assembly language

W.P.Fang Department of CSIE YUST, Copy Right Reserved

1

Assembly Language

W.P.fang


2

Reference

• Ytha Yu, Charles Marut, Assembly Language Programming and Organziationof the IBM PC


3

requirement

• Exercise+ Program homework 40%• Midterm examination 30%• Final examination 30%• Ps:

– Cheat and late hand out is not acceptable


4

Outline

• Microcomputer Systems• Representation of Numbers and

Characters• Organization of the IBM Personal

Computers• Introduction to IBM PC Assembly

Language• The Processor Status and the FLAGS

Register


5

Outline

• Flow Control Instructions• Logic, Shift, ad Rotate Instructions• The Stack and Introduction to Procedures• Multiplication and Division Instructions• Arrays and Addressing Modes• The String Instructions• Text Display and Keyboard Programming


6

Outline

• Macros• Memory Management• BIOS and DOS Interrupt• Color Graphics• Recursion• Advanced Arithmetic• Disk and File Operations• Intel’s Advanced Microprocessors6


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schdular

987654321

備註進度日期


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schdular

181716151413121110

備註進度日期


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Microcomputer Systems


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Microcomputer Systems

• Overview– Introduction to the architeture of

microcomputers in general and to the IBM PC in particular.


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Microcomputer System(component)

• The components o a microcomputer system– System unit– Keyboard – display screen– Disk drives


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• Computer circuits– Central processing unit (CPU)– I/O circuits

• Microcomputer– CPU is a single-chip processor

(microprocessor)

• System board– Microprocessor and memorycircuirs


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• Bytes – eight bits• Address – identified by a number • Content– the data stored in a memory byte.

content:

0 1 1 0 0 0 0 10 1 0 1 1 1 1 00 0 0 0 0 0 0 00 1 1 0 1 1 1 0

Address:3210


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• Example 1.1 Suppose a processor uses 20 bits for an address. How many memory bytes can be accessed?


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• Word – typical microcomputer, two bytes form a word

• Bit position

Byte bit position

Word bit position

7 6 5 4 3 2 1

15 14 13 12 11 10 9 8 7 6 5 4 3 2 1

0

08

High byte Low byte


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Microcomputer system(component)

• RAM and ROM• Bus


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• In any case, each instruction that the CPU executes is a bit string

• This language of 0’s and 1’s is called machine language

• The instructions performed by a CPU are called instruction set.

• The instruction set for each CPU is unique.• To keep the cost of computers down, machine

language instructions are designed to be simple


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• EU (Execution Unit)– Arithmetic

• +-*/

– Logic unit• AND, OR, NOT

– Registers• AX,BX,CXDX,SI,DI,BP,SP


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• EIU ( Bus Interface Unit)– Communication between the EU and the

memory or I/O circuits– Registers

• CS,DS,ES,SS,IP


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• I/O ports


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Microcomputer system(instruction execution)

• Instruction Execution• Machine instruction

– Opcode : the type of operation– Operand : given as memory address to he

data to be operated on.


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Microcomputer system(instruction execution)

• Fetch-execute cycle– Fetch

• Fetch an instruction from memory• Decde he inctruction to determine the operation• Fetch data from memory if necessary.

– Execute• Perform the operation on the data• Store the result in memory if needed.


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Microcomputer system(I/O Devices)

• Magnetics disks• Keyboard• Display monitor• printers


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Microcomputer system(programming languages)

• Machine language

Store the content of AX in memory word 0

10100011 00000000 00000000

Add 4 to AX00000101 00000100 00000000

Fetch the content of memory word to and put it in register AX

10100001 00000000 00000000

operationMachine instruction


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• Assembly language

move the content of AX into location A

MOV A,AX

Add 4 to AXADD AX,4

Fetch the content of location A and put it in register AX

MOV AX,A

operationAssembly language instruction


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• Assembler• compiler• High-level language


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• Advantage of high-level languages– Closer to natural language– Reduce Coding time– Executed on any machine

• Advantage of assembly language– Efficiency– Some operations can be done easily in

assembly language but may be impossible at a higher level


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Microcomputer system(assembly languages)

ch1.asmTITLE PGM1_1: SAMPLE PROGRAM.MODEL SMALL.STACK 100H.DATAA DW 2B DW 5SUM DW ?.CODEMAIN PROC;initialize DS

MOV AX,@DATAMOV DS,AX

;add the numbersMOV AX,AADD AX,BMOV SUM,AX

;exit to DosMOV AX,4C00HINT 21H

MAIN ENDPEND MAIN


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Microcomputer system(Debug)

• assemble A [address]• compare C range address• dump D [range]• enter E address [list]• fill F range list• go G [=address] [addresses]• hex H value1 value2• input I port• load L [address] [drive] [firstsector] [number]• move M range address• name N [pathname] [arglist]• output O port byte


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Microcomputer system(Debug)

• proceed P [=address] [number]• quit Q• register R [register]• search S range list• trace T [=address] [value]• unassemble U [range]• write W [address] [drive] [firstsector] [number]• allocate expanded memory XA [#pages]• deallocate expanded memory XD [handle]• map expanded memory pages XM [Lpage] [Ppage] [handle]• display expanded memory status XS


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Microcomputer system(tasm)

• Syntax: TASM [options] source [,object] [,listing] [,xref]• /a,/s Alphabetic or Source-code segment ordering• /c Generate cross-reference in listing• /dSYM[=VAL] Define symbol SYM = 0, or = value VAL• /e,/r Emulated or Real floating-point instructions• /h,/? Display this help screen• /iPATH Search PATH for include files• /jCMD Jam in an assembler directive CMD (eg. /jIDEAL)• /kh# Hash table capacity # symbols• /l,/la Generate listing: l=normal listing, la=expanded listing• /ml,/mx,/mu Case sensitivity on symbols: ml=all, mx=globals, mu=none• /mv# Set maximum valid length for symbols• /m# Allow # multiple passes to resolve forward references• /n Suppress symbol tables in listing• /os,/o,/op,/oi Object code: standard, standard w/overlays, Phar Lap, or IBM• /p Check for code segment overrides in protected mode• /q Suppress OBJ records not needed for linking• /t Suppress messages if successful assembly• /uxxxx Set version emulation, version xxxx• /w0,/w1,/w2 Set warning level: w0=none, w1=w2=warnings on• /w-xxx,/w+xxx Disable (-) or enable (+) warning xxx• /x Include false conditionals in listing• /z Display source line with error message• /zi,/zd,/zn Debug info: zi=full, zd=line numbers only, zn=none


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Microcomputer system(tlink)

• Turbo Link Version 3.01 Copyright (c) 1987, 1990 Borland International• Syntax: TLINK objfiles, exefile, mapfile, libfiles• @xxxx indicates use response file xxxx• Options: /m = map file with publics• /x = no map file at all• /i = initialize all segments• /l = include source line numbers• /s = detailed map of segments• /n = no default libraries• /d = warn if duplicate symbols in libraries• /c = lower case significant in symbols• /3 = enable 32-bit processing• /v = include full symbolic debug information• /e = ignore Extended Dictionary• /t = create COM file• /o = overlay switch• /ye = expanded memory swapping• /yx = extended memory swapping


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Glossary

• Add-in board or card• Address• Address bus• Arithmetic and logic unit, ALU• Assembler• Assembly language• Binary digit• Bit


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Glossary

• Bus• Bus interface unit, BIU• Byte• Central processing unit, CPU• Clock period• Clock pulse• Clock rate• Clock speed


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Glossary

• Compiler• Contents• Control bus• Data bus• Digital circuits disk drive• Execution unit ,EU• Expansion slots• Fetch-execute cycle


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Glossary

• Firmware• Fixed disk• Floppy diskhardcopy• Hard disk• I/O devices• I/O ports• Intruction pointer,IP• Instruction set• Kilobyte, KB


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Glossary

• Machine language• Mega• Megabyte, MB• Megahertz, MHz• Memory byte (circuit)• Memory location• Memory word• microprocessor


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Glossary

• Motherboard• Opcode• Operand• Peripheral (device)• Random access memory• RAM• Read-only memory (ROM)• register


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Glossary

• System board• Video adapter• word


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Representation of numbers and characters


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• Please refer to bcc


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• Binary number system


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• Ex:• 8A2Dh=?• 11101=?• 2BD4h=?• Convert 95 to binary• Convert 2B3Ch to binary• Convert 1110101010 to hex


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• Add• Subtraction• Find one’s complement of 5 (16 bits)• Find the two’s complement of 5• Show -97 in 8bit,16 bit


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ASCII code

• American Standard Code for Information Interchange


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Homework

• Chap 2– 2.– 7.d– 8.c– The rule of ascii


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• Number system– Binary– Hexadecimal

• Conversion• Addition and subtraction• Character representation


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Representation of numbers and characters (Glossary)

• ASCII (American Standard Code for Information Interchange) code

• Binary number system• Hexadecimal number system• Least significant bit lsb• most significant bit msb• One’s complement of a binarynmber• Scan code


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Representation of numbers and characters (Glossary)

• Signed integer• Two’s complement of a binary number• Unsigned integer


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Organization of the IBM Personal Computers


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Organization of the IBM Personal Computers

• Overall structure of the IBM PC– The memory organization– I/O ports– DOS routines– BIOS routines


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The intel x86 family

• 1978 8086 16 bit• 80186 extended instruction set• 80286 real address mode,protected virtual

address mode,multitasking,moreaddressable memory,2^30 bytes

• 80386 32 bit virtual 8086 mode/386 SX à16bit bus


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Register

• Data register• Address register• Status register• Segment• Pointer• Index register


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• Ax• Bx• Cx• Dx• Cs• Ds• Ss• Es• Si• Di• Sp• Bp• Ip• flag


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Data register

• AX(accumulator)• BX(base register)• CX(count register)• DX(data register)


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Segment register

• CS -- code segment• DS – data segment• SS – stack segment• ES – access second data segment (extra

segment)


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• Memory segment• Segment:offset address

– Because 20bit can not represent in 16 bit


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Logical address

• Ex:• A4FB:4872h• A4FB0+4872h=A9822h(20-bit physical

address)


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• For the memory location whose physical address is specified by 1256Ah,give the address in segment:offset form for segments 1256h and 1240h


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• 1256Ah=12560h+X• 1256Ah=12400h+Y• X=Ah,Y=16Ah• 1256Ah=1256:000A=1240:016A


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• A memory location has physical address 80FD2h. In what segment does it have offset BFD2h


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• Physical address=segmentx10h+offset• Segmentx10h=physical address-offset• 80FD2h-BFD2h=75000h• àsegment 7500h


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Pointer and index register

• SP – stack pointer– Used in conjunction with SS for accessing

stack segment.

• BP – Base pointer– Used primarily to access data on the stack.– Unlike SP, we can also use BP to access data

in the other segment


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• SI – Source Index– Point to memory locations in the data

segment addressed by DS.– By incrementing the contents of SI, we can

easily access consecutive memory locations• DI – Destination Index

– Same functions as SI– For string operation– Access memory locations addressed by ES


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Instruction Pointer

• IP – Use register CS ad IP to access instruction– CS à segment number of next instruction– IP à offset– Can not be directly manipulated by an

instruction


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Flags

• Status flags– Ex:ZF(zero flag)

• Controls– Ex: IF(interrupt flag) =0 à input from the

keyboard are ignored by the processor.


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The operating system

• Reading and executing the commands typed by the user

• Performing I/O operations• Generating error messages• Managing memory and other resources


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• Dos routine that services user commands is called COMMAND.COM


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BIOS

• Machine specific• DOS I/O operations are ultimately carried

out by the BIOS routines• Addresses of the BIOS routine – interrupt

vectors


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Memory

00000hIntreeupt vectors00400hBios and dos data

Dos

Application program area

A0000hVideoB0000hVideoC0000hReservedD0000hReservedE0000hReservedF0000hBios


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Serial port (COM1)3f8h-3ffh

CGA3d0h-3dfh

EGA3c0h-3cfh

Parallel printer port 1378h-37fh

Hard disk320h-32fh

Serial port (COM2)2f8h-2ffh

Game controller200h-20fh

Keyboard controller60h-63h

Interrupt controller20h-21h

descriptionPort address


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Start-up operation

• Power up• CS register is set to FFFFh• IP is set to 0000h• à FFFF0h (in ROM)• First check for system and memory errors• Initialize the interrupt vectors and BIOS

data• Boot program


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Introduction to IBM PC Assembly Language


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Introduction to IBM PC Assembly Language

• Syntax– Name operation operand(s) comment

• Ex:– START: MOV CX,5 ;initialize counter– MAIN PROC


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• Name Field– 1 to 31 characters long– Letters, digits, special characters ?.@_$%– Embeded blanks are not allowed


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Operation feilds

• Operands field– An instruction may have zero, one , or two

operand– Ex:– NOP– INC AX– ADD WORD1,2


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Comment fields

• A semicolon marks the beginning of this field

• The assembler ignores anything typed after the semicolon.

• Ex:– MOV CX,0 ;move 0 to CX

• Permissible to make an entire line a comment


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Program data

• The processor operates only on binary data

• The assembler must translate all data representation into binary number


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Hex0FFFFH

Illegal hex number – doesn’t begin with a decimal digit

FFFH

Illegal hex number – doesn’t end in “H”

1B4DHex1B4DHIllegal– contains a nondigit character1,234Decimal-21843DDecimal64223Binary11011BDecimal11011ＴYPENUMBER


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characters

Define tenbytesDT

Define quawordDQ

Define double wordDD

Define wordDW

Define byteDB

Stands forPseudo-op


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• Byte variable– Name DB initial_value

• Ex:– ALPHA DB 4


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• Word variable– Name DW initial_value

• Ex:– WRD DW -2


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• Array– B_ARRAY DB 10H,20H,30H

• String– LETTERS DB ‘ABC’– LETTERS DB 41H,42H,43H– MSG DB ‘HELLO’,0AH,0DH,’$’


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• Named Constanted– EQU (Equates)– Name EQU constant

• Ex:– LF EQU 0AH– PROMPT EQU ‘TYPE YOUR NAME’


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variables

• Byte variables• Name DB initial_value• Ex:• ALPHA DB• BYT DB ?


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Word variables

• Name DW initial_value• Ex:

– WRD DW -2


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array

• B_ARRAY DB 10H,20H,30H

30h202hB_ARRAY

20h201hB_ARRAY+1

10h200hB_ARRAY

contentsAddressSymbol


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Character string

• LETTERS DB ‘ABC’• LETTERS DB 41H,42H,43H• MSG DB ‘HELLO’,0AH,0DH,’$’• MSG DB

48H,45H,4CH,4CH,4FH,0AH,0DH,24H


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Names constants

• Name EQU constant• Ex:

– LF EQU 0AH– MOV DL,0AH– MOV DL,LF– Prompt EQU ‘TYPE YOUR NAME’– MSG DB PROMPT

• Note:no memory is allocated for EQU names


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A few basic instructions

• MOV• XCHG• ADD• SUB• INC• DEC• NEG


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• MOV destination,source

noYesNoYesConstant

NoNoYesYesMemory location

NoYesNoYesSegment register

NoYesYesyesGeneral register

constantMemory location

Segment register

General registerDestination

operandSourceoperand


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• XCHG destination, source

noYesMemory location

YesYesGeneration register

Memory location

General register

Destination operand

Source operand


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• XCHG AH,BL• XCHG AX,WORD1

1A 00

00 05

AH AL

BH BL

05 00

00 1A

AH AL

BH BL


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Restrictions on MOV and XCHG

• MOV or XCHG between memory location is not allowed

• Ex: MOV WORD1,WORD2 ; illegal• MOV AX,WORD2• MOV WORD1,AX


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• ADD destination, source• SUB destination, source

yesYesConstant

NoYesMemory location

YesYesGeneral register

Memory locationGeneral registerDestination operand

Source operand


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• INC destination• DEC destination• Ex:

– INC WORD1– DEC BYTE1


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• NEG destination• By two’s complement• Ex:

– NEG BX


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Type agreement of operands

• The operands of the preceding two-operand instruction must be of the same type.


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Translation of High-Level language to assembly labguage

• B=A– MOV AX,A– MOV B,AX

• A=5-A– MOV AX,5– SUB AX,A– MOV A,AX

– NEG A– ADD A,5


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• A=B-2*A– MOV AX,B– SUB AX,A– SUB AX,A– MOV A,AX


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• .MODEL memory_model


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Program structure (Memory model)

Code in more than one segmentData in more one segmentArrays may be larger than 64k bytes

HUGE

Code in more than one segmentData in more than one segmentNo array larger than 64k bytes

LARGE

Code one segmentData in more than one segment

COMPACT

Code in more than one segmentData in one segment

MEDIUM

Code in one segmentData in one segment

SMALL

DescriptionModel


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Data segment

• Contains all the variable definitions• .DATA• WORD1 DW2• WORD2 DW 5• MSG DB ‘THIS IS A MESSAGE’• MASK EQU 10010010B


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Stack segment

• .STACK size• Ex:

– .STACK 100H

• Ps: if size is omitted 1KB is set aside for the stack data


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Code segment

• .CODE name• There is no need for a name in a SMALL

program

• Name PROC• ;body of the procedure• Name ENDP


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• Ex:– .CODE– MAIN PROC– ;main procedure instructions– MAIN ENDP– ;other procedures go here


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Putting it together

• .MODEL SMALL• .STACK 100H• .DATA• ;data definitions go here• .CODE• MAIN PROC• ;instructions go here• MAIN ENDP• ;other procedures go here• END MAIN


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Input and output instructions

• IN• OUT• Most applications programs do not use IN

and OUT because– Port addresses vary among computer model– It’s much easier to program I/O with the

service routines provided by the manufacturer


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INT instruction

• INT interrupt_number


110

INT 21h

Character string outpur9

Single-character output2

Single-key input1

routineFunction number


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INT 21hFunction 1:

• Single-key input• Input:

– Ah=1

• Output:• AL=ASCII code if character key is pressed

=0 if non-character key is pressed


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• MOV AH,1• INT 21h


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INT 21hFunction 2:

• Display a character or execute a control function: – Ah=2– DL=ASCII code of the display character or

control character

• Output:• AL=ASCII code of the display character or

control character


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• MOV Ah,2• MOV DL,’?’• INT 21h


115Carriage return (start of current

CRD

Line feed (new line)

LFA

TabHT9

BackspaceBS8

Beep (sounds a tone)

BEL7

FunctionSymbolASCII code (HEX)


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INT 21hFunction 9:

• Display a string: – DX=offset address of string– The string must end with a ‘$’character.


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• Ex:– MSG DB ‘Hello!$’


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LEA

• INT21h, function 9, expects the offset address of the character string to be DX

• LEA destination, source• EX:

– LEA DX,MSG


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Program segment prefix

• When a program is loaded in memory, DOS prefaces it with a 256 byte program segment prefix(PSP)

• PSP contains information about the program,

• DOS places its segment number in both DS and ES before executing the program

• The result is that DS does not contain the segment number of the data segment


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• MOV AX,@DATA• MOV DS,AX


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Terminating a program

• INT 32h function 4Ch


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TITLE PGM4_1:ECHO PROGRAM.MODEL SMALL.STACK 100H.CODEMAIN PROC

MOV AH,2MOV DL, '?'

INT 21HMOV AH,1INT 21HMOV BL,ALMOV AH,2MOV DL,0AHINT 21HMOV DL,BLINT 21hMOV AH,4CHINT 21h

MAIN ENDPEND MAIN


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TITLE PGM4_2:PRINT STRING PROGRAM.MODEL SMALL.STACK 100H.DATAMSG DB ‘HELLO!$’.CODEMAIN PROC

MOV AX,@DATAMOV DS,AXLEA DX,MSGMOV AH,9INT 21hMOV AH,4CHINT 21h

MAIN ENDPEND MAIN


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TITLE PGM4_3: CASE CONVERSION PROGRAM.MODEL SMALL.STACK 100H.DATA

CR EQU 0DHLF EQU 0AH

MSG1 DB ‘ENTER A LOWER CASE LETTER: $’MSG2 DB 0DH,0AH,’IN UPPER CASE IT IS: ‘CHAR DB ?,’$’.CODEMAIN PROC

MOV AX,@DATAMOV DS,AXLEA DX,MSG1INT 21hMOV AH,1INT 21hSUB AL,20h


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MOV CHAR,ALLEA DX,MSG2MOV AH,9INT 21hMOV AH,4CHINT 21h

MAIN ENDPEND MAIN


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Exercise

• Using only MOV,ADD,SUB,INC,DEC and NEG translate below into assembly (A,B,C are word)

• A=B-A• A=-(A+1)• C=A+B• B=3*B+7• A=B-A-1


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homework

• Ch4– 4– 8– 12

• If programming exercise please print out code and result (hard copy)

• If more than one page , please bind in left upper

• Use A4 paper


128

• Write a program to – Display a “?”– Read two decimal digits whose sum is less

than 10– Display them and their sum on the next line,

with an appropriate message

• ?27• THE SUM OF 2 AND 7 is 9


129

• Write a program to – Display “?”– Read three initials– Display them in the middle of an 11x11 box of

asterisks– Beep the computer


130

The processor status and the FLAGS Register

CF

0

PFAFZFSFTFIFDFOF

123456789101112131415


131

• The status flags are located in bits 0,and 11

• Control flags are located in bits 8,9, and 10

• The other bits have no significance


132

Status flags

• Reflect the result of an operation• Ex:

– If SUB AX,AX is executed, the zero flag become 1


133

Status Flags

OFOverflow flag11

SFSign flag7

ZFZero flag6

AFAuxiliary carry flag

4

PFParity flag2

CFCarry flag0

symbolnamebit


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Control Flags

DFDirection flag10

IFInterrupt flag9

TFTrap flag8

Symbolnamebit


135

Carry flag

• CF=1 if– There is a carry out from the most significant

bit (msb) on addition– There is a borrow into themsb on subtraction– Affected by shift and rotate instruction

• Otherwise CF=0


136

Parity flags

• PF=1 if– The low byte of a result has an even number

of one bits


137

AuxiliRy carry flag

• AF=1 if– There is a carry out from bit 3 on addition– Borrow into bit 3 on substraction– Used in binary-coded decimal (BCD)

operations


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Zero flag

• ZF=1 for a zero result• ZF=0 for a nonezero result


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Sign Flag

• SF=1 if the msb of a result is 1• Means the result is negative if giving a

signed interpretation• SF=0 if the msb=0


140

Overflow Flag

• OF=1 if signed overflow occurred


141

Overflow

• Is associated with the fact that the range of numbers that be represented in a computer is limited.


142

Overflow examples

• Four possible outcome– No overflow– Signed overflow– Unsigned overflow– Both signed and unsigned overflow


143

• Unsigned overflow but not signed overflow– AX=FFFFh– BX=0001h– ADD AX,BX

1111 1111 1111 1111+ 0000 0000 0000 0001

-----------------------------------1 0000 0000 0000 0000

Signed : -1 + 1=0 ,unsigned 65536 àoverflow


144

• Signed but not unsigned overflow– AX=BX=7FFFh– ADD AX,BX

0111 1111 1111 1111+ 0111 1111 1111 1111---------------------------------------

1111 1111 1111 111032767+32767=65534 à -2


145

How the processor indicates overflow

• OF=1 for signed overflow• CF=1 for unsigned overflow• In determining overflow, the processor

does not interpret the result as either signed or unsigned.

• Programmer shall interpreting the result


146

How the processor determines that overflow occurred

• Limit the discussion to addition and substraction

• Unsigned overflow– On addition,Means that the correct answer is

larger than the biggest unsigned number– On substraction, means that the correct

answer is smaller than 0


147

• Signed overflow– On addition

• Signed overflow occurs when the sum has a different sign

– On substraction• With different signed like adding number of the same sign.• Ex:• A-(-B)=A+B • -A-(+B)=-A+ -B• Occur if the result has a different sign than expected


148

• In addition of number with different signs, overflow is impossible

• In substraction of numbers with the same sign cannot give overflow


149

• Processor uses the following method to set the OF– If the carries into and out of the msb don’t

match– There is a carry out but no carry in– Then signed overflow has occurredà OF is

set to 1


150

How Instructions affect the flags

All (CF=1 unless result is 0OF=1 if word operand is 8000hOr byte operand is 80h

NEG

All except CFINC/DEC

AllADD/SUB

NoneMOV/XCHG

Affects flagsinstruction


151

example

• ADD AX,BX, where AX contains FFFFh,BX contains FFFh


152

example

• ADD AL,BL, where AL contains 80h,BL contains 80h


153

example

• SUB AX,BX, where AX contains 8000h and BX contains 0001h


154

example

• INC AL, where AL contain FFh


155

example

• MOV AX,-5


156

example

• NEG AX, where AX contains 8000h


157

The DEBUG Program

• An environment in which a program may be tested.


158

TITLE PGM5_1:Check flags.MODELSMALL.STACK 100H.CODEMAIN PROC

MOV AX,4000HADD AX,AXSUB AX,0FFFFHNEG AXINC AX

MOV AH,4CHINT 21H

MAIN ENDPEND MAIN


159

-r

AX=0000 BX=0000 CX=000F DX=0000 SP=0100 BP=0000 SI=0000 DI=0000 DS=0B7E ES=0B7E SS=0B8F CS=0B8E IP=0000 NV UP EI PL NZ NA PO NC 0B8E:0000 B80040 MOV AX,4000 -r

AX=0000 BX=0000 CX=000F DX=0000 SP=0100 BP=0000 SI=0000 DI=0000 DS=0B7E ES=0B7E SS=0B8F CS=0B8E IP=0000 NV UP EI PL NZ NA PO NC 0B8E:0000 B80040 MOV AX,4000 -t

AX=4000 BX=0000 CX=000F DX=0000 SP=0100 BP=0000 SI=0000 DI=0000 DS=0B7E ES=0B7E SS=0B8F CS=0B8E IP=0003 NV UP EI PL NZ NA PO NC 0B8E:0003 03C0 ADD AX,AX


160

-t

AX=8000 BX=0000 CX=000F DX=0000 SP=0100 BP=0000 SI=0000 DI=0000 DS=0B7E ES=0B7E SS=0B8F CS=0B8E IP=0005 OV UP EI NG NZ NA PE NC 0B8E:0005 2DFFFF SUB AX,FFFF -t

AX=8001 BX=0000 CX=000F DX=0000 SP=0100 BP=0000 SI=0000 DI=0000 DS=0B7E ES=0B7E SS=0B8F CS=0B8E IP=0008 NV UP EI NG NZ AC PO CY 0B8E:0008 F7D8 NEG AX -t

AX=7FFF BX=0000 CX=000F DX=0000 SP=0100 BP=0000 SI=0000 DI=0000 DS=0B7E ES=0B7E SS=0B8F CS=0B8E IP=000A NV UP EI PL NZ AC PE CY 0B8E:000A 40 INC AX


161

AX=7FFF BX=0000 CX=000F DX=0000 SP=0100 BP=0000 SI=0000 DI=0000 DS=0B7E ES=0B7E SS=0B8F CS=0B8E IP=000A NV UP EI PL NZ AC PE CY 0B8E:000A 40 INC AX -t

AX=8000 BX=0000 CX=000F DX=0000 SP=0100 BP=0000 SI=0000 DI=0000 DS=0B7E ES=0B7E SS=0B8F CS=0B8E IP=000B OV UP EI NG NZ AC PE CY 0B8E:000B B44C MOV AH,4C -g

Program terminated normally-q


162

homework

• Ch5– 1.e– 2.b– 3.d– 4.e


163

Flow Control Instructions


164

Flow Control Instructions

• The jump and loop instructions transfer control to another part of the program.

• This transfer can be– Unconditional– Depend on a particular combination of status

flag settings


165

Jump example• TITLE PGM6_1:IBM CHARACTER DISPLAY• .MODEL SMALL• .STACK 100H• .CODE• MAIN PROC• MOV AH,2• MOV CX,256• MOV DL,0• PRINT_LOOP:• int 21h• INC DL• DEC CX• JNE PRINT_LOOP• MOV AH,4CH• INT 21h• MAIN ENDP• END MAIN


166


167

• JNZ:jump if not zero• CX:loop counter• PRINT_LOOP:label


168

Conditional jump

• Jxxx destination_label


169

Conditional jumps

• Range of a conditional jump– Precede no more than 126 bytes– Follow it by no more than 127 bytes


170

How the CPU implements a conditional jump

• CPU looks at the flags register• If the conditions for the jump are true, the

CPU adjusts the IP to point to the destination label.

• If the condition is false, the IP is not altered


171

• JNZ– Inspecting ZF

• If ZF=0,transfer to label• If ZF=1, next line


172

Conditional jumps

• Three categories– The signed jumps– Unsigned jumps– Single-flag jumps

• Ps: jump instruction do not affect the flags


173

Signed jumps

ZF=1 or SF<>OFJump if less than or equalJump if not great than

JLE/JNG

SF<>OFJump if less thanJump if not greater than or equal

JL/JNGE

SF=OFJump if greater than or equal toJump if not less than or equal to

JGE/JNL

ZF=0 and SF=OFJump if greater thanJump if not less than or equal to

JG/JNLE

Condition for jumpsdescriptionsymbol


174

Unsigned conditional jumps

CF=1 or ZF=1Jump if equalJump if not above

JBE/JNA

CF=1Jump if belowJump if not above or equal

JB/JNAE

CF=0Jump if above or equalJump if not below

JAE/JNB

CF=0 and ZF=0Jump if aboveJump if not below or equal

JA/JNBE

Condition for jumpsdescriptionSymbol


175

Single-Flag jumps

PF=0Jump if parity oddJNP/JPO

PF=1Jump if parity evenJP/JPE

SF=0Jump if nonnegative signJNS

SF=1Jump if sign negativeJS

OF=0Jump if no overflowJNO

OF=1Jump if overflowJO

CF=0Jump if no carryJNC

CF=1Jump if carryJC

ZF=0jump if not equalJump if not zero

JNE/JNZ

ZF=1Jump if equalJump if equal to zero

JE/JZ

Condition for jumpsdescriptionsymbol


176

The CMP instruction

• CMP destination , source• Compares destination and source by

computing (destination)-(source)• the result is not stored, but flags are

affected• May not both memory locations• Destination may not be a constant


177

• CMP AX,BX• JG BELOW• Where AX=7,BX=0001• Result is 7FFFh-0001=7FFEh• ZF=SF=OF=0


178

• DEC AX• JL THERE


179

Signed versus unsigned jumps

• Each of the signed jumps corresponds to an analogous unsigned jump

• Using the wrong kind of jump can lead to incorrect results


180

• Ex:– If AX=7FFFh– BX=8000h– CMP AX,BX– JA BELOW

• Because 7FFFh<8000h in an unsigned sense àdoes not jump to BELOW

• Ps: in signed 7FFFh>8000h


181

Working with Characters

• Example. Suppose AX and BX contain signed numbers, write some code to put the biggest one in CX

• Solution:– MOV CX,AX– CMP BX,CX– JLE NEXT– MOVE CX,BX– NEXT:


182

The JMP instruction

• JMP destination• JMP can be used to get around the range

restriction of a conditional jump


183

• TOP:• <<if too far>>• DEC CX• JNZ TOP• MOV AX,BX


184

• TOP:– DEC CX– JNZ BOTTOM– JMP EXIT

• BOTTOM:– JMP TOP

• EXIT:– MOV AX,BX


185

High-Level Language structure

• IF-THEN• IF-THEN-ELSE• CASE• Branch with compound conditions• AND conditions• OR conditions


186

IF-THEN

• IF condition is true• THEN

– Execute true-branch statements

• END_IF


187

• Ex :replace the number in AX by its absolute value


188

• IF AX<0• THEN

– Replace AX by –AX

• END_IF


189

• CMP AX,0– JNL END_IF– NEG AX

• END_IF:


190

IF-THEN-ELSE

• IF condition is true– THEN

• Execute true-branch statements

– ELSE• Execute false-branch statements

– END_IF


191

• Ex: suppose AL and BL contain extended ASCII characters. Display the one the comes first in the character sequence


192

• IF AL<=BL– THEN

• Display the character in AL

– ELSE• Display the character in BL

– END_IF


193

– MOV AH,2– CMP AL,BL– JNBE ELSE_– MOV DL,AL– JMP DISPLAY

• ELSE_:– MOV DL,BL

• DISPLAY:– INT 21h

• END_IF• Ps: the label ELSE_ is used because ELSE is a reserved

word.


194

Exercise• Write assembly code for each of the following decision

structures• a.

– IF AX<0– THEN

• PUT -1 IN BX– END_IF

• b.• IF AL<0

– THEN put FFh in AH• Put FFh in AH

– ELSE• Put 0 in AH

– END_IF


195

Exercise

• C.• Suppose DL contains the ASCII code oa a

character• (IF DL>=“A”) AND (DL<= ‘Z’)• THEN

– Display DL

• END_IF


196

• D.– If AX<BX

• THEN– IF BX<CX

» THEN» Put 0 in AX

– ELSE» Put 0 in BX

– END_IF

• END_IF


197

• E• IF(AX<BX) OR (BX<CX)

– THEN• Put 0 in BX

– ELSE• Put 1 in DX

– END_IF


198

HW

• Ch6. 3a,12


199

CASE

• Multiway branch structure• CASE expression

– Value1:statement 1– Value2:statement 2– :– :– Valuen:statement n

• END_CASE


200

ex

• If AX contains a negative number, put -1 in BX;if AX contains 0, put 0 in BX;if AX contains a positive number,put 1 in BX.


201

• CASE AX– <0:put -1 in BX– =0:put 0 in BX– >0:put 1 in BX


202

– CMP AX,0– JL NEGATIVE– JE ZERO– JG POSITIVE

• NEGATIVE:– MOV BX,-1– JMP END_CASE

• ZERO:– MOV BX,0– JMP END_CASE

• POSITIVE:– MOV BX,1

• END_CASE:


203

ex

• If AL contains 1 or 3, display “0”; if AL contains 2 or 4, display “e”


204

• CASE AL– 1,3:display ‘o’– 2,4:display ‘e’


205

– CMP AL,1– JE ODD– CMP AL,3– JE ODD– CMP Al,2– JE EVEN– CMP AL,4– JE EVEN– JMP END_CASE

• ODD:– MOV DL,’o’– JMP DISPLAY

• EVEN:– MOV DL,’e’

• DISPLAY:– MOV AH,2– INT 21H

• END_CASE:


206

Branches with compound conditions

• Condition_1 AND condition_2• Comdition_1 OR condition_2


207

AND conditions

• Ex:read a character, and if it’d an uppercase letter,display it


208

• Read a character (into AL)• IF (‘A’<= character) and (character <=‘Z’)• THEN

– Display character

• END_IF


209

– MOV AH,1– INT 21H– CMP AL.’A’– JNGE END_IF– CMP AL,’Z’– JNLE END_IF– MOV DL,AL– MOV AH,2– INT 21H

• END_IF


210

OR conditions

• Read a character. IF it’s “y”or “Y”, display it; otherwise, terminate the program.


211

• Read a character (into AL)• IF (character=‘y’) OR (character=‘Y’)

– THEN• Display it

– ELSE• Terminate the program

– END_IF


212

– MOV AH,1– INT 21H– CMP AL.’y’– JE THEN– CMP AL,’Y’– JE THEN– JMP ELSE

• THEN:– MOV AH,2– MOV DL,AL– INT 21H– JMP END_IF

• ELSE_:– MOV AH,4CH– INT 21H

• END_IF


213

Looping structures

• FOR LOOP• WHILE LOOP• REPEAT LOOP• WHILE versus REPEAT


214

FOR LOOP

• FOR loop_count times DO– Statements

• END_FOR


215

• LOOP destination_label• The counter for the loop is the register CX which

is initialized to loop_count.Execution of the LOOP instruction causes CX to be decremented automatically

• If CX is not 0, control transfers to destination_label

• If CX=0, the next instruction after LOOP is done.• Destination_label must precede the LOOP

instruction by no more than 126 bytes


216

• TOP:– LOOP TOP


217

ex

• Write a count-controlled loop to display a row of 80 stars


218

• FOR 80 times DO– Display ‘*’

• END_FOR


219

– MOV CX,80– MOV AH,2– MOV DL,’*’

• TOP:– INT 21h– LOOP TOP


220

• Notice:– FOR loop implement with a LOOP instruction,

is executed at least once– If CX contains 0 when the loop is entered, the

LOOP instruction cause CX to be decremented to FFFFh

– To prevent this, the instruction JCXZ(jump if CX is zero) may be used before the loop


221

• JCXZ destination_label


222

– JCXZ SKIP

• TOP:– LOOP TOP

• SKIP:


223

WHILE LOOP

• WHILE condition DO– Statements

• END_WHILE


224

ex

• Write some code to count the number of characters in an input line


225

• Initialize count to 0• Read a character• WHILE character<> carriage_return DO

– Count=count+1– Read a character

• END_WHILE


226

– MOV DX,0– MOV AH,1– INT 21H

• WHILE_:– CMP AL,0DH– JE END_WHILE– INC DX– INT 21H– JMP WHILE_

• END_WHILE


227

REPEAT LOOP

• REPEAT• Statements• UNTIL condition


228

ex

• Write some code to read characters until a blank is read


229

• REPEAT– Read a character

• UNTIL character is a blank


230

– MOV AH,1

• REPEAT:– INT 21H– CMP AL.’‘– JNE REPEAT


231

WHILE Versus REPEAT

• Use of a WHILE loop or a REPEAT loop is a matter of personal preference

• The advantage of a WHILE is that the loop can be bypassed if the terminating condition is initially false, whereas the statements in a REPEAT must be done at least once.


232

ex

• Type a line of text• THE QUICK BROWN FOX JUMPED• First capital =B last capital =X


233

• Display the opening message• Read and process a line of text• Display the results


234

TITLE PGM6_2:FIRST AND LAST CAPITALS.MODEL SMALL.STACk 100H.DATAPROMPT DB 'Type a line of text',0DH,0AH,'$'NOCAP_MSG DB 0DH,0AH,'No capitals $'CAP_MSG DB 0DH,0AH,'First capital = 'FIRST DB ']'

DB ' Last capital = 'LAST DB '@ $'.CODEMAIN PROC

MOV AX,@DATAMOV DS,AXMOV AH,9LEA DX,PROMPTINT 21HMOV AH,1INT 21H

WHILE_:


235

CMP AL,0DHJE END_WHILECMP AL,'A'JNGE END_IFCMP AL,'Z'JNLE END_IFCMP AL,FIRSTJNL CHECK_LASTMOV FIRST,AL

CHECK_LAST:CMP AL,LASTJNG END_IFMOV LAST,AL

END_IF:INT 21HJMP WHILE_

END_WHILE:


236

MOV AH,9CMP FIRST,']'JNE CAPSLEA DX,NOCAP_MSGJMP DISPLAY

CAPS:LEA DX,CAP_MSG

DISPLAY:INT 21HMOV AH,4CHINT 21H

MAIN ENDPEND MAIN


237

hw

• Ch6.3a,12


238

Logic, Shift ,and Rotate Instructions


239

Logic, Shift,and Rotate Instructions

• Logic instructions– AND– OR– XOR– NOT


240

Logic instructions

• Perform the following logic operations– 10101010 AND 11110000– 10101010 OR 11110000– 10101010 XOR 11110000– NOT 10101010


241

Truth tables(0=false ,1=true)

01111

11001

11010

00000

A XOR bA OR bA AND bba


242

instructions

• AND destination, source• OR destination, source• XOR destination, source• -------------------------------------------• Effect on flags

– SF,ZF,PF reflect the result– AF is undefined– CF,OF=0


243

• One use of AND, OR, and XOR is to selectively modify the bits in the destination.

• mask


244

b XOR 1=~b(compelemet of b)

b XOR 0=b

b OR 1 =1b OR 0 =b

b AND 0=0b AND 1 =b


245

• The AND instruction can be used to clear specific destination bits while preserving the others

• The OR instruction can be used to set specific destination bits while preserving the others.

• The XOR instruction can be used to complement specific destination bits while preserving the others.


246

ex

• Clear the sign bit of AL while leaving the other bits unchanged


247

• AND AL,7Fh

• Ps:• 01111111b=7Fh


248

Ex

• Set the most significant and least significant bits of AL while preserving the other bits


249

• OR AL,81h

• Ps:10000001b=81h


250

ex

• Change the sign bit of DX


251

• XOR DX,8000h


252

Converting an ASCII digit to a Number

• “5”à 5• Method 1:SUB AL,20h• Method 2:AND AL,0Fh


253

exercise

• Converting a stored decimal digit to its ASCII code


254

Converting a lowercase letter to upper case

01011010Z01111010z

::::

01000010B01100010b

01000001A01100001a

codecharactercodecharacter


255

• To convert lower to upper case we need only clear bit 5

• à AND with mask 11011111b (0DFh)• AND DL,0DFh


256

Clearing a Register

• MOV AX,0– Machine needs three bytes– Because prohibition on memory-on-memory

operations, must be used to clear a memory location

• SUB AX,AX– Machine needs two bytes

• XOR AX,AX– Machine needs two bytes


257

Testing a register for zero

• OR CX,CX• CMP CX,0• Because 1 OR 1 =1, 0 OR 0 =0• CX unchanged• But affect ZF and SF


258

NOT instruction

• NOT destination


259

ex

• Complement the bits in AX


260

• NOT AX


261

TEST instruction

• The TEST instruction performs and an AND operation of the destination with the source but does not change the destination contents

• The purpose of the TESt instruciton is to set the status flags


262

• TEST destination,source• ------------------------------------• Effect on flags

– SF,ZF, PF reflect the result– AF is undefined– CF,OF=0


263

Examining bits

• TEST destination,mask• If destination has 0’s in all the etsted

position, the result will be 0 and so ZF=1


264

ex

• Jump to label BELLOW if AL contains an even number


265

• Even numbers have a 0 in bit 0. thus, the mask is 00000001b=1

• TESt AL,1• JZ BELLOW


266

Shift instructions

• Opcode destination,CL


267

Left Shift instructions

• SHL destination,1• SHL destination, CL• ---------------------------------• Effect on flags

– SF,PF,ZF reflect the result– AF is undefined– CF=last bit shifted out– OF=1 if result changes sign on last shift


268

ex

• Suppose DH contain 8Ah and CL contains 3, what are the values of DH and CF after the instruction SHL DH,CL is executed?


269

• DH=10001010• à• 01010000b=50h


270

Multiplication by Left Shift

• A left shift on a binary number multiplies it by 2.

• Ex:• 00000101b = 5d• 00001010b =10d• 00010100b = 20d


271

SAL

• Shift arithmetic left• the same machine code with SHL


272

• Negative number can also be multiplied by powers of 2 by left shifts

• Ex:– FFFFh(-1) à FFF8h(-8)


273

Overflow

• The overflow flags are not reliable indicators for a multiple left shift,

• A multiple shift is really a series of single shifts, and CF,OF only reflect the result of the last shift


274

ex

• BL=80h• CL=2• SHL BL,CL• à CF=OF=0• (this occur signed and unsigned overflow)


275

ex

• Write some code to multiply th evalue of AX by 8


276

• MOV CL,3• SAL AX,CL


277

Right Shift instruction

• SHR destination,1• SHR destination, CL• --------------------------------• Effect on the flags is the same as for SHL


278

ex

• Suppose DH contains 8AH and CL contains 2. what are the values of DH and CF after the instruction SHR DH,CL is executed?


279

• 10001010• à• 00100010b=22h


280

SAR

• Shift arithmetic right• SAR destination,1• SAR destination, CL• The effect on flags is the same as for SHR• Different with SHR :

– The msb retains its original value


281

Division by right shift

• 00000101b=5• 00000010b=2


282

Signed and unsigned division

• Signed : SAR• Unsigned: SHR


283

ex

• Use right shifts to divide the unsigned number 65143 by 4. put the quotient in AX


284

• MOV AX,65143• MOV CL,2• SHR AX, CL


285

ex

• If AL contains -15, give the decimal value of Al after SAR AL,1 is performed


286

• Execution of SAR AL,1 divides the number by 2 and rounds down

• -15 /2 = -7.5• We get -8• Ps:• -15=11110001b• -8=11111000b


287

General multiplication and division

• MUL• IMUL• DIV• IDIV


288

Rotate instructions

• Rotate left• Shift bits to left. • The msb is shifted into the rightmost bit• ROL destination,1• ROL destination, CL


289

• Rotate right• ROR destination,1• ROR destination, CL


290

ex

• Use ROL to count the number of 1 bits in BX, without changing BX. Put the answer in AX


291

– XOR AX,AX– MOV CX,16

• TOP:– ROL BX,1– JNC NEXT– INC AX

• NEXT:– LOOP TOP


292

Rotate Carry left

• Rotate through Carry Left• The msb is shifted into CF• The previous value of CF is shifted into the

rightmost bit• RCL destination ,1• RCL destination, CL


293

Rotate carry right

• RCR destination,1• RCR destination, CL


294

ex

• Suppose DH contains 8Ah, CF=1, and CL contains 3.what are the alues of DH and CF after the instruction RCR DH, CL is executed.


295

10110001=B1h0After 3 right rotations

011000101After 2 right rotation

110001010After 1 right rotation

100010101Initial values

DHCF


296

• Effect of the rotate instructions on flags– SF,PF,ZF reflect the result– AF is undefined– CF=last bit shifted out– OF=1 if result changes sign on the last

rotation


297

reversing a bit pattern

• 11011100à00111011– MOV CX,8

• REVERSE:– SHL AL,1– RCR BL,1– LOOP REVERSE– MOV AL,BL


298

Binary and Hex I/O

• Reads in a binary number from the keyboard, followed by a carriage return

• Convert it to a bit value, and collect the bits in a register


299

• Clear BX• Input a character• WHILE character <> CR DO

– Convert character to binary value– Left shift BX– Insert value int lsb of BX– Input a character

• END_WHILE


300

– XOR BX,BX– MOV Ah,1– INT 21H

• WHILE_:– CMP AL,0DH– JE END_WHILE– AND AL,0FH– SHL BX,1– OR BL,AL– INT 21H– JMP WHILE_

• END_WHILE:


301

Binary output

• FOR 16 times DO– Rotate left BX

• Put msb nito CF

– IF CF=1– THEN

• Output ‘1’

– ELSE• Output ‘0’

– END_IF

• END_FOR


302

Hex input

• Clear BX• Input hex character• WHILE character<> CR DO

– Convert character to binary value– Left shift BX 4 times– Insert value into lower 4 bits of BX– Input a character

• END_WHILE


303

– XOR BX,BX– MOV CL,4– MOV AH,1– INT 21H

• WHILE_:– CMP AL,0DH– JE END_WHILE– CMP AL,39H– JG LETTER– AND AL,0FH– JMP SHIFT

• LETTER:– SUB AL,37H

• SHIFT:– SHL BX,CL– OR BL,AL– INT 21H– JMP WHILE_

• END_WHILE:


304

Hex output

• For 4 times DO– MOV BH to DL– Shift DL 4 times to the right– IF DL<10

• THEN– Convert to character in ‘0’.. ‘9’

• ELSE– Convert to character in ‘A’.. ‘F’

• END_IF• Output character• Rotate BX left 4 times

– END_FOR


305

homework

• Ch7.2,7.4,14


306

The Stack and Introduction to procedures


307

The stack

• A stack is one-dimensional data structure.• Items are added and removed from one

end of the structure.• Last-inn first-out• The most recent addition to the stack is

called the top of the stack.


308

• Declaring a stack segment• .STACK 100H


309

• When the program is assembled and loaded in memory, SS will contain the segment number of the stack segment.

• For the preceding stack declaration, SP, the stack pointer, is initialized to 100h.

• When the stack is not empty, SP contains the offset address of the top of the stack.


310

• PUSH source• Ex:

– PUSH AX


311

• SP is decreased by 2• A copy of the source content is moved to

the address specified by SS:SP. The source is unchanged


312

• PUSHF:– Pushes the contents of the FLAGS register

onto the stack.


313

• Initially, SP contains the offset address of the memory location immediately following the stack segment

• The first PUSH decreases SP by 2• Making it point to the last word in the stack

segment.


314

• POP destination• Ex:

– POP BX


315

• The content of SS:SP is moved to the destination

• SP is increased by 2


316

• POPF:– Pop the top of the stack into the FLAGS

register.


317

• There is no effect of PUSH,PUSHF,POP,POPF on the flags

• Note:– PUSH and POP are word operations, so a

byte instruction is illegal– Ex: (illegal) àbut this legal for 80186/80486

• PUSH DL• PUSH 2


318

A stack application

• Read a sequence of characters and display them in reverse order on the next line


319

• Display a ‘?’• Initialize count to 0• Read a character• WHILE character is not a carriage return DO

– Push character onto the stack– Increment count– Read a character

• END_WHILE• Go to a new line• FOR count times DO

– Pop a character from the stack– Display it

• END_FOR


320

• TITLE PGM8_1:REVERSE INPUT• .MODEL SMALL• .STACK 100H• .CODE• MAIN PROC

– MOV AH,2– MOV DL.’?’– INT 21H– XOR CX,CX– MOV AH,1– INT 21H


321

• WHILE_:– CMP AL,0DH– JE END_WHILE– PUSH AX– INC CX– INT 21H– JMP WHILE_

• END_WHILE:– MOV AH,2– MOV DL,0DH– INT 21H– MOV DL,0AH– INT 21H– JCXZ EXIT


322

• TOP:– POP DX– INT 21H– LOOP TOP

• EXIT:• MOV AH,4CH• INT 21H

• MAIN ENDP– END MAIN


323

Terminology of procedures

• Procedure declaration– Name PROC type

• RET

– Name ENDP


324

• Name is the user-defined name of the procedure.• The optional operand type is NEAR or FAR (if

type is omited, NEAR is assumed)• Ps:

– NEAR – the statement that calls the procedure is in the same segment as the procedure itself.

– FAR – the calling statement is in a different segment


325

• RET– Return– Causes control to transfer back to the calling

procedure– Every procedure should have a RET – Except main procedure


326

communication

• Up to the programmer to devise a way for procedures to communicate

• Unlike high-level language, do not have parameter lists


327

Procedure documentation

• ;(describe what the procedure does)• ; input: (where it receives information from

the calling program)• ;output: (where it delivers results to the

calling program)• ;users: (a list of procedures that it calls)


328

CALL and RET

• CALL name• CALL address_expression


329

• Executing a CALL instruction causes the following to happen:– The return address to the calling program is

saved on the stack (CS:IP)– IP get the offset address of the first instruction

of the procedure


330

• RET pop_value• NEAR : pop IP• FAR : pop CS:IP


331

0100

00FE

00FC

Offset Stack segmentaddress

Code segmentOffset address

MAIN PROC

CALL PROC1Next instruction

PROC1 PROCFirst instruction

RET

00100012

0200

IP

SP


332

0100

00FE

00FC



MAIN PROC



RET

00100012

0200IP

SP0012


333

0100

00FE

00FC



MAIN PROC



RET

00100012

0200

IPSP0012

0300


334

0100

00FE

00FC



MAIN PROC



RET

00100012

0200

IP

SP0300


335

example

• Finding the product of two positive integers A and B by addition and bit shifting


336

• Product=0• REPEAT

– IF lsb of B is 1– THEN

• Product=Product +A– END_IF– Shift left A– Shift right B

• UNTIL B=0


337

• Ex:– A=111b– B=1101b

111bx 1101b

---------------------111

000111

111---------------------

1011011b


338

• TITLE PGM8_2:REVERSE INPUT• .MODEL SMALL• .STACK 100H• .CODE• MAIN PROC• CALL MULTIPLY• MOV AH,4CH• INT 21H• MAIN ENDP• MULTIPLY PROC• PUSH AX• PUSH BX• XOR DX,DX• REPEAT:• TEST BX,1• JZ END_IF• ADD DX,AX• END_IF:• SHL AX,1• SHR BX,1• JNZ REPEAT• POP BX• POP AX• RET• MULTIPLY ENDP• END MAIN


339

-u0B87:0000 E80400 CALL 0007 0B87:0003 B44C MOV AH,4C 0B87:0005 CD21 INT 21 0B87:0007 50 PUSH AX 0B87:0008 53 PUSH BX 0B87:0009 33D2 XOR DX,DX 0B87:000B F7C30100 TEST BX,0001 0B87:000F 7402 JZ 0013 0B87:0011 03D0 ADD DX,AX 0B87:0013 D1E0 SHL AX,1 0B87:0015 D1EB SHR BX,1 0B87:0017 75F2 JNZ 000B 0B87:0019 5B POP BX 0B87:001A 58 POP AX 0B87:001B C3 RET0B87:001C 03D1 ADD DX,CX 0B87:001E E6D1 OUT D1,AL


340

-rAX=0000 BX=0000 CX=001C DX=0000 SP=0100 BP=0000 SI=0000 DI=0000 DS=0B77 ES=0B77 SS=0B89 CS=0B87 IP=0000 NV UP EI PL NZ NA PO NC 0B87:0000 E80400 CALL 0007 -dss:f0 ff0B89:00F0 23 05 00 00 0B 00 87 0B-7A 05 0D 00 07 00 00 00

#.......z.......-raxAX 0000:7-rbxBX 0000:d-rAX=0007 BX=000D CX=001C DX=0000 SP=0100 BP=0000 SI=0000 DI=0000 DS=0B77 ES=0B77 SS=0B89 CS=0B87 IP=0000 NV UP EI PL NZ NA PO NC 0B87:0000 E80400 CALL 0007 -tAX=0007 BX=000D CX=001C DX=0000 SP=00FE BP=0000 SI=0000 DI=0000 DS=0B77 ES=0B77 SS=0B89 CS=0B87 IP=0007 NV UP EI PL NZ NA PO NC 0B87:0007 50 PUSH AX -dss: f0 ff0B89:00F0 23 05 00 00 07 00 00 00-07 00 87 0B 7A 05 03 00

#...........z...


341

-gbAX=0007 BX=000D CX=001C DX=0000 SP=00FA BP=0000 SI=0000 DI=0000 DS=0B77 ES=0B77 SS=0B89 CS=0B87 IP=000B NV UP EI PL ZR NA PE NC 0B87:000B F7C30100 TEST BX,0001 -dss: f0 ff0B89:00F0 23 05 00 00 0B 00 87 0B-7A 05 0D 00 07 00 03 00

#.......z.......-g17AX=000E BX=0006 CX=001C DX=0007 SP=00FA BP=0000 SI=0000 DI=0000 DS=0B77 ES=0B77 SS=0B89 CS=0B87 IP=0017 NV UP EI PL NZ NA PE CY 0B87:0017 75F2 JNZ 000B -tAX=000E BX=0006 CX=001C DX=0007 SP=00FA BP=0000 SI=0000 DI=0000 DS=0B77 ES=0B77 SS=0B89 CS=0B87 IP=000B NV UP EI PL NZ NA PE CY 0B87:000B F7C30100 TEST BX,0001 -g17AX=001C BX=0003 CX=001C DX=0007 SP=00FA BP=0000 SI=0000 DI=0000 DS=0B77 ES=0B77 SS=0B89 CS=0B87 IP=0017 NV UP EI PL NZ NA PE NC 0B87:0017 75F2 JNZ 000B


342

-tAX=001C BX=0003 CX=001C DX=0007 SP=00FA BP=0000 SI=0000 DI=0000 DS=0B77 ES=0B77 SS=0B89 CS=0B87 IP=000B NV UP EI PL NZ NA PE NC 0B87:000B F7C30100 TEST BX,0001 -g17AX=0038 BX=0001 CX=001C DX=0023 SP=00FA BP=0000 SI=0000 DI=0000 DS=0B77 ES=0B77 SS=0B89 CS=0B87 IP=0017 NV UP EI PL NZ NA PO CY 0B87:0017 75F2 JNZ 000B -tAX=0038 BX=0001 CX=001C DX=0023 SP=00FA BP=0000 SI=0000 DI=0000 DS=0B77 ES=0B77 SS=0B89 CS=0B87 IP=000B NV UP EI PL NZ NA PO CY 0B87:000B F7C30100 TEST BX,0001 -g17AX=0070 BX=0000 CX=001C DX=005B SP=00FA BP=0000 SI=0000 DI=0000 DS=0B77 ES=0B77 SS=0B89 CS=0B87 IP=0017 NV UP EI PL ZR NA PE CY 0B87:0017 75F2 JNZ 000B -tAX=0070 BX=0000 CX=001C DX=005B SP=00FA BP=0000 SI=0000 DI=0000 DS=0B77 ES=0B77 SS=0B89 CS=0B87 IP=0019 NV UP EI PL ZR NA PE CY 0B87:0019 5B POP BX


343

-tAX=0070 BX=000D CX=001C DX=005B SP=00FC BP=0000 SI=0000 DI=0000 DS=0B77 ES=0B77 SS=0B89 CS=0B87 IP=001A NV UP EI PL ZR NA PE CY 0B87:001A 58 POP AX -tAX=0007 BX=000D CX=001C DX=005B SP=00FE BP=0000 SI=0000 DI=0000 DS=0B77 ES=0B77 SS=0B89 CS=0B87 IP=001B NV UP EI PL ZR NA PE CY 0B87:001B C3 RET-dss:f0 ff0B89:00F0 70 00 70 00 07 00 00 00-1B 00 87 0B 7A 05 03 00

p.p.........z...-tAX=0007 BX=000D CX=001C DX=005B SP=0100 BP=0000 SI=0000 DI=0000 DS=0B77 ES=0B77 SS=0B89 CS=0B87 IP=0003 NV UP EI PL ZR NA PE CY 0B87:0003 B44C MOV AH,4C -gProgram terminated normally-q


344

HW

• Ch8.9


345

Multiplication and division instruction


346

MUL and IMUL

• The process of multiplication and division is different for signed and unsigned number

• There are byte and word forms


347

ex

• 10000000 x 1111111– Unsigned:

128x255=32640(0111111110000000– Signed:

-128x-1=128(0000000010000000


348

• MUL (multiply)• IMUL (integer multiply)


349

• MUL source• IMUL source


350

Byte Form

• One umber is contained in the source• The other is assumed to be in AL• 16-bit production will be in AX• Ps:

– Source may be a byte register or memory byte, but not a constant


351

Word form

• One number is contained in the source• The other is assumed to be in AX• The most significant 16 bits of the double

word product will be in DX• The least significant 16 bits will be in AX• DX:AX


352

• For positive number ( 0 in the most significant bit) MUL and IMUL give the same result


353

Effect of MUL/IMUL on the status flags

• SF,ZF,AF,PF: undefined• CF/OF:

– After MUL,CF/OF • =1 it he upper half of the result is zero• =1 otherwise

– After IMUL, CF/OF• =0 if the upper half of the result is the sign

extension of the lower half ( the bits of the upper half are the same as the sign bit of the lower half)

• =1 otherwise


354

• If CF/OF=1 – The product is too big to fit in the lower half of

the destination (ALor AX)


355

ex

• Suppose AX contains 1 and BX contains FFFFh


356

0FFFFFFFFFFFFFFFF

-1IMUL BX

0FFFF00000000FFFF65535MUL BX

CF/OFAXDXHex product

Decimal product

Instruction


357

ex

• Suppose AX contains FFFFh and BX contains FFFFh


358

000010000000000011IMUL BX

10001FFFEFFFE0001

4294836225

MUL BX


Decimal product

Instruction


359

ex

• Suppose AX contains 0FFFh


360

100100FF00FFE00116769025IMUL AX

1E00100FF00FFE00116769025MUL AX


Decimal product

Instruction


361

ex

• Suppose AX contains 0100h and CX contains FFFFh


362

0FF00FFFFFFFFFF00

-256IMUL CX

1FF0000FF00FFFF0016776960MUL CX


Decimal product

Instruction


363

ex

• Suppose AL contains 80h and BL contains FFh


364

180000080128IMUL BL

1807F7F80128MUL BL

CF/OFALAHHex product

Decimal product

Instruction


365

application

• Ex : translate the high-level language assignment statement A=5xA-12xB into assembly code. Let A and B be word variables, and suppose there is no overflow. Use IMUL for multiplication


366

• MOV AX,5• IMUL A• MOV A,AX• MOV AX,12• IMUL B• SUB A,AX


367

ex

• Write a procedure FACTORIAL that will compute N! for a positive integer N. The procedure should receive N in CX and return N! in AX. Suppose that overflow does not occur


368

• N!=1 if N=1• =Nx(N-1)x(N-2)x… x1 if N>1


369

• Product =1• Term=N• FOR N times DO

– product=product x term– term=trem-1

• ENDFOR


370

• FACTORIAL PROC– MOV AX,1

• TOP:– MUL CX– LOOP TOP– RET

• FACTORIAL ENDP• Ps: CX=N,AX=N!


371

DIV and IDIV

• When division is performed, we obtain two results– Quotient– remainder


372

• DIV divisor• IDIV divisor


373

Byte form

• Divisor is an 8-bit register or memory byte• 16-bit dividend is assumed to be in AX• 8-bit quotient is in AL• 8-bit remainder is in AH• Ps:

– Divisor may not be a constant


374

Word form

• The divisor is a 16-bit register or memory• 32-bit dividend is assumed to be in DX:AX• 16-bit quotient is in AX• 16-bit remainder is in DX• For signed division, the remainder has the

same sign as the dividend


375

• The effect of DIV/IDIV on the flags is that all status flags are undefined


376

Divide overflow

• The quotient will be too big to fit in the specified destination

• Devisor is much smaller than dividend


377

ex

• Suppose DX contains 0000h, AX contains 0005h , and BX contains 0002h


378

0001000212IDIV BX

0001000212DIV BX

DXAXDecimal remainder

Decimal quotient

instruction


379

ex

• Suppose DX contains 0000h, AX contains 0005h, and BX contains FFFEh


380

0001FFFE1-2IDIV BX

0005000050DIV BX


Decimal quotient

instruction


381

• Suppose DX contains FFFFh, AX contains FFFBh, and BX contains 0002


382

DIVIDE OVERFLOW

IDIV BX

FFFFFFFE-1-2DIV BX


Decimal quotient

instruction


383

ex

• Suppose AX contains 00FBh and BL contains FFH


384

DIVIDE OVERFLOW

IDIV BL

00FB2510DIV BL


Decimal quotient

instruction


385

Sign extension of the dividend

• In word division, the dividend is in DX:AX even if the actual dividend will fit in AX.– For DIV,DX should be cleared– For IDIV,DX should be made the sign

extension of AX.the instruction CWD(convertword to doubleword) will be the extension


386

ex

• Divide -1250 by 7


387

• MOV AX,-1250• CWD• MOV BX,7• IDIV BX


388

Byte division

• In byte division, the dividend is in AX.• If the actual dividend is a byte, then AH

should be prepared as follows– For DIV, AH should be cleared– For IDIV, AH should the sign extension of AL.

the instruction CBW (convert byte to word) will do the extension


389

ex

• Divide the signed value of the byte variable XBYTE by -7


390

• MOV AL,XBYTE• CBW• MOV BL,-7• IDIV BL


391

• There is no effect of CBW and CWD on the flags


392

Decimal input and output procedures

• Decimal output• Decimal input


393

Decimal output

• IF AX<0 – THEN

• Print a minus sign• Replase AX by its two’s complemet

• END_IF• Get the digits in AX’s decimal

representation• Convert these digits to characters and

print them


394

Get the digits in AX’s decimal representation

• Count=0• ERPEAT

– Divide quotient by 10– Push remainder on the stack– Count=count+1

• UNTIL quotient=0


395

Convert these digits to characters and print them

• FOR count times DO– Pop a digit from the stack– Convert it to a character– Output the character

• END_FOR


396

OUTDEC PROCPUSH AXPUSH BXPUSH CXPUSH DXOR AX,AXJGE @END_IF1PUSH AXMOV DL,’-’MOV AH,2INT 21HPOP AXNEG AX

@END_IF1:XOR CX,CXMOV BX,10D

@REPEAT1:XOR DX,DXDIV BXPUSH DXINC CXOR AX,AXJNE @REPEAT1

PGM9_1.asm


397

MOV AH,2@PRINT_LOOP:

POP DXOR DL,30HINT 21HLOOP @PRINT_LOOPPOP DXPOP CXPOP BXPOP AXRET

OUTDEC ENDP


398

INCLUDE Pseudo-op

• INCLUDE filespec• Ex:

– INCLUDE A:PGM9_1.asm


399

• TITLE PGM9_2: DECIMAL OUTPUT• .MODEL SMALL• .STACK 100H• .CODE• MAIN PROC

– CALL OUTDEC– MOV AH,4CH– INT 21H

• MAIN ENDP• INCLUDE A:PGM9_1.asm

– END MAIN


400

• Debug pgm9_2.exe• -rax• AX 0000• :9C71• -g• -25487• Program terminated normally• -rax• AX 9C71• :28E• -g• 654


401

Decimal input

• Convert a string of ASCII digits to the binary representation of a decimal integer


402

• Print a question mark• Total=0• Negative =false• Read a character• CASE character OF

– ‘-’: negative=true– Read character– ‘+’:read a character

• END_CASE• REPEAT

– IF character is not between ‘0’and ‘9’• THEN

– Go to beginning– ELSE

• Convert character to a binary value• Total =10xtotal +value

– END_IF– Read a character

• UNTIL character is a carriage return• IF negatiev = true

– THEN• Totla=-total

– ENDIF


403

exercise

• A=5xA-7• B=(A-B)x(B+10)• A=6-9xA


404

A=5xA-7

• MOV AX,5• IMUL A• SUB AX,7• MOV A,AX


405

B=(A-B)x(B+10)

• MOV AX,A• SUB AX,B• MOV BX,B• ADD BX,10• IMUL BX• MOV B,AX


406

A=6-9xA

• MOV AX,9• IMUL A• NEG AX• ADD AX,6• MOV A,AX


407

– MOV AX,A– IMUL AX– MOV A,AX– MOV AX,B– IMUL AX– MOV B,AX– MOV AX,C– MUL AX,AX– SUB AX,A

• IF_:– CMP AX,B– JNE ELSE

• THEN:– STC– JMP EXIT

• ELSE:– CLC

• EXIT:


408

exercise

• IF A^2+B^2=C^2• THEN

– SET CF

• ELSE– Clear CF

• END_IF


409

• MOV AX,A• IMUL AX• MOV A,AX• MOV AX,B• IMUL AX• MOV B,AX• MOV AX,C• MUL AX,AX• SUB AX,A• IF_:

– CMP AX,B– JNE ELSE

• THEN:– STC– JMP EXIT

• ELSE:– CLC

• EXIT:


410

Decimal Input• Print a question mark• 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’and ‘9’– THEN

• Go to beginning

– ELSE• Convert character to a binary value• Total =10xtotal+value

– END_IF– Read a character

• UNTIL character is a carriage return• IF negative =true• THEN• Total=-total• END_IF


411

PGM9_3.asm


412

PGM9_4.asm


413

HW

• Ch9.7,11


414

Arrays and Addressing Modes


415

One-dimensional arrays

• An ordered list of elements, all of the same type.


416

• MSG DB ‘abcde’• W DW 10,20,30,40,50,60


417

• Base address of the array


418

DUP (duplicate)

• Define arrays whose elements share a common initial value

• repeat_count DUP (value)


419

ex

• GAMMA DW 100 DUP(0)• DELTA DB 212 DUP(?)• LINE DB 5,4, 3 DUP(2,3, DUP (0), 1)• LINE DB 5,4,2,0,0,0,1,2,0,0,0,1,2,0,0,0,1


420

Location of array elements

• A=(N-1)xS• S: number of bytes in an element• A:array


421

ex

• exchange the 10th and 25th elements in a word array W.


422

• MOV AX,W+18• XCHG W+48,AX• MOV W+18,AX

• Ps:• W[10] is located at address W+9x2=18• W[25] is located at address

W+24x2=W+48


423

Address modes

• Register mode• Immediate mode• Direct mode


424

Register indirect mode

• [register]• The register is BX,SI,DI or BP• Segment is contained in DS• MOV AX,[SI] ; not same as MOV AX,SI• (suppose SI contain 0100h))• The CPU

– Examines SI and obtains the offset address 100h– Use the address DS:0100h to obtain the value 1234h– Move 1234h to AX


425

ex

Offset 3000h contains 031DhDI contains 1000h

Offset 2000h contains 20FEhSI contains 1000h

Offset 1000h contains 1BAChBX contains 1000h

Tell which of the following instruction are legal, give the source offset address, and the result or number movesa.MOV BX,[BX]b.MOV CX,[SI]c.MOV BX,[AX]d.ADD [SI],[DI]e.INC [DI]


426

031Eh3000he

Illegal memory-memory addition

d

(must be BX,SI or DI)

Illegal source registerc

20FEh2000hb

1BACh1000ha

resultSource offset


427

ex

• Write some code to sum in AX the elements of the 10-element array W define by

• W DW 10,20,30,40,50,60,70,80,90,100


428

– XOR AX,AX– LEA SI,W– MOV CX,10

• ADDNOS:– ADD AX,[SI]– ADD SI,2– LOOP ADDNOS


429

Ex

• Write a procedure REVERSE that wullreverse an array of N words.

• The procedure is entered with SI pointing the array, and BX has the number or words N


430

• The idea is to exchange the 1st and N thwords, the 2nd and (N-1)st words, and so on. The number of exchanges will be N/2(rounded down to the nearest integer if N is odd) .

• Ps:– Nth element in a word array A has address

A+2x(N-1)


431

REVERSE PROCPUSH AXPUSH BXPUSH CXPUSH SIPUSH DIMOV DI,SIMOV CX,BXDEC BXSHL BX,1ADD DI,BXSHR CX,1

XCHG_LOOP:MOV AX,[SI]XCHG AX,[DI]MOV [SI],AXADD SI,2SUB DI,2LOOP XCHG_LOOPPOP DIPOP SIPOP CXPOP BX

POP AXRET

REVERSE ENDP


432

Based and Indexed addressing mode

• Displacement may be any of the following:– The offset address of a variable– A constant (positive or negative)– The offset address of a variable plus or minus

a constant


433

• Example of displacements• (if A is a variable)

– A (offset address of a variable)– -2 (constant)– A+4 (offset address of a variable plus a

constant)


434

• [register+displacement]• [displacement+register]• [register]+displacement• Displacement+[register]• Displacement[register]


435

• The register must be BX,BP,SI or DI• If BX,SI or DI is used, DS contains the segment

number of the operand’s address• If BP is used,SS has the segment number.• The addressing model is called based if BX

(base register) or BP(base pointer) is used• The addressing model is called indexed if

SI(source index) or DI(destination index) is used


436

ex

• MOV AX,W[BX]• MOV AX,[W+BX]• MOV AX,[BX+W]• MOV AX,W+[BX]• MOV AX,[BX]+W• MOV AX,[SI+2]• MOV AX,[2+SI]• MOV AX,[SI}+2• MOV AX,2[SI]


437

ex

• Write some code to sum in AX the element of 10-element array W defined by

• W DW 10,20,30,40,50,60,70,80,90,100


438

– XOR AX,AX– XOR BX,BX– MOV CX,10

• ADDNOS:– ADD AX,W[BX]– ADD BX,2– LOOP ADDNOS


439

ex• Suppose ALPHA is declared as • ALPHA DW 0123h,0456h,0789h,0ABCDh• In the segment address by DS, suppose that • BX contains 2 offset 0002 contains 1084h• SI contains 4 offset 0004 contains 2BACh• DI contains 1• Tell which of the following instruction are legal, if legal, give the source

offset address and the number moveda. MOV AX,[ALPHA+BX]b. MOV BX,[BX+2]c. MOV CX,ALPHA[SI]d. MOV AX,-2[SI]e. MOV BX,[ALPHA+3+DI]f. MOV AX,[BX]2g. ADD BX,[ALPHA+AX]


440Illegal source registerg

Illegal form of source operand

F

0789hALPHA+3+1=ALPHA+4E

1084h-2+4=2D

0789hALPHA+4C

2BACh2+2=4B

0456hALPHA +2A

Number movedSource offset


441

ex

• Replace each lowercase letter in the following string by its upper case equivalent. Use index addressing mode

• MSG DB ‘this is a message’


442

– MOV CX,17– XOR SI,SI

• TOP:– CMP MSG[SI],’‘– JE NEXT– AND MSG[SI],0DFh

• NEXT:– INC SI– LOOP TOP


443

The PRT operator and LABEL pseudo-op

• Operands of an instruction must be of the same type

• If one operand is a constant, the assembler attempts to infer the type from the other operand


444

• MOV [BX],1 ;illegal ,can’t assemble

• Shall be• MOV BYTE PTR [BX],1• MOV WORD PTR [BX],1


445

ex

• LEA SI,MSG• MOV BYTE PRT [SI],’T’


446

ex

• Using index mode– XOR SI,SI– MOV MSG[SI],’T’– Ps:here it is not necessary to use th PTR

operator, because MSG is a byte variable


447

Use PTR to override a type

Type PTR address_expressionEx:

DOLLARS DB 1AhCENTS DB 52hMOV AX,DOLLARS ;illegalMOV AX,WORD PRT DOLLARS;AL=dollars,AH=cents


448

The LABEL Pseudo-Op

• Declaration types MONEY as a word variable, and the components DOLLARS and CENTS as byte variables– MONEY LABEL WORD– DOLLAR DB 1Ah– CENTS DB 52h


449

• MOV AX,MONEY; AL=dollars,AH=cents


450

Segment override

• Specify an relative to one of the other segment registers

• Segment_register:[pointer_register]


451

• MOV AX,ES:[SI]


452

Accessing the stack

• Move the top three words on the stack into AX,BX, and CX without changing the stack


453

• MOV BP,SP• MOV AX,[BP]• MOV BX,[BP+2]• MOV,[BP+4]


454

Selection algorithm

• i=N• FOR N-1 times DO

– Find the position k of the largest elemnt• Among A[1]..A[i]

– (*) swap A[k] and A[i]– i=i-1

• END_FOR


455

• PGM10_2.asm


456

Two-dimensional arrays

• Memory is one-dimensional ,the elements of a two-dimensional array must be stored sequentially

• Two common used ways– Row-major order– Column-major order


457

• B DW 10,20,30,40• DW 50,60,70,80• DW 90,100,110,120


458

• MxN– A[i][j] è A+((i-1)xN+(j-1))xS– A[i][j] è A+((i-1)+(j-1)xM)xS


459

Based indexed addressing mode

• Offset address of the operand is the sum of– The contents of a base register (BX or BP)– The contents of an index register (SI or DI)– Optionally, a variable’s offset address– Optionally, a constant (positive or negative)


460

• Variable[base_register][index_register]• [base_register+index_register+variable+co

nstant]• Variable[base_register+index_register+co

nstant]• Constant[base_register+index_register+va

riable]


461

ex

• Suppose A is a 5x7 word array stored in row-major order.write some code to (1) clear row 3,(2) clear column4 use based indexed mode


462

• (1)– MOV X,28– XOR SI,SI– MOV CX,7

• CLEAR:– MOV A[BX][SI],0– ADD SI,2– LOOP CLEAR


463

• (2)– MOV SI,6– XOR BX,BX– MOV CX,5

• CLEAR:– MOV A[BX][SI],0– ADD BX,1– LOOP CLEAR


464

The XFLAT instruction

• A no-operand instruction that can be used to convert a byte value into another value that comes from a table

• The byte to be convert must be in AL, and BX has the offset address of the conversion table


465

• The instruction– Add the contents of Al to the address n BX to

produce an address within the table– Replaces the contents of AL by the value

found at that address


466

ex

• TABLE DB 030h,031h,032h,033h,034h,035h,036h,037h,038h,039h

• DB 041h,042h,043h,044h,045h,046h• MOV AL,0Ch• LEA BX,TABLE• XLAT• àAL has ‘C’


467

HW

• Ch10.6


468

The string instructions

• Copy string• Search a string for a particular byte or

word• Store characters in a string• Compare strings of characters

alphabetically


469

Direction flag

• DF (direction flag)– determine the direction in which string

operations will proceed.– These operations are implemented by the two

index register SI and DI


470

• STRING1 DB ‘ABCDE’

• If DF=0,SI and DI proceed in the direction of increasing memory addresses from left to right across the string.

• If DF=1, SI and DI proceed in the direction of decreasing memory address: from right to left


471

CLD and STD

• CLD clear direction flag• STD set direction flag


472

Moving a string

• MOVSB • Copies the contents of the byte addressed

by DS:SI, to the byte address by ES:DI


473

ex• .DATA• STRING 1 DB ‘HELLO’• STRING2 DB 5 DUP(?)• MOV AX,@DATA• MOV DS,AX• MOV EX,AX• LEA SI,STRING1• LEA DI,STRING2• CLD

MOVSB• MOVSB


474

Move entire string

• REP MOVSB

• Ex:– CLD – LEA SI,STRING1– LEA DI,STRING2– MOV CX,5– REP MOVSB


475

Ex:

• Write instructions to copy STRING1 into STRING2 in reverse order


476

– LEA SI,STRING1+4– LEA DI,STRNIG2– STD– MOV CX,5

• MOVE:– MOVSB– ADD DI,2– LOOP MOVE


477

Move word string

• MOVSW • Moves a word from the source string to the

destination string


478

ex

• For the following array• ARR DW 10,20,40,50,60,?• Write instructions to insert 30 between 20

and 40.( assume DS and ES have been initialized to the data segment.)


479

• STD• LEAD SI,ARR+8h• LEA DI,ARR+Ah• MOV CX,3• REP MOVSW• MOV WORD PTR [DI],30


480

Stroe string

• STOSB– Move the contents of the AL register to the

byte addressed by ES:DI.

• STOSW– Move the contents of the AX register to the

byte addressed by ES:DI.


481

ex

• MOV AX,@DATA• MOV ES,AX• LEA DI,STRING1• CLD• MOV AL,’A’• STOSB• STOSB


482

Reading and storing a character string

• Use INT 21h, function 1 and STOSB


483

• Chars_read=0• Read a char• WHILE char is not a carriage return DO

– IF char is a backspace– THEN chars_read=chars_read-1

• Chars_read=chars_read-1• Remove previous char from string

– ELSE• Store char ni string• Chars_read=chars_read+1

– END_IF– Read a char

• END_WHILE


484

• READ_STR PROC NEAR– PUSH AX– PUSH DI– CLD– XOR BX,BX– MOV AH,1– INT 21H

• WHILE1:– CMP AL,0DH– JE END_WHILE1– CMP AL,8H– JNE ELSE1– DEC DI– DEC BX– JMP READ

• ELSE1:– STOSB– INC BX

• READ:– INT 21H– JMP WHILE1

• END_WHILE1:– POP DI– POP AX– RET

• READ_STR ENDP


485

Load string

• LODSB– Moves the byte addressed by DS:SI into

AL,SI is then incremented if DF=0 or decremented if DF=1

• LODSW– Moves the byte addressed by DS:SI into

AX,SI is then incremented by 2 if DF=0 or decremented by 2 if DF=1


486

Displaying a character string

• FOR count times DO– Load a string character into AL– Move it to DL– Output character

• END_FOR


487

• Let Si =offset of string,BX=no. of chars. To display– MOV CX,BX– JCXZ P_EXIT– CLD– MOV AH,2

• TOP:– LODSB– MOV DL,AL– INT 21H– LOOP TOP– P_EXIT:


488

Scan string

• SCANSB• Examine a string for a target byte• The target byte is contained in AL.• SCANSB subtracts the string byte pointed

to by ES:DI from contents of AL and uses the result to set the flags

• The result is not stored


489

• SCANSW– If not found ZF=0


490

• REPNZ (repeat while not zero)• REPNE (the same as above)• Ex:

– REPNE SCANSB


491

Ex:counting vowels and consonants

• PGM11_4


492

Compare string

• CMPSB– Subtracts the byte with address ES:DI from

the byte with address DS:SI, and sets the flags. the result is not stored

• CMPSW


493

• .DATA• STRING1 DB ‘ACD’• STRING2 DB ‘ABC’• MOV AX,@DATA• MOV DS,AX• MOV ES,AX• CLD• LEA SI,STRING1• LEA DI,STRING2• CMPSB• CMPSB

assembly language

Documents

fang department of csie

fang w

microcomputer systems

acceptable w

ibm pc w

content of ax

rom bus w

content of memory word