AR Chapter5 v2

8/11/2019 AR Chapter5 v2

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ECE 2211

Microprocessor and Interfacing

Br. Athaur Rahman Bin Najeeb

Room 2.105

Email: [email protected]

Website: http://eng.iiu.edu.my/~athaur

Consultation : Tuesday 10.00 am ( appointment)


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Announcement


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Chapter 5:P 8088/8086 Programming

Recall of debugging

Assembler

Instructions

Data transfer instructions

Arithmetic instructions

Shift instructions

Rotate instructions

Useful Link: http://oopweb.com/Assembly/Documents/Art

OfAssembly/Volume/Chapter_6/CH06-1.html


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Debug Debug is a little tool that comes with windows

Part of PCs Disk Operating System ( DOS)

Invoking debug

Start -> Run-> type cmd -> typedebug

Start -> Run ->type debug

Debug Commands / Monitor Commands

A

I

T

G H

M

E

F

R

C


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Structure of ASM file


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Mnemonic Meaning Format Operation Flagsaffected

Mov Move Mov D,S (s) (D) None

Data Transfer Instructions - MOV

Destination Source

Memory Accumulator

Register Register

Register Memory

Memory Register

Register Immediate

Memory Immediate

Seg reg Reg 16

Seg reg Mem 16

Reg 16 Seg reg

Memory Seg reg

NO MOV

Mem

ImmSeg Reg

Mem

Seg RegSeg reg

EX: MOV AL, BL


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Data Transfer Instruction -MOV

The MOV instruction Move data from source to destination; e.g. MOV [DI+100H], AH It does not modify flags Cannot transfer data between 2 external memory address

Eg: MOV [SI],[1234] Cannot move (load) directly to Segment Registers

Eg: MOV DS, 1000

Byte Operation / Word Operation Question [Example 5.1 , text book ? ]

What is the effect of executing MOV CX, [ SOURCE_MEMORY ] .. Where

SOURCE_MEMORY equal 20(H) and DS equal 1A00[H]


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Ex: Mov dx,cs


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Data Transfer Instructions - XCHG

Mnemonic Meaning Format Operation Flags affected

XCHG Exchange XCHG D,S (s) (D) None

Destination Source

Accumulator Reg 16

Memory Register

Register Register

Register Memory

Example: XCHG [1234h], BX

NO XCHGMEMsSEG REGs


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Example XCHG

Example:

For the data shown, what is the result of executing XCHG [SUM], Bx ?

[SUM] = 1234(16)

Solution:Execution of this instruction performs the operation

((DS)0 + SUM)(BX)

The PA is

= + =

Then,

13234(16)(BL)

13235(16)(BH)

Result: (BX) = 00FF(16)


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Your First Program Write a ASM program to load values 0011(H), 00AB(H) into regiter AX, BX and swap

them

Read from memory location 0200(H) and swap with register AX

Step 1: Flow Chart (? )

Step 2 Translate to Assembly Language

Step 3: write, test, debug and execute the code

Answer


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Data Transfer Instructions LEA, LDS, LES

Loading GP registers and segment registers with an address directly from memory

Ex: LEA SI,[EA]

LEA SI, [DI+BX+5H] ; DI=1000(H), Bx=20(H), EA ? PA = DS[0] +EA LEA SI, [200h] , if DS=1200 , calculate the PA

LEA Destination, Source

Transfers the offset address of source (must be a memory location) to the destination register

It does not modify flags

LES Destination Source It is identical to LDS except that the second two bytes are copied to ES It does not modify flags

Load 4-byte data (pointer) in memory to two 16-bit registers Source operand gives the memory location The first two bytes are copied to the register specified in the destination operand; the second

two bytes are copied to register DS It does not modify flags


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Data Transfer Instructions LEA, LDS, LES

Mnemonic

Meaning Format Operation Flagsaffected

LEA Load Effective

Address

LEA Reg16,EA EA (Reg16) None

LDS Load RegisterAnd DS

LDSReg16,MEM32

(MEM32)(Reg16)

None

(Mem32+2) (DS)

LES Load Registerand ES

LESReg16,MEM32

(MEM32)(Reg16)

(Mem32+2) (DS)

None

LEA SI DATA or MOV SI Offset DATA


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Examples for LEA, LDS, LES

DATAX DW 1000H DATAY DW 5000H .CODE LEA SI, DATAX MOVE DI, OFFSET DATAY; THIS IS MORE EFFICIENT LEA BX,[DI]; IS THE SAME AS MOV BX,DI; THIS JUST TAKES LESS CYCLES. LEA BX,DI; INVALID!

LES similar to LDS except that it loads ES

LDS BX, [DI];

127ABX

1000DI

3000DS

7A

12

00

30

11000

11001

11002

11003


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Data Transfer Instructions SAHF

LAHF

Store data in AH to the low 8 bits of the flag register

It modifies flags: AF, CF, PF, SF, ZF

Copies bits 0-7 of the flags register into AH


LDS Destination Source

Load 4-byte data (pointer) in memory to two 16-bit registers

Source operand gives the memory location

The first two bytes are copied to the register specified in the destination operand;

the second two bytes are copied to register DS


LES Destination Source

It is identical to LDS except that the second two bytes are copied to ES



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Arithmetric Instruction

Addition

Subtraction

Multiplication Division


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Arithmetic Instructions ADDDestination, Source

Destination + Source Destination

Destination and Source operands can not be memory locations at the same time

It modifies flags AF CF OF PF SF ZF

ADCDestination, Source

estination + ource + arry ag estination



INCDestination

Destination + 1 Destination

It modifies flags AF OF PF SF ZF (Note CF will not be changed)

DECDestination

Destination - 1 Destination

It modifies flags AF OF PF SF ZF (Note CF will not be changed)


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Arithmetic Instructions

ADD, ADC, INC, AAA, DAA


ADD Addition ADD D,S (S)+(D) (D)

carry (CF)

ALL

ADC Add withcarr

ADC D,S (S)+(D)+(CF) (D)

ALL

INC Incrementby one

INC D (D)+1 (D) ALL but CY

AAA ASCIIadjust foraddition

AAA If the sum is >9, AH

is incremented by 1

AF,CF

DAA Decimaladjust foraddition

DAA Adjust AL for decimalPacked BCD

ALL


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Arithmetic Instructions SUB, SBB, DEC, AAS, DAS,

NEG


SUB Subtract SUB D,S (D)-(S) (D)Borrow (CF)

All

SBB Subtractwith borrow

SBB D,S (D)-(S)-(CF) (D) All

DEC Decrementby one

DEC D (D)-1 (D) All but CF

NEG Negate NEG D All

DAS Decimaladjust for

subtraction

DAS Convert the result in AL topacked decimal format

All

AAS ASCIIadjust for

subtraction

AAS (AL) difference

(AH) dec by 1 if borrow

CY,AC


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SUBDestination, Source Destination - Source Destination



SBBDestination, Source

estination - ource - arry ag estination



CMPDestination, Source

Destination Source (the result is not stored anywhere)


It modifies flags AF CF OF PF SF ZF (if ZF is set, destination = source)


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Example 1

Write a program to add a data byte located at offset 0500(h) in 2000h data segmentWITH an other byte which is available in 0700h in the same segment


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Example 2

Write a program to move the contects of the ML 0500h to register BX and to registerCX. Add immediate byte 05h to the data residing in ML whose address computedusing ds-2000h and offset 0600h. Store the result of addition in 0700h. Assume alldata is located in ds=2000h

start

Move content to BX

Move content to CX

Move data , 05h to [0600h]

Store result

End


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MUL Source

Perform unsigned multiply operation

If source operand is a byte, AX = AL * Source

If source operand is a word, (DX AX) = AX * Source

Source operands can not be an immediate data

It modifies CF and OF (AF,PF,SF,ZF undefined)

IMUL Source

Perform signed binary multiply operation If source operand is a byte, AX = AL * Source

If source operand is a word, (DX AX) = AX * Source


It modifies CF and OF (AF,PF,SF,ZF undefined)

Examples:MOV AL, 20H

MOV CL, 80H

MUL CL

MOV AL, 20H

MOV CL, 80H

IMUL CL


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DIV Source

Perform unsigned division operation

If source operand is a byte, AL = AX / Source; AH = Remainder of AX / Source

If source operand is a word, AX=(DX AX)/Source; DX=Remainder of (DX AX)/Source


IDIV Source


Examples:MOV AX, 5

MOV BL, 2

DIV BL

MOV AL, -5

MOV BL, 2

IDIV BL

Perform signed division operation

If source operand is a byte, AL = AX / Source; AH = Remainder of AX / Source

If source operand is a word, AX=(DX AX)/Source; DX=Remainder of (DX AX)/Source



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Summary so far


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NEG Destination

0 Destination Destination (the result is represented in 2s complement)

Destination can be a register or a memory location


CBW Extends a signed 8-bit number in AL to a signed 16-bit data and stores it into AX

CWD

Extends a signed 16-bit number in AX to a signed 32-bit data and stores it into DX

and AX. DX contains the most significant word


Other arithmetic instructions:

DAA, DAS, AAA, AAS, AAM, AAD


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Example

What is the result of executing the following sequence of instruction ?

MOV AL, 0A1H

CBW

CWD


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Logical Instructions ( 3)

NOT Destination

Inverts each bit of the destination operand



AND Destination, Source

Performs logic AND operation for each bit of the destination and source; stores the

resu n o es na on

Destination and source can not be both memory locations at the same time It modifies flags: CF OF PF SF ZF

OR Destination, Source

Performs logic OR operation for each bit of the destination and source; stores the

result into destination Destination and source can not be both memory locations at the same time

It modifies flags: CF OF PF SF ZF


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Logic operation: a review


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Clearing Setting, Toggling bit of an Operand

Mask : clear a bit to zero

To clear a bit, we AND it with zero

AND with 1 : unchanged

Example: AND AX, 000F; if AX = FFFF Setting a bit to 1: OR with logic 1

XOR to reverse the login level

AND AL, 0FH (given AL=FF)

Application example:

A microprocessor is connected to read 8 LEDs into AL , if LED number 0 is on, thenturn on the alarm


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SHIFT INSTRUCTIONS


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A logical shift fills the newly created bit position with zero:

Logical VS Arithmetic Shifts

CF

0

An arithmetic shift fills the newly created bit position with a copy of the numberssign bit:

CF


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SHIFT

ARITHMETIC LOGICAL

Shift Instructions : 4 types

LEFT LEFTRIGHT RIGHT

SAL SAR SHL SHR


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Bit Manipulation Instructions

SHL(SAL) Destination, Count Left shift destination bits; the number of bits shifted is given by operand Count

During the shift operation, the MSB of the destination is shifted into CF and

zero is shifted into the LSB of the destination Operand Count can be either an immediate data or register CL



CF 0

SHR Destination, Count Right shift destination bits; the number of bits shifted is given by operand Count

During the shift operation, the LSB of the destination is shifted into CF and

zero is shifted into the MSB of the destination

Operand Count can be either an immediate data or register CL Destination can be a register or a memory location


CF0 Destination

Destination

LSBMSB

LSBMSB


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SAR Destination, Count

Right shift destination bits; the number of bits shifted is given by operand Count

The LSB of the destination is shifted into CF and the MSB of the destination remians

the same

Operand Count can be either an immediate data or register CL


It modifies flags: CF PF SF ZF

CFDestination

LSBMSB


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Fast Multipication


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ROTATE INSTRUCTIONS


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ROL Destination, Count

Left shift destination bits; the number of bits shifted is given by operand Count

The MSB of the destination is shifted into CF, it also goes to the LSB of the destination

Operand Count can be either an immediate data or register CL Destination can be a register or a memory location

It modifies flags: CF OF

CF Destination

LSBMSB

ROR Destination, Count


The LSB of the destination is shifted into CF, it also goes to the MSB of the destination



It modifies flags: CF OF

CFDestination

LSBMSB


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RCL Destination, Count

Left shift destination bits; the number of bits shifted is given by operand Count

The MSB of the destination is shifted into CF; the old CF value goes to the LSB

of the destination Operand Count can be either an immediate data or register CL


It modifies flags: CF OF PF SF ZFCF Destination

LSBMSB

RCR Destination, Count


The LSB of the destination is shifted into CF, the old CF value goes to the MSB

of the destination


Destination can be a register or a memory location It modifies flags: CF OF PF SF ZF

CFDestination

LSBMSB


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Example: Using ROL to Exchange Values.

.data

byteval db 0Fh

wordval dw 1234h

.code

byte_values:

,

ROL AL,4 ; AL = 62hROL byteval,4 ; byteval = F0h

word_values:

MOV AX,0203h

ROL AX,8 ; AX = 0302h

ROL wordval,8 ; wordval = 3412h


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Example:

What is the result in BX and CF after execution of

the following instruction?

RCR BX,CL

,


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Example:

Indicate the hexadecimal value of AL after each rotation

MOV AL,6BH

ROR AL,1

RCL AL,3


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ANSWER:

A B5H

B AEH

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