Trailing-Edge
-
PDP-10 Archives
-
decus_20tap1_198111
-
decus/20-0001/muldiv.for
There is 1 other file named muldiv.for in the archive. Click here to see a list.
C
C
C
C COPYRIGHT (c) 1977 BY
C DIGITAL EQUIPMENT CORPORTATION, MAYNARD, MASS.
C
C THIS SOFTWARE IS FURNISHED UNDER A LICENSE AND MAY BE USED AND COPIED
C ONLY IN ACCORDANCE WITH THE TERMS OF SUCH LICENSE AND WITH THE
C INCLUSION OF THE ABOVE COPYRIGHT NOTICE. THIS SOFTWARE OR ANY OTHER
C COPIES THEREOF MAY NOT BE PROVIDED OR OTHERWISE MADE AVAILABLE TO ANY
C OTHER PERSON. NO TITLE TO AND OWNERSHIP OF THE SOFTWARE IS HEREBY
C TRANSFERRED.
C
C THE INFORMATION IN THIS SOFTWARE IS SUBJECT TO CHANGE WITHOUT NOTICE
C AND SHOULD NOT BE CONSTRUED AS A COMMITMENT BY DIGITAL EQUIPMENT
C CORPORATION.
C
C DIGITAL ASSUMES NO RESPONSIBILITY FOR THE USE OR RELIABILITY OF ITS
C SOFTWARE ON EQUIPMENT WHICH IS NOT SUPPLIED BY DIGITAL.
C
C
C
C
C
C
C
C
C ++++++++++++++++++++++++++++++++++++++++++++++++++
C + +
C + CALC VERSION X01-01 +
C + +
C + PETER BAUM 1-SEP-77 +
C + DIGITAL EQUIPMENT CORPORATION +
C + 146 MAIN STREET +
C + MAYNARD, MASSACHUSETTS 01754 +
C + +
C ++++++++++++++++++++++++++++++++++++++++++++++++++
C
C
C **************************************************
C * *
C * SUBROUTINE MULDIV (PT1,PT2,RETCD,BASE) *
C * *
C **************************************************
C
C
C SUBROUTINE MULDIV PERFORMS MULTIPLE PRECISION DIVISION
C
C
C
C ARGUMENTS: (PT1,PT2,RETCD)
C
C PLACES STACK1( ,PT1)/STACK2( ,PT2) INTO STACK1( ,PT1)
C
C STACK 2 IS NOT CLEANED UP BY THIS ROUTINE
C
C ERROR RETURN 1 = NORMAL
C 2 = ERROR
C
C
C
C ENTRY INDICATES THE BASE: 8, 10, OR 16
C
C
C
C
C MODIFY CODES: M3, M10
C
C
C
C
C MULDIV CALLS ERRMSG TO PRINT ERROR MESSAGES.
C
C
C
C
C MULDIV IS CALLED BY CALBIN
C
C
C
C
C VARIABLE USE
C
C ANSWER HOLDS QUOTIENT
C BASE HOLDS THE BASE: 8, 10, OR 16
C CARRY USED TO "BORROW" WHEN SUBTRACTING
C DIVPT POINTS TO HIGHEST (NON-ZERO) DIGIT OF DIVISOR.
C I,M,K,L TEMPORARY VALUES.
C NPT HOLDS POSITION OF HIGHEST (NON-ZERO) DIGIT OF DIVIDEND.
C PT1 POINTER TO STACK1 ELEMENT (DIVIDEND)
C PT2 POINTER TO STACK2 ELEMENT (DIVISOR)
C RETCD RETURN CODE: 1=O.K., 2=ERROR.
C
C
C
C
C
C
SUBROUTINE MULDIV (PT1,PT2,RETCD,BASE)
INTEGER*2 RETCD,BASE,DIVPT,NPT,CARRY
INTEGER*2 ST1PT,ST2PT,ST1LIM,ST2LIM,PT1,PT2
INTEGER*2 ST1TYP(40),ST2TYP(40)
INTEGER*2 I,M,K,L
C
LOGICAL*1 ANSWER(100)
LOGICAL*1 STACK1(100,40),STACK2(100,40)
C
COMMON /STACK/STACK1,STACK2,ST1PT,ST2PT,ST1TYP,ST2TYP,
; ST1LIM,ST2LIM
C
C
C
C
C
C SET DEFAULT RETURN CODE AND ZERO ANSWER
RETCD=1
DO 110 I=1,100
110 ANSWER(I)=0
C
C
C DETERMINE SIGN OF ANSWER
IF (STACK1(100,PT1)+STACK2(100,PT2).EQ.1)ANSWER(100)=1
C
C
C FIND HIGH ORDER NON-ZERO DIGIT OF DIVISOR (HELD BY DIVPT)
DO 140 I=1,99
DIVPT=100-I
IF (STACK2(DIVPT,PT2).NE.0) GOTO 150
140 CONTINUE
C
C
C DIVISOR IS ZERO, PRINT OUT APPROPRIATE ERROR MESSAGE
RETCD=2
CALL ERRMSG (23)
RETURN
C
C
C FIND STARTING NON-ZERO DIGIT OF DIVIDEND (HELD BY NPT)
150 DO 170 I=1,99
NPT=100-I
IF (STACK1(NPT,PT1).NE.0) GOTO 200
170 CONTINUE
C
C
C ANSWER IS ZERO
GOTO 10000
C
C
C
C
C
C ************************************************
C ****** FIND OUT WHERE TO SUBTRACT DIVISOR ******
C ************************************************
C
200 IF (NPT.LT.DIVPT) GOTO 10000
C
C M POINTS TO A DIGIT OF DIVIDEND ALIGNED WITH THE LOW ORDER DIGIT OF
C DIVISOR.
M=NPT-DIVPT+1
C
C K INDEXES DIVIDEND FROM HIGH ORDER END.
K=NPT
C
C L INDEXES DIVISOR FROM HIGH ORDER END.
L=DIVPT
DO 250 I=1,DIVPT
IF (STACK2(L,PT2).EQ.STACK1(K,PT1)) GOTO 240
C
C WHEN COMPARING DIGITS OF DIVISOR AND DIVIDEND, IF THE DIVISOR'S
C DIGIT IS SMALLER AND ALL PREVIOUS WERE EQUAL,
C THEN WE GO TO 300 TO SUBTRACT OFF THE DIVISOR.
IF (STACK2(L,PT2).LT.STACK1(K,PT1)) GOTO 300
C
C OTHERWISE THOSE DIGITS OF THE DIVIDEND REPRESENT A LARGER
C NUMBER. IF THE NUMBER OF DIGITS OF THE DIVISOR IS GREATER
C THAN OR EQUAL TO DIVIDEND THEN THE QUOTIENT IS 0 AND THE
C REMAINDER IS THE DIVIDEND. (GO TO 10000)
IF (NPT.LT.DIVPT+1) GOTO 10000
C
C OTHERWISE SHIFT THE POSITION OF THE DIVISOR BY 1.
M=M-1
GOTO 300
C
C
C K INDEXES DIVIDEND
C L INDEXES DIVISOR
240 K=K-1
L=L-1
250 CONTINUE
C
C
C
C
C **************************************
C ****** SUBTRACT OFF THE DIVISOR ******
C **************************************
300 CONTINUE
C
C K POINTS TO LOW ORDER DIGIT WHERE SUBTRACTION TAKES PLACE,
C CARRY TAKES CARE OF "BORROWS"
K=M-1
CARRY=0
DO 350 I=1,DIVPT
K=K+1
L=STACK1(K,PT1)-STACK2(I,PT2)-CARRY
IF (L.GE.0) GOTO 325
C
C IF SUBTRACTION RESULTS IN A "NEGATIVE DIGIT", ADD BASE AND SET CARRY
C INDICATOR.
CARRY=1
L=BASE+L
GOTO 340
C
C RESULT OF SUBTRACTION IS A POSITIVE NUMBER, SO
C CLEAR CARRY INDICATOR.
325 CARRY=0
340 STACK1(K,PT1)=L
350 CONTINUE
IF(CARRY.EQ.1) STACK1(K+1,PT1)=STACK1(K+1,PT1)-1
C
C
C THE QUOTIENT (ANSWER(M)) COUNTS THE NUMBER OF SUBTRACTIONS.
ANSWER(M)=ANSWER(M)+1
C
C
C RESET THE POINTER TO THE HIGH ORDER NON-ZERO DIGIT OF THE
C DIVIDEND IF NECESSARY.
370 IF (STACK1(NPT,PT1).NE.0) GOTO 200
NPT=NPT-1
IF (NPT.EQ.0) GOTO 10000
GOTO 370
C
C
C
C
C ***************************
C ****** COPY ANSWER ******
C ***************************
10000 DO 10010 I=1,100
10010 STACK1(I,PT1)=ANSWER(I)
RETURN
END