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August 1964

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Document:	F-64PX PDP-6 Programming Examples
Order Number:	XX-00828-0F
Revision:
Pages:	27
Original Filename:	http://bitsavers.org/pdf/dec/pdp6/F-64PX_PDP-6_Programming_Examples_Aug64.pdf

OCR Text

F-64PX
8/ 64

PROGRAMMING EXAMPLES

F64-PX
8/64

PDP-6 PROGRAMMING EXAMPLES

DIGITAL EQUIPMENT CORPORATION

•

MAYNARD, MASSACHUSETTS

INTRODUCTION

This manual contains examples of programming for the PDP-6 Type 166 Processor. They have
been chosen to illustrate both the arithmetic and logical capabilities of the processor.
explanation of the instructions shown see the PDP-6 Handbook (F-65).

For an

The examples use the

same instruction mnemonics as the MACRO-6 assembler. The language is described in the
MACRO-6 Manual (F-64MAS).
Times based on design estimates are shown in some of the examples.
have been conservatively calculated.

All of the instruction times

For example, no attempt has been made to take advan-

tage of speed gains due to memory overlapping. Two of the examples show how time ma), be
saved by moving to fast memory a short program which executes a large number of iterations.
One of these, Character Manipu lation, is programmed in both a straightforward manner and by
being moved to fast memory in order to show the break-even point between the time gained
and the increased overhead time. The second, Two-bit Testing, was programmed for 500 iterations. In this case there is a considerable gain in time by moving the program to fast memory.
The last example, Any Radix Print, demonstrates the use of recursion to shorten programs.
Sixteen examples are contained in this booklet:

Single Precision Integer Arithmetic

Double Prec is ion Integer Arithmetic

Floating Point Arithmetic

Fix a Floating Point Number

Float a Fixed Point Integer

Repetitive Calculation

Subscripts

Exponentiation

Character Manipulation

10.

Character Translation

1l.

Character Addition

12.

Fifteenth Degree Polynom ia I

13.

Evaluation of Complex Polynomial

14.

Matrix Inversion

15.

Two-Bit Testing and Depositing of Data

16.

Any Radix Print

SINGLE PRECISION INTEGER ARITHMETIC
Assume:

1) A, B, C, D, E, F, G, H, J, K, L, M, N, and P are arbitrary memory locations.
2) Arguments and instructions are in the same memory module.
3) No scaling is required.
4) Y indicates 1 of the 16 accumulators
Time in microseconds
A=B+C

MOVE Y, B
ADD Y, C
MOVEM Y, A

4
4
4
Total

D=E+F+G
MOVE Y, E
ADD Y, F
ADD Y, G
MOVEM Y, D

4
4
4
4
Total

Total

4
13.6
4
21.6

Total

4
13.6
4
4
25.6

H=JxK
MOVE Y, J
IMUL Y, K
MOVEM Y, H
L=MxN+P
MOVE Y, M
IMUL Y, N
ADD Y, P
MOVEM Y, L

DOUBLE PRECISION INTEGER ARITHMETIC

Assume:

A, B, C, 0, E, F, G, H, J, K, L, M, N, and P are arbitrary memory locations,
each denoting a block of two consecutive memory registers.

Each integer is stored in two consecutive memory locations with the high order integer in the first location and the low order integer in the second.

Instructions and arguments are in same memory module.
Time in dcroseconds
A=B+C

J FC L 16, . + I,
MOVE 0, B

CLEAR STRAY FLAGS

MOVE 1, B + 1
ADD I, C+l,
JFCL2, 01,
02: ADD 0, C,
MOVEM 0, A,
MOVEM I, A+l

ADD LOW ORDER PARTS
DID LOW ORDER PARTS OFLO
ADD HIGH ORDER PARTS
STORE RE SU LTS

01: AOJA 0, 02

2.1

4.0
4.0
4.0
2.1

4.0
4.0
4.0

COMPENSATE FOR OVERFLOW
3.3
Total 28.2-31.5
D=E+F+G
2.1

JFCL16, .+1,
MOVE 0, E
MOVE I, E + 1
ADD I, F + 1,
JFCL 2, DOl,
DOl A: ADD I, G + I,
JFC L 2, 002,
DD2A: ADD 0, F
ADD 0, G
MOVEM 0, 0,
MOVEM I, 0+1

CLEAR STRAY FLAGS

001: AOJA 0, 001 A,
002: AOJA 0, DD2A,

COMPENSATE FOR OVERFLOW
3.3
COMPENSATE FOR OVERFLOW
3.3
Total 38.3-44.5

ADD lOW ORDER PARTS
DID LOW ORDER PARTS OFLO
ADD lOW ORDER PARTS
OVERFLOW?
ADD HIGH ORDER PARTS
STORE ANSWERS

4.0
4.0
4.0
2.1

4.0
2.1

4.0
4.0
4.0
4.0

DOUBLE PRECISION INTEGER ARITHMETIC (continued)
Time in microseconds
K=JxK
MOVE 0, J
MUL 0, K,
MOVE 2, J + 1
MUL 2, K,
JFCL16, .+1,
ADD 1,2,
JFCL 2, M1
M1 A: MOVE 2, J,
MUL 2, K + 1
ADD 1, 2,
JFCL 2, M2
M2A: MOVEM 1, H + 1,
MOVEM, H

M1: AOJA 0, M1A
M2: AOJA 0, M2A

MULTIPLY HIGH ORDER PARTS
MULTIPLY LOW {J)i HIGH (K)
CLEAR STRAY FLAGS
SUM PRODUCTS
OVERFLOW?
MULTIPLY HIGH (J), LOW (K)
SUM PRODUCTS
OVERFLOW?
STORE RESU LTS

COMPENSATE OVERFLOW
COMPENSATE OVERFLOW

4.0
14.0
4.0
14.0
2.1
4.0
2.1
4.0
14.0
4.0
2.1
4.0
4.0

3.3
3.3
Total 76.3-82.9

FLOATING POINT ARITHMETIC
Assume:

1) A, B, C, D, E, F, G, H, J, K, L, M, N, and P are arbitrary memory locations.

2) Arguments and instructions are in the same memory module.
3) Y indicates 1 of the 16 accumulators.

A=B+C
MOVE Y, B
FAD Y, C
MOVEM Y, A

4.0

5.8
4.0
Total 13.8
D=E+F+G
4.0

MOVE Y , E
FAD Y, F
FAD Y, G
MOVEM Y, D

5.8
5.8
4.0
Total 19.6
H=JxK

MOVE Y, J
FMP Y, K
MOVEM Y, H

4.0
12.6
4.0
Total 20.6
L=MxN+P

MOVE Y , M
FMP Y, N
FAD Y, P
MOVEM Y, L

PROBLEM:

4.0
12.6

5.8
4.0
Total 26.4

Consider an eight block table with 100 entries in each block. Let A, B, C, D,
E, F, G, and Hdenote the first location of each block. For each entry find:
G = {A_B)2 + (C_D)2

H = (G/E) x F
Assume:

1) All entries in normalized floating point.
2) Argument and instructions are in the same memory module.

FLOATING POINT ARITHMETIC (continued)

Time in microseconds

Program
BEGIN:

2
4

MOVSI1,+D100
MOVE 2, A(l)
FSB 2, B(l)
FMP 2, 2
MOVE 3, C(l)
FSB 3, D(l)
FMP 3, 3
FAD 2, 3
MOVEM 2, G(l)
FDV 2, E(l)
FMP 2, F(l)
MOVEM 2, H(l)
AOBJN 1, BEGIN+1

6.2
12.3
4

6.2
12.3
5.5
4
18.0
12.8

4.0
3.3
The total time for 100 repetitions:
92.6 x 100 + 2 = 9.26 mill iseconds

FIX A FLOATING POINT NUMBER
Assume:

1) A floating point number in any accumulator from 1-15 designated by F. The
resu It is returned in F+ 1 modu Ie 16.

MUll F, 400,
TSC F, F,
ASH F+ 1, -243(F),

EXPONENT IN F, FRACTION IN F+1
COMPLEMENT EXPONENT IF NEGATIVE
TRUNCATE TO GREATEST INTEGER
FLOAT A FIXED POINT NUMBER

Assume:

1) A fixed point integer less than 227 in magnitude in accumulator C.
TLC C,233000,
FAD C, 0,

XOR INTO WORD
FLOATING ADD ZERO TO NORMALIZE

2) A fixed point integer I, _235:s. I < 235 , in accumulator F.
Note: Accumulator F+1 is used in the calculation.
IDIVI
SKIPE
TLC
TLC
FAD

F,400,
F-,

F,243000,
F+1,233000,
F, F+1,

DIVIDE WORD INTO TWO PIECES
SKIP IF NORMALIZED ZERO
XOR EXPONENT INTO F
XOR EXPONENT INTO F+l
COMBINE AND NORMALIZE

REPETITIVE CALCULATION

The following are repeated 10000 times:
A=B+C, D=E+F+G, H=Jx K

Assume:

1) A, B, C, D, E, F, G, H, J, K, L, M, N, and P are arbitrary memory locations.
2) Arguments are in floating point.
3) Arguments and instructions are in the same memory module.
Time in microseconds

B1 :

MOVE I 2," D1 0000,
MOVE 0, B
FAD 0, C
MOVEM 0, A
MOVE 0, E
FAD 0, F
FAD 0, G
MOVEM 0, D
MOVE 0, J
FMP 0, K
MOVEM 0, H
SOJN 2, B1,

IN ITIALIZE COUNTER

COUNT

The total time for 10000
repetitions:
55.2 x 10000 + 2 = 0.552 sec.

2
4
5.2
4
4
5.2
5.2
4
4
12.6
4
3
55.2

SUBSCRIPTS
Compute for 1=1, 100
A(I)=B(I)+C(I)
D(I)=E (I)+F (I)+G( I)
H(I)=A(I)xD(I)
Assume:

1) The data is arranged in memory as follows:
Bl, B2, ---Bl00, Cl, C2, ---Cl00, El, Fl, Gl,
E2, F2, G2 ---El 00, Fl00, Gl00, A1, D1, A2,
D2 ---A100, Dl00, Hl, H2, ---H100
Time in microseconds

C1:

CLEARB 3, 2,
HRLZI 1, -+D100,
MOVE 4, E(3)
FAD 4, F(3)
FAD 4, G(3)
MOVEM 4, D(2),
MOVE 0, B(1)
FAD 0, C(1)
MOVEM 0, A(2)
FMP 0, 4
MOVEM 0, H(1)
ADDI 2, 2,
ADDI 3, 3,
AOBJN 1, C1,

IN ITIALIZE INDEX REGISTERS
IN ITIAUZE INDEX COUNTER

4.0
2.0

4.0
6.0
6.0
D=E+F+G

4.2
4.0
6.0

4.2
12.3

4.2
INCREMENT 2 STEP INDEX
INCREMENT 3 STEP INDEX
INCREMENT 1 STEP INDEX
AND COUNT

3.0
3.0
3.3

Total Time required is 0.062 seconds

EXPONENTIATION
,FLOATING POINT NUMBER TO A FIXED POINT POWER
,COMPUTE X+-I USING ACCUMULATORS Al, A2, AND T.
,STORE THE RESULT IN Y. IF X IS ZERO, RETURN ZERO
FEXP: MOVE Al, X
MOVSI T, 201400
SKIPGE A2, I
FDVM T, Al
MOVMS A2
JUMPN A1, FEXP2
CLEARB T, A2
FEXP1: FMP A1, A1
LSH A2, -1
FEXP2: TRZE A2, 1
FMP T, A1
JUMPN A2, FEXP1
MOVEM T, Y

iMOVE X TO A1
ill = 1.0

iMOVE I TO A2, SKIP IF NON-NEGATIVE
iTAKE RECIPROCAL OF X (NEGATIVE POWER)
iTAKE ABSOLUTE VALUE OF I
iGO TO MAIN LOOP (IF NON ZERO BASE)
iZERO EXPONENT AND RESULT FOR QUICK EXIT
iSQUARE BASE TERM
iSHIFT RIGHT FOR NEXT BIT OF EXPONENT
ilS POWER A FACTOR? TURN OFF BIT
i YES
iMORE FACTORS?
iNO, STORE RESULT

CHARACTER MANIPULATION
PROBLEM:

Assume:

There is a string of 7 bit ASC II characters beginning at memory location A and
ending with a slash. Transfer the characters, excluding the slash to a block
beginning at location B. Count the number of characters and leave the result
in an index register.
1) The code for a slash is 74 8 ,

Program

Time in microseconds

4
4

MOVE 3, [POINT 7, BJ
MOVE 2, [POINT 7, AJ
MOVEI 1, 0
LOBI 0, 2
CAIN 0, "/11
JRST EXIT
DPBI 0, 3
AOJA 1, C

2.0
5

2.6
2.1
.5
3.4

Total time is 18 + 16 x N where
N is the number of charac ters.
MOVSI AI, CMOVP
BLT Al, Al-I,
JRST CMOV
CMOVP: PHASE 0
B1
0
PTA: POINT7,A
PTB:
POINT 7, B
CMOV: LOBI A1, PTB
CAIN A1, 74
JRST EXIT
DPBI A1, PTB
CM:
AOJA B1, CMOV
DEPHASE
Al = CM + 1

2.0
MOVE TO FAST MEMORY

17.6

3.0

5.0
2.0
2.0
5.0
2.0

The time for this case is 31 + 14 x N.

CHARACTER TRANSLATION
Assume:

1) That the number in accumulator A is a 6-bit code read from the card reader.
The program must translate the card code into the equivalent 7-bit ASC II code.
A translation table begins at location TAB consisting of 7-bit ASCII characters
packed five to a word.
2) The characters in th is table are in order of the ir appearance in the card code.
Because characters are packed five to a word, the quotient of the card code
divided by 5 gives the word in wh ich the ASC II character is found. The remainder gives the character position. An auxilliary table of five byte pointers,
one pointing to each character position, allows retrieval of the proper ASCII
with a single LOB instruction.

TRANSL:

A,5

IDIVI
LOB
JRST

A, BT AB (A+ 1)
EXIT

BTAB:

POINT
POINT
POINT
POINT
POINT

7, TAB (A), 6
7, TAB (A),13
7, TAB (A), 20
7, TAB (A),27
7, TAB (A), 34

TAB:

ASCII
ASCII
ASCII
ASCII
ASCII
ASCII
ASCII
ASCII
ASCII
ASCII
ASCII
ASCII
ASCII

.+-1234.
.56789.
.0=@t'.
.\ 1ST.

.UVWXy.
.Z;,(".
.#7.-JK.

.LMNOP.
.QR:$*.
.[>&+A.
.B CDEF.
IGHI?/
.))<1.

CHARACTER ADDITION
PROBLEM:

Add two 5 digit numbers expressed at 7-bit ASCII characters.

Calling Sequence:
,ASCIAD:

,
,

JSP AC3, ASCIAD

A ROUTINE TO ADD OR SUBTRACT FIVE DIGIT ASCII NUMBERS (7 BIT
CHARACTERS) .
1. CHARACTERS MUST BE RIGHT JUSTIFIED
2. TO ENTER ROUTINE:
A. MOVE ACO, (ADDEND)
B. MOVE AC1, (AUGEND)/ MOVN AC1, (SUBTRAHEND)
C. JSP AC3, ASCIAD
3. ON RETURN THE SUM OR DIFFERENCE IS IN AC2 AND THE CONTENTS
OF ACO AND AC1 ARE UNCHANGED
4. THE ROUTINE IS A RING COUNTER; FOR EXAMPLE
99999+2=00001 and 3-6=99997
5. NOTE THAT TWO NEGATIVE NUMBERS CANNOT BE COMBINED AND
THAT IF ONE IS NEGATIVE IT MUST APPEAR IN AC1 ON ENTRY.

,
ASCIAD:

AND
lOR
TLZN
ADD
ADD
AND
MOVE
AND
ASH
SUBM
lOR
JRST

ACO,M2
AC1,M4
AC1,400000
AC1,M1
ACO,AC1
ACO,M3
AC2, M4
AC2,ACO
AC2,-3
ACO,AC2
AC2,M4
(AC3)

M1 :
M2:
M3:
M4:

BYTE (l) (7) 6, 6, 6, 6, 6
BYTE (1) (7) 17, 17, 17, 17, 17
BYTE (1) (7) 77, 77, 77, 77, 77
BYTE (1) (7) 60, 60, 60, 60, 60
ACO=O
AC1=1
AC2=2
AC3=3
END

FIFTEENTH DEGREE POLYNOMIAL

Assume:

1) P denotes a block of memory conta ining the 16 coeffic ients; X is a memory location containing the argument; the answer is stored in location Z.
Time in microseconds
RADIX 10,
MOVE 3, X,
MOVE 12, 15
MOVE 0, P + 15,
IMUL 0, 3,
ADD 0, P-1 (2)
SOJ GE 2, . -2,
MOVEM 0, Z,

SET ASSEMBLER RADIX TO 10
MOVE ARG TO FAST MEMORY
INITIALIZE INDEX COUNTER
INITIALIZE VALUE
MULTIPLY BY ARGUMENT
ADD NEXT LOWER COEFFICIENT
INCREMENT AND COUNT
STORE ANSWER

4.0
2.0
4.0
13.2

4.0
2.8
4.0

Total time required is 314 microseconds

EVALUATION OF COMPLEX POLYNOMIAL
1
2
n
Y=Po +P 1 X +P2 X + ..... PnX

WHERE Y, X, and P are complex numbers.
The real parts of the coefficients, P, are stored in an array, the first location labeled P. The
imaginary parts are stored in another array, PI. The argument is X (real part) and XI (imaginary
part), the answer is placed in Yand YI, and the order is in N.
DATA STRUCTURE
P: BLOCK 14,
PI: BLOCK 14,

X: a
XI: a

Y: a
YI:O
N: a

REAL COEFFICIENT PARTS
IMAGINARY COEFFICIENT PART
REAL PART OF ARGUMEN T
IMAG. PART OF ARGUMENT
REAL PART OF ANSWER
IMAG. PART OF ANSWER
ORDER OF POLYNOMIAL
Time in microseconds

MOVEI4, N,
MOVE a, P (4),
MOVE 1, PI(4),
MOVE 2, X,
MOVE 3, XI,
P13: MOVE 5, 1
FMP 5, 3,
MOVE 6, a
FMP 6, 3,
FMP a, 2,
FSB a, 5,
FMP 1, 2,
FAD 1, 6,

IN ITIALIZE INDEX COUNTER
INITIALIZE ANSWER
MOVE ARGUMENT TO
FAST MEMORY
PI * XI
P * XI

P*X
P * X - PI * X I = REAL PART
PI * X
P * XI + PI * X = IMAGINARY PART

FAD a, P-1 (4),
FAD 1, PI-1 (4)
SOJGE 4, P13

ADD NEXT LOWER COEFFICIENT

MOVEM a, Y,
MOVEM 1, YI

STORE ANSWER

2.0
4.0
4.0
4.0
4.0
2.4
12.2
2.4
12.2
12.2

5.6
12.2
5.4

6.0
6.2
4.0

TIME = 28 + 80.8N flsec.
Example: A 13th Degree Polynomial Requires 1.04 milliseconds.

4.0
4.0

MATRIX INVERSION
PROBLEI\~:

To invert an NxM matrix, stored row-wise in sequential locations beginning
with A.

,CALLING SEQUENCE:
,CALL: JSP 17, INVER
EXP A
,
JRST ERROR
, THE ORDER OF THE MATRIX IS IN A, WITH THE NUMBER OF ROWS IN THE LEFT HALF,
,AND THE NUMBER OF COLUMNS IN THE RIGHT HALF. THE ELEMENTS ARE STORED
,ROW-WISE BEGINNING IN A+l

,
,IF THE INVERSION WAS SUCCESSFUL IT WILL RETURN TO CALL +3, AND IF A ZERO
,PIVOT ELEMENT OR OVERFLOW OCCURRED, IT WILL RETURN TO CALL +2

,
,ACCUMULATOR ASSIGNMENTS
T=15
J=14
K=13
P=12
PT=11
LC=10
LCS=7

,PIVOT ELEMENT
,COLUMN SUBSCRIPT
,ROW SUBSCRI PT
,INDEX POINTING TO PIVOT ELEMENT
,MULTIPLIER
,STOP COUNTER
,ROW COUNTER

INVERT:

HRRZ @ (17)
MOVEM, ROWS#
HLRZ @ (17)
MOVEM, COLS#
MOVE [XWD ROWPRG, ROW)
BLT ROWLi
HRR ROW, (17)
ADDI ROW,1
HRR ROW+2, (17)
HRR ROW+3, ROW
ADDI ROW+2, 2
HRRM ROW, INZl+1
HRRM ROW, DIV+2
HRRM ROW, DIV+6
HRRM ROW, INZROW+1
HRRM ROW+2, DI V

;GET COLUMN COUNT
iGET ROW COUNT
iMOVE ROW SUBROUTINE
INTO FAST MEMORY
iSET UP PROGRAM ADDRESSES

MOVE I P, 0
MOVN T, COLS
MOVE T, 1 (T)
HRRM T, INZ1

INZSTP:

MOVEI K, 0
MOVE LCS, ROWS
MOVE J, P

;INITIALIZE INVERSION STEP

INZ1:

HRLI J, 0
SKIPN T, A(P)
JRST 1 (17)

;GET PIVOT ELEMENT
;IF IT IS ZERO, EXIT AS ERROR

DIV:

MOVE A+1 (J)
FDV T
MOVEM A(J)
AOBJN J, DIV
MOVSI 1.0B53
FDV T
MOVEM A(J)

;DIVIDE PIVOT ROW THROUGH BY
;PIVOT ELEMENT

MOVE J, P
MOVE PT, A(K)
CAMN K, P
JRST ROWSKP
JRST ROW
MOVN@ ROW
FMP PT
MOVEM@ROW+2

;INITlALlZE TO PROCESS A ROW

INZROW:

ROWOUT:

CTX:

i LAST ELEMENT OF PIVOT ROW

;IF THE ROW IS THE PIVOT ROW,
iSKIP IT
;GO TO PROGRAM IN FAST MEMORY
;HANDLE FINAL ELEMENT OF THE ROW

SOJN LCS, INZROW
ADD P, COLS
SOJN LC, INZSTP

;IS STEP FINISHED?

JRST 2(17)

;RETURN

;IS JOB FINISHED?

ROWSKP:

ADD K, COLS
;SKIP PIVOT ROW DURING
JRST CTX
ilNVERSION STEP
, THIS PROGRAM FOR PROCESSING THE ELEMENTS IN A ROW IS MOVED INTO FAST MEMORY
ROWPRG:
PHASE 1
ROW:

ROWL:
DEPHASE
END

MOVN A(J)
FMP PT
FAD A+1 (K)
MOVEM A(K)
ADDI J, 1
AOBJN K, ROW
JRST ROWOUT

lWO-BIT TESTING AND DEPOSITING OF DATA
PROBLEM:

Consider four tables with 500 registers a table. The entries of the first table
contain a 2-bit item, ITEM zeros, in bits 13 and 14. The entries of the second
table contain ITEM ones in bits 1-6; the third table contains ITEM twos in bits
1-9; and the fourth table contains ITEM threes in bits 1-10.

For' n = 1, 500
If ITEM a = a
n

Set:

ITEM 1n = 108
ITEM 2n = 1008
ITEM 3n = 300 8

If ITEM a = 1
n

Set:

ITEM 1n = 20 8
ITEM 2n = 200 8
ITEM 3n = 400 8

If ITEM a = 2
n

Set:

ITEM 1n = 308
ITEM 2n = 300 8
ITEM 3n = 5008

If ITEM 0

Set:

ITEM 1n = 40 8
ITEM 2n = 400 8
ITEM 3 n = 600 8

Program:

For 500 cases, moving the program to fast memory results in a time saving of
approximately 5000 microseconds.

HRLZIO, A
BLT 0, 17

JRST 2

XWD+D-500, a
LDB 0, 14
ROT 0, 3
ADDIO, 10
DPB 0, 15
ROT 0, 3
19

lWO-BIT TESTING AND DEPOSITING OF DATA {continued}
DPB 0, 16
ADDIO, 200
DPB 0, 17
AOBJN1,2
JRST EXIT
POINT 2, TABO (1), 14
POINT 6, TABI (1), 6
POINT 9, TAB2 (1), 9
POINT 10, TAB3 (1), 10

ANY RADIX PR INT
PROBLEM:
Assume:

To Print out a signed number in an arbitrary radix.
1) TOUT is the first location of an I/O Routine which exits by POPJ P,O. The argument to tout is in accumulator B.
2) The output radix is stored in the address part of RADIX. The output radix in
this example is R.
3) Place the number to be converted in accumulator A and call RADPTwith PUSHJ P,
RADPT. Th is routine suppresses lead ing zeros.

RADPT:

RADIX:

RADPTl:

JUMPGE B, RADIX
MOVEI B, "_"
PUSHJ P, TOUT
MOVNSA
IDIVI A, R
HRLM A+1, (P)
SKIPE A
PUSHJ P, RADIX
HLRZB, (P)
ADDI B, 260
JRST TOUT

;IS NUMBER NEGATIVE?
;YES, GET ASCII MINUS SIGN
;OUTPUT THE MINUS SIGN
;TAKE ABSOLUTE VALUE OF ARGUMENT
;QUOTIENT GOES TO A, REMAINDER TO A+1
;SAVE REMAINDER IN LEFT SIDE OF LAST
; ITEM ON PUSH DOWN LIST
;IS QUOTIENT = a?
;NO, GO BACK FOR ANOTHER DIGIT
;GET THE DIGIT OFF THE PUSHDOWN LIST
;CONVERT THE DIGIT TO ASCII
;GO TO THE I/O ROUTINE. TOUT EXECUTES
;A POPJ P, BACK TO RADPTl OR (FINALLY)
;TO THE PLACE WHERE RADPT WAS CALLED.

15106

PRINTED IN U.S.A.

20-8164