input is keyboard
ttyout: .priout ; Primary output to TTY
base: 12 ; Input base, 10 decimal
mask: ^B1 ; Compare with LSB for HIGH

getmsg: asciz /Enter a decimal integer to examine: / ; User prompt
himsg: asciz /HIGH
/
lomsg: asciz /LOW
/ ; Announce high or low for bits msg

start: hrroi ac1,getmsg ; Load prompt for getting integer
psout% ; Print to TTY

; nin% wants device, radix to 1, 3 and puts results in ac2
move ac1,ttyin ; Indicate keyboard is input
move ac3,base ; Request base 10 decimal
nin% ; Get number
erjmp done ; Can't continue if no number

; nout% wants device, number, base in accum 1,2,3 and erjmp
strip: move ac1,ttyout ; Print to TTY
movei ac3,2 ; Indicate binary base
nout% ; Output number
erjmp done ; Just exit on error
movei ac1,15 ; Print \r\n
pbout%
movei ac1,12
pbout%
move ac4,ac2 ; Don't clobber ac2 - we need it
and ac4,mask ; Test current LSB
skipe ac4 ; Is it zero?
jrst high ; No, it's one

low: hrroi ac1,lomsg ; Load "low" message
psout% ; Print to TTY
jumpa next ; One message per condition

high: hrroi ac1,himsg ; Load "high" message
psout% ; Print it; fall-thru to next LSB

next: lsh ac2,-1 ; Logical shift right 1: look at LSB
jumpn ac2,strip ; Num not zero yet? Strip more bits

done: haltf% ; Exit to monitor
jrst start ; Restartable program
end start ; Assembly has concluded
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ch6u4

More bit testing programs. The text didn't mention IDIVI but I knew it
from the TOPS-10 assembly boot. Thankfully I didn't have to use
subtraction to do division/mod. Still no way mentioned to clear the
input buffer. There's got to be a better way than calling PBIN% twice.

Input a character, count bits, and set parity (see comment below at top
of program). I'm going to try out the FAIL assember next. So far this
has been MACRO-20.

@exec ch6u4
LINK: Loading
[LNKXCT CH6U4 execution]
Enter a character to check for parity: a
1100001
Number of bits set: 3
Odd, setting bit 28
11100001
@cont
Enter a character to check for parity: c
1100011
Number of bits set: 4
Even, clearing bit 28
1100011
@cont
Enter a character to check for parity: #
100011
Number of bits set: 3
Odd, setting bit 28
10100011
@

There's no answer key, but that looks correct because '#' = 35 ASCII
decimal which is

Python 3.12.13 (main, Mar 3 2026, 15:06:31) [GCC 15.2.0] on linux
Type "help", "copyright", "credits" or "license" for more information.

print( bin( 35 ) )

0b100011

comment $
PBIN% a character. Count the number of bits set in the word. If the number
of bits is odd, set bit 28. If the number is even, clear 28. This program illustrates even parity - the parity bit is available for checking errors
when transferring information over phone lines.
$
title ch6u4
search monsym

ac1=1 ; Label used accumulators
ac2=2
ac3=3
ac4=4 ; Keep track of each bit to test
count=5 ; Track number of set bits
rem=6 ; Remainder for division

ttyout: .priout ; Primary output to TTY
getnum: asciz /Enter a character to check for parity: /
bitset: asciz /Number of bits set: / ; User prompts and messages
crlf: 15B6+12B13 ; Bit-pack a \r\n
b28msk: ^B10000000 ; Bit 28 mask
tstmsk: ^B1 ; Test a bit mask
char: block 1 ; Save input char for prg end
base10: 12 ; 10 in decimal - display base
base2: 2 ; Base to display char in so
; so we can verify which bits set
subttl input

start: hrroi ac1,getnum ; Load user prompt message
psout% ; Print to TTY
pbin% ; Get character to ac1
movem ac1,char ; And save for later

subttl calculate

setz count, ; Clear count (total)
more: move ac4,ac1 ; Don't clobber ac1 - we need it
and ac4,tstmsk ; Test current LSB
skipe ac4 ; Is it zero?
addi count,1 ; No, count it in total

lsh ac1,-1 ; Look at next LSB
jumpn ac1,more ; More bits to look at?

subttl output

; By now, we have total number of set bits in 'count' and the
; original character saved to 'char'.
pbin% ; Clear input buffer. Eat \r
pbin% ; and \n

; nout% wants device, number, base in accum 1,2,3 and erjmp
move ac2,char ; Reload original character
move ac1,ttyout ; Point to TTY device
move ac3,base2 ; Indicate binary display
nout% ; Output original char as bits
erjmp done ; Handle error

hrroi ac1,crlf ; Emit newline
psout%
hrroi ac1,bitset ; Load "bits set" message
psout% ; Print to TTY

move ac1,ttyout ; Print to TTY the
move ac2,count ; number of bits that are set
move ac3,base10 ; Print number in decimal
nout% ; Output to TTY
erjmp done ; Done either way
hrroi ac1,crlf ; Emit newline
psout%

; Now we look at 'count' and see if it's even or odd. Use mod for this.
div: idivi count,2 ; Divide 2 and check remainder
jumpe rem,even ; No remainder? It's even

odd: hrroi ac1,[asciz /Odd, setting bit 28
/] ; Else it's odd
psout% ; Print to TTY
move ac1,char ; Fetch original character
ior ac1,b28msk ; At odd, set bit 28
jumpa endmsk ; Skip "even" code block

even: hrroi ac1,[asciz /Even, clearing bit 28
/] ; Note even bit number
psout%
move ac1,char ; Fetch orginal character
andcm ac1,b28msk ; At even, clear bit 28

endmsk: move ac2,ac1 ; Display final bits to show
move ac1,ttyout ; the results of masking
move ac3,base2 ; As bits
nout% ; Output to TTY
erjmp done ; Handle error (or not)

done: haltf% ; Exit to monitor
jrst start ; Restartable program
end start ; Assembly is finished
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ch6u5.fai

This is the FAIL assembler. It seems to be a drop-in replacement - at
least for the programming that I'm doing. This program requires to check
a user-input number, look at bit 30, then output bits 30-35 on a special condition (see comment section).

@exec ch6u5
<PROGRAMMING>CH6U5.FAI.6
FAIL: ch6u5
LINK: Loading
[LNKXCT CH6U5 execution]
Enter a number to check bits 30-35: 1230
Your number in bits is 10011001110
@cont
Enter a number to check bits 30-35: 1454
Your number in bits is 10110101110
30-bit is set
New number in octal: 56
@cont
Enter a number to check bits 30-35: 1455
Your number in bits is 10110101111
30-bit is set
New number in octal: 57

comment $
Write a program that accepts a number from the terminal. If the number
sets bit 30, then treat bits 30 to 35 as an octal number and display
this value (I am assuming 30-35 inclusive).
This is the 6 right-most bits: 10110101110
$
title ch6u5
search monsym ; Use monitor's symbols

ac1=1 ; Label used accumulators
ac2=2
ac3=3

ttyout: .priout ; TTY/keyboard input/outputs
ttyin: .priin
mask: ^B111111 ; Mask for bits 30-35 inclusive
mask30: ^B100000 ; Mask for evaluating bit 30
base10: 12 ; Input base
base8: 10 ; Output base
base2: 2 ; Output base for user bits num
crlf: 15B6+12B13 ; Bit-pack a \r\n
char: block 1 ; Save orginal number
getnum: asciz /Enter a number to check bits 30-35: /
bitmsg: asciz /Your number in bits is /
bitset: asciz /
30-bit is set / ; User prompts and messages
newchr: asciz /
New number in octal: /

start: hrroi ac1,getnum ; Load user prompt
psout% ; Display to TTY

subttl input

; nin% wants device, radix to 1, 3 and puts results in ac2
move ac1,ttyin ; Get input from keyboard
move ac3,base10 ; User integer is in decimal
nin% ; Get number
erjmp done ; Or error if can't

movem ac2,char ; Save original number, need later
hrroi ac1,bitmsg ; Load "bit message"
psout% ; Display it
move ac1,ttyout ; Display num to TTY
move ac3,base2 ; Display as bit string
nout% ; Print it
erjmp done ; Or error if can't

and ac2,mask30 ; Match num against 30bit
jumpe ac2,done ; Zero? Not a special number

subttl output

set: hrroi ac1,bitset ; Signal bit 30 set
psout% ; Print to TTY
move ac2,char ; Fetch original number
and ac2,mask ; Mask off bits 30-35
hrroi ac1,newchr ; Load new char message
psout% ; Display to TTY
move ac1,ttyout ; ditto
move ac3,base8 ; Output as octal
nout% ; Send ac2
erjmp done ; Exit either way

done: haltf% ; Exit to monitor
jrst start ; Restartable program
end start ; Tell assembler it's done
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ch6u6.mac

This program wants you prove that two bit strings entered will be:
char1 XOR char2 = ( (NOT char1) AND char2 )
IOR ( char1 AND (NOT char2) )

The confusing part is that the Longo text speaks about NOT as if it is
an opcode. It's not (pun not not intended). The display font for it in
the text even matches XOR and AND. The Gorin text does mention an opcode
to get a complement, but it's SETCM.

@exec ch6u6
LINK: Loading
[LNKXCT CH6U6 execution]
Enter binary string one: 10101
Enter binary string two: 101
The results using AND, IOR, NOT: 10000
The results using only XOR: 10000
@cont
Enter binary string one: 1010
Enter binary string two: 111
The results using AND, IOR, NOT: 1101
The results using only XOR: 1101

comment $
Accept two binary words using NIN% base 2. Perform AND, IOR, and NOT
operations yielding the XOR of the words. Display (NOUT%) the results
using base 2. Using the same words, XOR them and display the results
to see they are the same as the previous calculations.
There's no "NOT" opcode. SETCM is what you're likely supposed to
use.
$
title ch6u6
search monsym ; Use monitor's symbols

ac1=1 ; Label used accumulators
ac2=2
ac3=3

char1: block 1 ; Storage for the first word
char2: block 1 ; and the second word
ttyout: .priout ; TTY/keyboard input/outputs
ttyin: .priin
base2: 2 ; Output base for user bits num
get1: asciz /Enter binary string one: / ; User prompts and messages get2: asciz /Enter binary string two: /
andres: asciz /The results using AND, IOR, NOT: /
xorres: asciz /
The results using only XOR: /

start: hrroi ac1,get1 ; Load 'string one' message
psout% ; Print it

subttl input
; nin% wants device, radix to 1, 3 and puts results in ac2
move ac1,ttyin ; Get from keyboard
move ac3,base2 ; base 2
nin% ; the number
erjmp done ; Handle error (or not)
setam ac2,char1 ; Save number 1

hrroi ac1,get2 ; Load 'string two' message
psout% ; and print it
move ac1,ttyin ; Get from keyboard
move ac3,base2 ; base 2
nin% ; the number
erjmp done ; Handle error (or not)
setam ac2,char2 ; Save number 2

subttl calculate and output
hrroi ac1,andres ; Print 'and' result before we
psout% ; need to use those accumulators

; How nice of the text to give us the algorithm for this. We have plenty
; of registers on the PDP-10, let's use them so as not to clobber memory
; because we need the orginal values later for the XOR part of this.
; char1 XOR char2 = ( (NOT char1) AND char2 ) IOR ( char1 AND (NOT char2) )
setcm 4,char1 ; Complement mem at char1
move 5,char2 ; Fetch char2 from memory
and 4,5 ; First part is done, in 4
move 6,char1 ; Fetch char1 from memory
setcm 7,char2 ; Complement mem at char2
and 6,7 ; Second part is done, in 6
ior 4,6 ; Final result in 4

; nout% wants device, number, base in accum 1,2,3 and erjmp
move ac2,4
move ac1,ttyout
move ac3,base2
nout%
erjmp done

hrroi ac1,xorres ; Load 'xor' message
psout% ; Print to TTY
setm 4,char1 ; Fetch char1 again
setm 5,char2 ; Fetch char2 again
xor 4,5 ; Do xor part of algorithm
move ac2,4 ; Read to print it
move ac1,ttyout ; Print to TTY
move ac3,base2 ; Print as binary
nout%
erjmp done ; Handle error (or not)

done: haltf% ; Exit to monitor
jrst start ; Restart program?
end start ; Assembler is done
--
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ch6u7.mac

Demo a jump table. This is the last program in chapter 6 then onto
chapter 7.

@exec ch6u7
MACRO: ch6u7
LINK: Loading
[LNKXCT CH6U7 execution]
Input a 1, 2, or 3 to demo a jump table: 1
ONE
@cont
Input a 1, 2, or 3 to demo a jump table: 2
TWO
@cont
Input a 1, 2, or 3 to demo a jump table: 3
THREE
@cont
Input a 1, 2, or 3 to demo a jump table: 4
@

comment $
Accept a character from the terminal. If the character is a 1, 2, or
3, then index into the appropriate position in a jump table that
points to locations that print ONE, TWO, or THREE.
$
title ch6u7
search monsym

ac1=1
ac2=2
ac3=3

ttyin: .priin ; Input from keyboard
base10: 12 ; Base 10 in octal
inmesg: asciz /Input a 1, 2, or 3 to demo a jump table: /
crlf: 15B6+12B13 ; Bit-pack a \r\n

start: hrroi ac1,inmesg ; Prompt user for integer
psout% ; Display message

subttl input

; nin% wants device, radix to 1, 3 and puts results in ac2
move ac1,ttyin ; Load input device
move ac3,base10 ; Indicate radix
nin% ; Get integer
erjmp done ; Handle error

caile ac2,3 ; If greater than 3 this is
jrst done ; an error. Exit.
caige ac2,1 ; Same with less than 1
jrst done
subi ac2,1 ; The jump table is zero-indexed
jrst tbl(ac2) ; Valid option, use as idx 2 tbl

subttl jump table

tbl: jrst tbl.1 ; Jump table via index
jrst tbl.2
jrst tbl.3


tbl.1: hrroi ac1,[asciz /ONE/] ; Table entry for ONE
psout%
jumpa tbl.4 ; Skip rest of the output
tbl.2: hrroi ac1,[asciz /TWO/] ; Table entry for TWO
psout%
jumpa tbl.4 ; Skip other text outputs
tbl.3: hrroi ac1,[asciz /THREE/] ; Table entry for THREE
psout% ; Fall-thru on this choice
tbl.4: hrroi ac1,crlf ; Emit newline for proper
psout% ; output after table message

done: haltf% ; Exit to monitor
jrst start ; Restart program?
end start ; Assembler is done
--
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From Newsgroup: alt.lang.asm

ch7u1.mac
ch7u2.mac

Chapter 7 is about pointers. Yup, even in assembler we suffer
pointers. Also literals, and moving bytes in and out of memory according
to pointers.

The first program locates the letter "A" in a string and tells you the position. The second program simulates RDTTY%.

@exec ch7u1
LINK: Loading
[LNKXCT CH7U1 execution]
Found "A" at position 3

comment $
Write a program that will print the numerical positions of the
letter "A" in a string. Use a 7-bit pointer to gain access to
each character. You can write the string into the program via ASCIZ.

This solution starts counting at postion 1, not zero, and accounts
for the character not being in the string at all.
$
title ch7u1
search monsym

ac1=1 ; Label used accumulators
ac2=2
ac3=3
iter=4 ; Loop iteration counter

strptr: point 7,[asciz /BCADEF/] ; String, move "A" around to test fndmsg: asciz /Found "A" at position /
base: 12 ; Base 10 decimal

start: setz iter, ; Clear counter
loop: addi iter,1 ; Track position as we move
ildb ac1,strptr ; Load next byte
skipn ac1 ; If it's asciz's 0 we're at end
jrst notfnd ; and thus didn't find it
caie ac1,"A" ; Is it "A"
jrst loop ; No, keep checking

hrroi ac1,fndmsg ; Load found message
psout% ; Print to TTY
move ac2,iter ; Set up for NOUT%
move ac1,[.priou] ; Set primary output w/literal
move ac3,base ; Decimal base output
nout% ; Print position
erjmp [ move 1,[point 7,[asciz /Error./]]
psout%
haltf% ] ; Error-handling dance

notfnd: haltf% ; Exit to monitor, found or not
lit ; Debugger expand literals
end start ; Assembler is finished


The next program was rough, because it deals with walking around a
pointer to simulate a DEL and a ctrl-U with string input. ctrl-U cancels
the entire input line on TOPS-20 so that's what my handling of ctrl-U
("2" in the program) does in my program.

@exec ch7u2
LINK: Loading
[LNKXCT CH7U2 execution]
Enter a string, 1=DEL, 2=ctrl-U, Enter=END: ab1cd1e

The rendered string: ace
@cont
Enter a string, 1=DEL, 2=ctrl-U, Enter=END: abcdef1

The rendered string: abcde
@cont
Enter a string, 1=DEL, 2=ctrl-U, Enter=END: abcdef2

The rendered string:
@

comment $
Write a program that simulates an RDTTY% by looping over a PBIN%. Use
1 to delete a character, 2 as a ctrl-U, and a carriage return to
terminate the input loop. Print the message after accepting it.
Ex:
AB1CD1E
ACE
$
title ch7u2
search monsym ; Use monitor's symbols

ac1=1 ; Label used accum. General input
ac2=2 ; Generally used for pointers
ac3=3 ; Generally used for ADJBP
count=4 ; Track number of entered chars

size==12 ; Don't show 'size' in debugger
buffer: block size ; 10 words * 5 = 50 ASCII chars
; Each block is 5 chars
ptrbuf: point 7,buffer ; Pointer to buffer
etrmsg: asciz /Enter a string, 1=DEL, 2=ctrl-U, Enter=END: /
renmsg: asciz /
The rendered string: / ; User prompts and messages

start: hrroi ac1,etrmsg ; Prompt user for string
psout% ; Write to TTY

move ac2,ptrbuf ; Get pointer to buffer for input
setz ac1, ; Zero out ac1, it will get chars
setz count, ; No characters input to start
getchr: pbin% ; Main char input loop, to ac1
cain ac1,15 ; \r? If so, it signals end
jrst lf
cain ac1,"1" ; Is it the delete signal (1)?
; Once the user inputs "1", back up the pointer by moving a -1 to an
; accumulator then using ADJBP on the accumulator that has the buffer-pointer. ; After, the result is left in ac3 - move this result back to where the
; buffer-pointer points, thus updating the contents. Express as a literal.
jrst [ movni ac3,1
adjbp ac3,ac2
movem ac3,ac2
jumpa getchr] ; Back up pointer in buffer-pointer
; using -1, then re-enter getchar loop

cain ac1,"2" ; Is it the ctrl-u/blank line char?
jrst [ movn ac3,count
adjbp ac3,ac2
movem ac3,ac2
jumpa getchr] ; Like delete, move pointer back in
; buffer-pointer, but this time
; according to total chars entered

idpb ac1,ac2 ; Stuff char to buff, inc pointer
addi count,1 ; Record that we entered a character
jrst getchr ; Get next character

lf: pbin% ; Eat \n, after \r from above
setz ac1, ; Zero out ac1 to put 0
idpb ac1,ac2 ; Null terminator for PSOUT%

hrroi ac1,renmsg ; Load rendered string message
psout% ; Display it
move ac1,ptrbuf ; Load pointer to buffer
psout% ; Print buff contents to TTY
haltf% ; End program
jrst start ; Restart program
lit ; Debugger to show literals
end start ; Assembler is done
--
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From Newsgroup: alt.lang.asm

ch7u3.mac
ch7u4.mac

The first program emulates MID$ (from BASIC). TOPS-20 has a BASIC so I
could see how it worked. The second program reads from the terminal and
stores strings. Then you read from that memory and convert to integers.

@exec ch7u3
LINK: Loading
[LNKXCT CH7U3 execution]
Enter a short string to demo MID$: Where is Paris?
Enter an integer position to begin the substring: 10
Enter an integer reflecting the substring length: 5
Paris
@cont
Enter a short string to demo MID$: Hello world
Enter an integer position to begin the substring: 0
Invalid start position

comment %
Write a program that simulates a MID$ (in BASIC) function. The program
should accept a string and two numbers and then print the partial string.
Ex:
@basic

READY
10 a$ = "Where is Paris?"
20 print mid$( a$, 10, 5 )
run

NONAME.B20
Thursday, June 25, 2026 11:54:54

Paris
%
title ch7u3
search monsym

ac1=1 ; Label used accumulators
ac2=2
ac3=3
count=4 ; How many chars in substring?

size==36 ; 30 dec, don't show sym in debugger buffer: block size ; 30 words * 5 = 150 ASCII chars ptrbuf: point 7,buffer ; Pointer to input buffer
base10: 12 ; 10 decimal, base we in/output with pos: 0 ; Starting postion of substring
ttyout: .priou ; Display device
ttyin: .priin ; Keyboard input device

strmsg: asciz /Enter a short string to demo MID$: /
inimsg: asciz /Enter an integer position to begin the substring: /
endmsg: asciz /Enter an integer reflecting the substring length: /
outmsg: asciz /
The substring is / ; User prompts and messages

subttl input
start: hrroi ac1,strmsg ; Load user prompt
psout% ; Display
move ac1,ptrbuf ; Set up buffer for RDTTY%
movei ac2,size-1 ; Num chars to read, allow for null end
setz ac3, ; No ctrl-r, not needed on modern TTY
rdtty% ; Read the user's test string
erjmp error ; Handle error

getn1: hrroi ac1,inimsg ; Load prompt to get start pos
psout% ; Display it
; nin% wants device, radix to 1, 3 and puts results in ac2
move ac1,ttyin ; Ready NIN%
move ac3,base10 ; Input base is decimal
nin% ; Get starting pos number
erjmp error ; Handle error
caig ac2,0 ; Don't allow zero or less
jrst [move ac1,[point 7,[asciz /Invalid start position/]]
psout%
jumpa done] ; Exit w/error if not > 1
subi ac2,1 ; Adjust start position from zero
; indexing to one, which is what
; a user would expect
movem ac2,pos ; Save starting position

getn2: hrroi ac1,endmsg ; Load prompt to get num chars
psout% ; Display it
move ac1,ttyin ; Read from TTY
move ac3,base10 ; Still using decimal
nin% ; Get num chars to read
erjmp error ; Handle error
move count,ac2 ; Save num of chars in substring

subttl substring
subs.1: move ac2,ptrbuf ; Load pointer to char buffer
move ac3,pos ; Fetch starting substring position
adjbp ac3,ac2 ; Move pointer to new starting pos
movem ac3,ac2 ; Save result into memory
subs.2: ildb ac1,ac3 ; Load next byte in ptr ac3 for ac1
cain ac1,15 ; \r? If so, we're end-of-string
jrst done ; with nothing more to print, exit
pbout% ; Print ac1's character
sojg count,subs.2 ; Loop if still chars (count)

done: haltf% ; Exit to monitor
jrst start ; Restartable program

; The error message is very generic because this is only an exercise
error: hrroi ac1,[point 7,[asciz /Error, can't demo MID$/ ] ]
haltf% ; Exit to monitor
lit ; Display literals on debug
end start ; Assembler's job is done


This program was harder because the text doesn't spend much time
explaining using NIN% pointed at memory. It gets numbers as strings,
then ORs them and outputs them as numbers. Think atoi() in C. This
program also uses a reprompter (ctrl-r), though outside a paper TTY it's
not that useful.

@exec ch7u4
LINK: Loading
[LNKXCT CH7U4 execution]
Enter first 5-bit binary number: 00101
Enter second 5-bit binary number: 01010
Final result: 01111

comment $
Set up an RDTTY%/NIN% to accept two 5-bit binary numbers. IOR the
numbers; then NOUT% the result in a field of width 5, with leading
zeroes.
Additional info on using RDTTY%/NIN% in conjunction: https://www.bourguet.org/v2/pdp10/jsys-user/chap2 , section 2.9
$
title ch7u4
search monsym

ac1=1 ; Label used accumulators
ac2=2
ac3=3
ac4=4

num1: block 2 ; 2 * 5 = 10 chars space
num2: block 2 ; 2 * 5 = 10 chars space
ptrnm1: point 7,num1 ; Pointer to first buffer
ptrnm2: point 7,num2 ; Pointer to second buff

msg1: asciz /Enter first 5-bit binary number: /
msg2: asciz /Enter second 5-bit binary number: /
finmsg: asciz /Final result: / ; User messages and prompts

start: hrroi ac1,msg1 ; Prompt for first number
psout% ; Display message to TTY
move ac1,ptrnm1 ; Set up buffer for RDTTY%
movei ac2,5+2 ; 5 chars plus \r\n
hrroi ac3,[asciz /First number? /] ; Reprompter message
rdtty% ; Get first number
erjmp [move ac1, [point 7, [asciz /Error getting first number /]]
psout%
haltf%] ; Handle error here

hrroi ac1,msg2 ; Prompt for second number
psout% ; Print to TTY
move ac1,ptrnm2 ; Set up buffer for RDTTY%
movei ac2,5+2 ; 5 chars plus \r\n
hrroi ac3,[asciz /Second number? /]
rdtty% ; Get second number
erjmp [move ac1, [point 7, [asciz /Error getting num 2 /]]
psout%
haltf%] ; Handle error on num2
hrroi ac1,finmsg ; Load results message
psout% ; Print to TTY

; This section uses NIN% to read the inputted data from memory and
; convert it. In this case, give ac1 a pointer to the buffer.
; nin% wants device, radix to 1, 3 and puts results in ac2
conv: move ac1,ptrnm1 ; Load pointer to buffer, num1
movei ac3,2 ; Convert to base binary
nin% ; Fetch number
erjmp [move ac1, [point 7, [asciz /Conversion error/ ]]
psout%
haltf%] ; Handle conv error

move ac4,ac2 ; Park this for a moment
move ac1,ptrnm2 ; Load pointer to buff/num1
movei ac3,2 ; Convert to base binary
nin% ; Fetch second number
erjmp [move ac1, [point 7, [asciz /Conversion error/ ]]
psout%
haltf%] ; Handle 2nd conv error

or: ior ac4,ac2 ; Perform the IOR, result in ac4
; nout% wants device, number, base in accum 1,2,3 and erjmp
move ac1,[.priou] ; Indicate TTY as output
move ac2,ac4 ; The resulting number
movei ac3,2 ; Indicate binary base output

; The left half of ac3 controls the display of the number with NOUT%
; Each bit sets a feature. Note "NO%*" type values do not seem to work
; even though the text mentions them. Use a binary string instead.
hrli ac3,^B1110000000000101 ; Leading 0's, 5 places
nout% ; Print resulting number
erjmp done ; Handle error (or not)

done: haltf% ; Exit to monitor
jrst start ; Restart program
lit ; Debugger expands literals
end start ; Assembler is done
--
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