Arithmetic Operations Multiplication – Division - PowerPoint Presentation

Slide1

Arithmetic OperationsMultiplication – Division

System Programming

Lab

Computer Engineering

Department

College

of

Engineering

ObjectivesWrite programs in 8086 assembly language for multiplication and

division

by

using different size of data as bytes or words.

Use arithmetic instructions to accomplish simple binary, BCD, and ASCII arithmetic.

TheoryAnother arithmetic instructions found in any microprocessor includes multiplication and division. Also

shown are their uses in manipulating register and memory data.

These

operations used any addressing mode with 8-bit or 16

bit.

These arithmetic

instructions can be divided into two

types of instructions

:

First

Group

:

MUL, IMUL, TEST

Second group:

DIV, IDIV, CBW, CWD

TheoryThese types of operands are supported:REG

Memory

REG

:

AX, BX, CX,

DX,

DI, SI, BP, SP

.

Memory

:

[BX], [BX+SI+7], variable, etc...

Multiplication is performed on bytes, words, or double words, and can be signed integer or

unsigned.

TheoryAfter operation between operands, result is always stored in AL, AX.

The

product after a multiplication is always a double-width product

.

Some flag bits (overflow and carry) change when the multiply instruction

executes

CF

,

OF

Multiplication MUL – unsigned multiply When operand is a byte AX = AL * operand (8 bit) When operand is a word (AX,DX) = AX * operand (8 or 16 bit)IMUL

– signed

multiply

When operand is a byte

AX

= AL * operand (8 bit)

When operand is a word (AX,DX) = AX * operand

(

8 or 16 bit)

Unsigned MultiplyInstr.OperandsDescription 

MUL

REG

Memory

Algorithm:

when operand is a byte:

AX = AL * operand.

when operand is a word:

(DX AX) = AX * operand.

Example:

MOV AL, 200 ; AL = 0C8h

MOV BL, 4

MUL BL ; AX = 0320h (800)

RET

Flags:

C

,

O

Signed MultiplyInstr.OperandsDescription 

IMUL

REG

Memory

Algorithm:

when operand is a byte:

AX = AL * operand.

when operand is a word:

(DX AX) = AX * operand.

Example:

MOV AL, -2

MOV BL, -4

IMUL BL ; AX = 8

RET

Flags

:

C

,

O

TEST Instruction Instr.OperandsDescription 

TEST

REG , Memory

Memory

, REG

REG

, REG

Memory ,

Immed

.

REG ,

Immed

.

Logical

AND

between all bits of two operands for flags only.

MOV AL, 00000101bTEST AL, 1 ; ZF = 0.TEST AL, 10b ; ZF = 1.RETFlags: Z, S, P

1

AND

1 = 1

0

AND

1 = 0

1

AND

0 = 0

0

AND

0 = 0

DivisionDIV – unsigned division When operand is a byte AL = AX / operand (8 bit) AH = remainder (modulus) When operand is a word AX = (AX,DX) / operand (8 or 16 bit)

DX = remainder (modulus

)

IDIV

– signed division

When operand is a byte AL = AX / operand (8 bit)

AH

= remainder (modulus)

When operand is a word AX = (AX,DX) / operand

(

8 or 16 bit)

DX = remainder (modulus)

Unsigned DivideInstr.OperandsDescription 

DIV

REG

M

emory

Algorithm:

when operand is a byte:

AL = AX / operand

AH = remainder (modulus)

when operand is a word:

AX = (DX AX) / operand

DX = remainder (modulus)

Example:

MOV AL, 203 ; AX = 00CBh

MOV BL, 4

DIV BL ; AL = 50 (32h), AH = 3

RET

Flags: NON

Signed DivideInstr.OperandsDescription 

IDIV

REG

M

emory

Algorithm:

when operand is a byte:

AL = AX / operand

AH = remainder (modulus)

when operand is a word:

AX = (DX AX) / operand

DX = remainder (modulus)

Example:

MOV AX, -203 ; AX = 0FF35h

MOV BL, 4

IDIV BL ;

AL = -50 (0CEh), AH = -3 (0FDh)

RETFlags: NON

Convert Byte into WordInstr.OperandsDescription

 

CBW

No operands

Algorithm:

If high bit of

AL = 1

then:

AH = 255 (0FFh)

else

AH = 0

Example:

MOV AX, 0 ; AH = 0, AL = 0

MOV AL, -5 ; AX = 000FBh (251)

CBW ; AX = 0FFFBh (-5)

RET

Flags: NON

Convert Word to Double wordInstr.OperandsDescription 

CWD

No operands

Algorithm:

if high bit of

AX = 1

then:

DX = 65535 (0FFFFh)

else

DX = 0

Example:

MOV DX, 0 ; DX = 0

MOV AX, 0 ; AX = 0

MOV AX, -5 ; DX AX = 00000h:0FFFBh

CWD ; DX AX = 0FFFFh:0FFFBh

RET

Flags

: NON

ExamplesMultiply two 8-bit unsigned numbers the first in the register BL, and the second in CL, the result is stored in DX (assumed BL=05, CL=10)ORG 100MOV BL, 05H

MOV CL, 10

MOV AL, CL

MUL BL

MOV DX, AX

HLT

ExamplesDivide the unsigned byte contents of memory location NUMB by the contents of memory location NUMB1. Store the quotient in location ANSQ and reminder in location ANSR (assume [NUMB]=15 and [NUMR]=4).ORG 100MOV AL, NUMB

MOV AH, 00H

DIV NUMB1

MOV ANSQ,AL

MOV ANSR,AH

HLT

NUMB DB 15H

NUMB1 DB 04H

ANSQ DB 0H

ANSR DB 0H

ExamplesDivide two 16-bit signed numbers, 1st number= -100 the 2nd = +9, store the result in DX, AX.ORG 100hMOV AX, -100

MOV CX, 9

CWD

IDIV CX

HLT

After the division the results appear in DX, AX

AX = -11 quotient

DX = -1 reminder

ExamplesWrite 8086 program that generate a byte size integer in the memory location define as RESULT, the value of integer is to be calculated from the following equation:(RESULT) = ( AL . NUM1 ) + ( NUM2 . AL ) + BL

MOV AL, 01H

MOV DL, AL

MOV

CL, NUM1

MUL CL ; ( AL . NUM1 )

MOV SI, AX

MOV

AL, DL

MUL NUM2 ;

( AL . NUM2 )

ADD AX, SI

MOV BX, 0001H

ADC

AX, BX

MOV RESULT, AX

HLT

NUM1 DB 02h

NUM2 DB 02h

RESULT DW ?

ExamplesWrite the program to find the value of AX=((3*A+B ))/2 ;A = 2H, B = 2HA) By using successive addition.B) By using multiplication and division instruction.

MOV

AL, A

ADD

AL

, AL

ADC

AL

, A

ADC AL, B

SUB AL,02H

SUB AL,02H

HLT

A

DB 02H

B DB 02H

MOV AL, A

MOV BL, 03H

MUL BL

ADD AL, B

MOV BL, 02H

DIV BL

HLT

A DB 02H

B DB 02H

10+10+10 =30

30 + 6 = 36

36 – 2

–

2

–

2

–

2

–

2

–

2

–2 –2 = 16

2+2+2 =6

6 + 2 = 8

8 – 2

–2 = 4

ExamplesWrite 8086 program to compute SUM

+

2

Where Xi & Yi are signed 8-bit numbers, N=10, assume no overflow

MOV SI, OFFSET

X

MOV DI, OFFSET Y

MOV BX, 00H

MOV DX, 00H

MOV CX, 0BH

L1

:

MOV

AL, [SI+BX]

IMUL [DI+BX]

ADD AX, 0002H

ADC

DX, AX

INC BX

LOOP L1

MOV

SUM

,

DX

HLT

X

DB

11 DUP (+2)

Y

DB 11

DUP (+3)

SUM DW ?

MOV SI, OFFSET

X

MOV DI, OFFSET Y

MOV BX, 00H

MOV DX, 00H

MOV CX, 0BH

L1

:

MOV

AL, [SI+BX]

IMUL [DI+BX]

ADD AX, 0002H

ADC

DX, AX

INC BX

LOOP L1

MOV

SUM

,

DX

HLT

X

DB

11 DUP (+2)

Y

DB 11

DUP (+3)

SUM DW ?