this Presentation is prepared to demonstrate Multiplication and Division Instruction In microprocessor 8086

2. TOPIC:
MULTIPLICATI
ON AND
DIVISION
Group member:
M Hamza Nasir
(12063122-067)
M Usaman Ali
(12063122-086)
Syed Farhan Abbas
(12063122-009)
M Faran Ali
(12063122-055)
Ateeb Saeed
(12063122-094)
University Of Gujrat

3. Multiplication
MUL(unsinged
)
IMUL(singed)

4. MUL INSTRUCTION
(UNSIGNED MULTIPLY)
Multiplies an 8-, 16-, or 32-bit operand by
either AL, AX
Syntax:MUL AL R/M8
Syntax: MUL AX R/M16

5. MUL INSTRUCTION
Note that the product is stored in a register (or
group of registers) twice the size of the
operands.
The operand can be a register or a memory
operand

6. MUL INSTRUCTION

7. MUL EXAMPLES
Mov al, 5h
Mov bl, 10h
Mul bl ; AX = 0050h, CF = 0
 No overflow - the Carry flag is 0 because the
upper half of AX is zero

8. IMUL INSTRUCTION
(SIGNED MULTIPLY)
same syntax
uses the same operands as the MUL
instruction
 preserves the sign of the product
Opcode=IMUL

9. IMUL INSTRUCTION
Suppose AX contains 1 and BX contains FFFFh
Mov AX, 1h
Mov BX, FFFFh
IMUL BX ;
Decimal product=-1,
Hex Product =FFFFFFFFh
DX = FFFFh, AX=FFFFh CF/OF=0
DX is a sign extension of AX for this CF/OF=0

10. IMUL INSTRUCTION
IMUL sets the Carry and Overflow flags if the
high-order product is not a sign extension of
the low-order product
Mov al, 48
Mov bl, 4
Imul bl ;AX = 00C0h, OF = 1
AH is not a sign extension of AL, so the
Overflow flag is set

11. IMUL INSTRUCTION
Suppose AX contains FFFFh and BX contains FFFFh
Mov AX, FFFFH
Mov BX, FFFFh
IMUL BX
Decimal product=1
Hex Product = 00000001 h
DX = 0000h, AX=0001h CF/OF=0
DX is a sign extension of AX for this CF/OF=0

12. IMUL INSTRUCTION
Suppose AX contains 0FFFh.and BX contains 0FFFh
Mov AX, 0FFFH
Mov BX, 0FFFh
IMUL BX
Decimal product=16769025
Hex Product = 00FFE001 h
DX = 00FFh, AX=E001h CF/OF=0
DX is a not sign extension of AX for this CF/OF=1

13. IMUL INSTRUCTION
Suppose AL contains 80h.and BL contains FFh
Mov AL, 80H
Mov BL, FFh
IMUL BL
Decimal product=128,
Hex Product = 0080 h
AH = 00h, AL=80h CF/OF=01
DX is a not sign extension of AX for this CF/OF=1

14. APPLICATION OF MUL AND IMUL
Translate the high level language assignment
statement
A=5×A-12×B
Let A and B be word variables, and suppose
there is no overflow.
Use IMUL for multiplication

15. SOLUTION:
Mov AX,5 ;AX=5
IMUL A ;AX=5*A
MOV A,AX ;A=5*A
MOV AX,12 ;AX=12
IMUL B ;AX=12*B
SUB A,AX ;5*A-12*B

16. DIVIDE
DIV(unsinged) IDIV(singed)

17. DIVIDE AND IDIVIDE
When division is performed we obtain two
results
The quotient and
The remainder.
Similarly like Multiplication there are separate
instructions for unsigned and signed division

18. CONT..
Syntax:
DIV divisor
IDIV divisor
Byte Form:
The divisor is eight bit register or memory byte
The 16 – bit dividend is assumed to be in AX. After
division 8-bit quotient is in AL and 8-bit remainder in
AH.

19. CONT..
Word Form:
The divisor is a 16-bit register or memory word
The 32-bit dividend is assumed to be in DX:AX
After division, the 16-bit quotient is in AX and 16-
bit remainder is in DX
Effect on flags:
All status flags are undefined

20. CONT..
Divide Overflow:
It is possible that the quotient s too big to fit
in the specified destination(AL or AX).
This Happens Because the divisor is much
smaller than the dividend.
When this Happens the program terminates
and system displays message “divide overflow”

21. CONT..
Examples:
EXAMPLE 9.8:
Suppose DX contains 0000h , AX contains 0005h, and BX
contains 0002h.
EXAMPLE 9.9:
Suppose DX contains 0000h , AX contains 0005h and BX
contains FFFEh.
EXAmple 9.10:
Suppose AX contains 00FBh and BL contains FFh.

22. DECIMAL INPUT AND
OUTPUT
Computer represent every thing in binary
But it is convenient for user to represent
input and output in decimal
If we input 21543 character string then it
must to converted internally
Conversely on output the binary contents of
R/M must be converted to decimal equivalent
before being printed

23. DECIMAL INPUT
Convert a string of ASCII digits to the binary
representation of decimal equivalent
For input we repeatedly multiply AX by 10
Algorithm (First version):
Total=0
Read an ASCII
REPEAT
convert character to number
Total=total*10+value
Read a character
Until character is carriage return

24. CONT..
Example: input of 123
Total =0
Read ‘1’
Convert ‘1’ to 1
Total=10*0 +1=1
Read ‘2’
Convert ‘2’ to 2
Total=10*1 +2=12
Read ‘3’
Convert ‘3’ to 3
Total=10*12 +3=123

25. CONT..
Range: -32768 to 32767
Optional sign followed by string of digits
& carriage return
Outside ‘0’ to ‘9’
jumps to new line and ask for input
again

26. CONT..
Algorithm(second version):
total=0
Negative=false
Read a character
Case character of
• ‘-’: negative=true
read a character
• ‘+’: read a character
End_case
Repeat
If character is not between ‘0’ to ‘9’
Then
Go to beginning

27. CONT..
Else
Convert character to binary value
Total=10*total+value
End_if
Read a character
Until character is carriage return
If negative =true
Then
total=-total

28. CONT..
Program(source code):
INDEC PROC
;READ NUMBER IN RANGE -32768 TO 32767
PUSH BX
PUSH CX
PUSH DX
@BEGIN:
;total =0
XOR BX,BX ;BX hold total
;negative =false

29. CONT..
XOR CX,CX ;CX hold sign
;read char
MOV AH,1
INT 21H
;case char of
CMP AL,'-' ;minus sign
JE @MINUS ;yes,set sign
CMP AL,'+' ;plus sign
JE @PLUS ;yes,get another char

30. CONT..
JMP @REPEAT2 ;start processing char
@MINUS: MOV CX,1
@PLUS: INT 21H
;end case
@REPEAT2:
;if char. is between '0' and '9'
CMP AL,'0' ;char >='0'?
JNGE @NOT_DIGIT ;illegal char.

31. CONT..
CMP AL,'9' ;char<='9' ?
JNLE @NOT_DIGIT
;then convert char to digit
AND AX,000FH
PUSH AX ;save number
;total =total *10 +digit
MOV AX,10
MUL BX
POP BX ;retrieve number
ADD BX,AX ;total =total *10
+digit

32. CONT..
;read char
MOV AH,1
INT 21H
CMP AL,0DH ;CR
JNE @REPEAT2 ;no keep going
;until CR
MOV AX,BX ;store number
in AX
;if negative
OR CX,CX ;negative

33. CONT..
Jz @EXIT ;no,exit
;then
NEG AX ;yes,negate
;end if
@EXIT: ;retrieve registers
POP DX
POP CX
POP BX
RET

34. CONT..
;here if illegal char entered
@NOT_DIGIT:
MOV AH,2
MOV DL,0DH
INT 21H
MOV DL,0AH
INT 21H
JMP @BEGIN
INDEC ENDP

35. CONT..
Output:

36. INPUT OVERFLOW
AX:FFFFh
In decimal:65535
Range:-32768 to 32767
Anything out of range called input overflow
For example:
Input:32769
Total=327690

37. CONT..
Algorithm:
total=0
Negative=false
Read a character
Case character of
• ‘-’: negative=true
read a character
• ‘+’: read a character

38. CONT..
End_case
Repeat
If character is not between ‘0’ to ‘9’
Then
Go to beginning
Else
Convert character to binary value
Total=10*total

39. CONT..
If overflow
Then
go to beginning
Else
Total =total*10 +value
If overflow
Then
go to beginning

40. CONT..
End_if
End_if
End_if
Read a character
Until character is carriage return
If negative =true
Then
total=-total

41. CONT..
Code:
;total =total *10 +digit
MOV AX,10
MUL BX
CMP DX,0
JNE @NOT_DIGIT
POP BX
ADD BX,AX
JC @NOT_DIGIT

42. CONT..
Output:

43. Decimal Output
Algorithm for Decimal Output:
If AX < 0 /*AX holds output value */
THEN
Print a minus sign
Replace AX by its twos complement
End_IF
Get the digits in AX’s decimal representation
Convert these digits into characters and print
them

44. CONT..
To see what line 6 entitles, suppose the contents of
AX, expressed in decimal is 24168. To get the digits
in decimal representation , we can proceed as
follows,
Divide 24618 by 10, Quotient= 2461,
remainder=8
Divide 2461 by 10, Quotient= 246, remainder=1
Divide 246 by 10 , Quotient=24, remainder=6
Divide 24 by 10, Quotient=2, remainder=4
Divide 2 by 10, Quotient=0, remainder=2

45. CONT..
LINE 6:
Cout =0 /*will count decimal digit */
REPEAT
divide quotient by 10
Push remainder on the stack
Count= count +1
UNTILL
Quotient=0

46. CONT..
LINE 7:
FOR count times DO
Pop a digit from the stack
Convert it to a character
Output the character
END_FOR

47. CONT..
Program Listing PMG9_1.ASM
.MODEL SMALL
.STACK 100H
.CODE
OUTDEC PROC
;prints AX as a signed decimal integer
;input: AX
;output: none
PUSH AX ;save registers
PUSH BX
PUSH CX
PUSH DX

48. CONT..
;if AX < 0
OR AX,AX ;AX < 0?
JGE @END_IF1 ;NO >0
;then
PUSH AX ; save number
MOV DL,’-’ ;get ‘-’
MOV AH,2 ;print character function
INT 21H ;print ‘-’
POP AX ;get Ax back
NEG AX ;AX= -AX
@END_IF1:

49. CONT..
;get decimal digits
XOR CX,CX ;CX counts digits
MOV BX,10D ;BX has divisor
@REPEAT1:
XOR DX,DX ;prepare high word of dividend
DIV BX ;AX=quotient, DX=remainder
PUSH DX ;save remainder on stack
INC CX ;count = count +1
;until
OR AX,AX ;quotient = 0?
JNE @REPEAT ;no, keep going

50. CONT..
;convert digits to character and print
MOV AH,2 ;print character function
;for count time do
@PRINT_LOOP
POP DX ;digit in DL
OR DL,30H ;convert to character
INT 21H ;print digit
;end_for
POP DX ; restore registers
POP CX
POP BX
POP AX
OUTDEC ENDP