8086 segmentation and addressing modes

Addressing mode Mnemonic Segment for memory access

Symbolic representation

IMMEDIATE

IMMEDIATE MOV AX,1000

IMMEDIATE MOV AX,1000 CODE

IMMEDIATE MOV AX,1000 CODE AH<=10, AL<=00

REGISTER

REGISTER MOV DX,CX

REGISTER MOV DX,CX WITHIN THE CPU

DX<=CX

DIRECT

DX<=CX

DIRECT MOV AL,[2000]

DX<=CX

DIRECT MOV AL,[2000] DATA

DX<=CX

DIRECT MOV AL,[2000] DATA AL<=[2000]

DX<=CX

REGISTER INDIRECT

DX<=CX

REGISTER INDIRECT

MOV AX,[SI]

JMP [DI]

INC BYTEPTR[BP]

DEC WORDPTR[BX]

DX<=CX

REGISTER INDIRECT

MOV AX,[SI] DATA

JMP [DI] DATA

INC BYTEPTR[BP] STACK

DEC WORDPTR[BX] DATA

DX<=CX

REGISTER INDIRECT

MOV AX,[SI] DATA AL<=[SI]AH<=[SI+1]

JMP [DI] DATA

DX<=CX

REGISTER INDIRECT

JMP [DI] DATA IP <={[DI+1] [DI]}

DX<=CX

REGISTER INDIRECT

INC BYTEPTR[BP] STACK [BP]<=[BP]+1

DX<=CX

REGISTER INDIRECT

INC BYTEPTR[BP] STACK [BP]<=[BP]+1

DEC WORDPTR[BX] DATA {[BX+1][BX]}<={[BX+1][BX]} – 1

RELATIVE INDEXED(RELATIVE REGISTER INDIRECT)

MOV AX,[SI+6]

JMP [DI+6]

RELATIVE BASED(RELATIVE REGISTER INDIRECT)

MOV AX,[BP+2]

JMP [BX+2]

MOV AX,[SI+6] DATA

JMP [DI+6] DATA

MOV AX,[BP+2] STACK

JMP [BX+2] DATA

MOV AX,[SI+6] DATA AL<=[SI+6], AH<=[SI+7]

JMP [DI+6] DATA

MOV AX,[BP+2] STACK

JMP [BX+2] DATA

JMP [DI+6] DATA IP <={[DI+7] [DI+6]}

MOV AX,[BP+2] STACK

JMP [BX+2] DATA

MOV AX,[BP+2] STACK AL<=[BP+2], AH<=[BP+3]

JMP [BX+2] DATA

MOV AX,[BP+2] STACK AL<=[BP+2], AH<=[BP+3]

JMP [BX+2] DATA IP <={[BP+3] [BP+2]}

BASED AND INDEXED

MOV AX,[BX+SI]

JMP [BX+DI]

INC BYTEPTR[BP+SI]

DEC WORDPTR[BP+DI]

BASED AND INDEXED

MOV AX,[BX+SI] DATA

JMP [BX+DI] DATA

INC BYTEPTR[BP+SI] STACK

DEC WORDPTR[BP+DI]

BASED AND INDEXED

MOV AX,[BX+SI] DATA AL<=[BX+SI], AH<=[BX+SI+1]

JMP [BX+DI] DATA

DEC WORDPTR[BP+DI]

BASED AND INDEXED

JMP [BX+DI] DATA IP <={[BX+DI+1] [BX+DI]}

DEC WORDPTR[BP+DI]

BASED AND INDEXED

INC BYTEPTR[BP+SI] STACK [BP+SI]<=[BP+SI]+1

DEC WORDPTR[BP+DI]

BASED AND INDEXED

INC BYTEPTR[BP+SI] STACK [BP+SI]<=[BP+SI]+1

DEC WORDPTR[BP+DI]

STACK {[BP+DI+1][BP+DI]} <={[BP+DI+1][BP+DI]} - 1

BASED AND INDEXED WITH DISPLACEMENT (RELATIVE BASED AND INDEXED]

MOV AX,[BX+SI+5]

JMP [BX+DI+5]

INC BYTEPTR[BP+SI+5]

DEC WORDPTR[BP+DI+5]

MOV AX,[BX+SI+5] DATA

JMP [BX+DI+5] DATA

MOV AX,[BX+SI+5] DATA AL<=[BX+SI+5], AH<=[BX+SI+6]

JMP [BX+DI+5] DATA

JMP [BX+DI+5] DATA IP <={[BX+DI+6] [BX+DI+5]}

STACK [BP+SI+5]<=[BP+SI+5]+1

STACK {[BP+DI+6][BP+DI+5]} <={[BP+DI+6][BP+DI+5]} - 1

STRING MOVSB

IMPLICIT CLD

STRING MOVSB EXTRA, DATA

IMPLICIT CLD WITHIN THE CPU

STRING MOVSB EXTRA, DATA ES:[DI]<=DS:[SI]IF DF=0, INCREMENT SI AND DI BY 1IF DF =1, DECREMENT SI AND DI BY 1

IMPLICIT CLD WITHIN THE CPU

DF = 0

IDENTIFY THE ADDRESSING MODE

MOV AH,47MOV AH,[BP+2]MOV AL,[BP+SI]STDMOV AH,BLMOV DX,[2002]

Problem statement – clear the contents of three memory location

Memory content location62380 0062381 0062382 00

8086 Sample programs MOV AX,6000H move 6000h into AX MOV ES,AX move AX contents into ES MOV DI,2380H move 2380H into DI MOV CX,0003H move 0003H into CX MOV AL,00H move 00H into ALAgain MOV [DI],AL AL => ES:[DI} INC DI Increment the contents of DI LOOP AGAIN dec CX, and if CX is NZ go to again HLT Halt

8086 Sample programs MOV AX,6000H move 6000h into AX MOV ES,AX move AX contents into ES MOV DI,2380H move 2380H into DI CLD clear the direction flag MOV CX,0003H move 0003H into CX MOV AL,00H move 00H into ALREP STOSB AL => ES:[DI]and increment DI (rep this CX times) HLT Halt

(CX=0003h) (AL = 00h)(DI=2380h)(ES=6000h)

• 1st time REP STOSBAL =>ES:[DI] i.e. 00=> [62380]h and DI = 2381h

CX=0002h• 2ND time REP STOSB

AL =>ES:[DI] i.e. 00=> [62381]h and DI = 2382hCX=0001h

• 3RD time REP STOSBAL =>ES:[DI] i.e. 00=> [62382]h and DI = 2383h

CX=0000h• 4TH Time will not happen cos CX = 0000h

REP LOOPPREFIX INSTRUCTIONUSED ONLY BEFORE STRING INSTRUCTION

USED ANYWHERE

USED TO REPEAT ONLY THAT STRING INSTRUCTION

USED TO REPEAT SERIES OF INSTRUCTION

String instructionsSTOSB AL=>ES:[DI] After string opr, if DF = 0, then DI is incremented by 1and viceversa LODSB AL<=DS:[SI]After string opr, if DF = 0, then SI is incremented by 1and viceversa MOVSB DS:[SI]=>ES:[DI]After string opr, if DF = 0, then SI and DI is incremented by 1and viceversa SCASB AL – ES:[DI]After string opr, if DF = 0, then DI is incremented by 1and viceversa CMPSB DS:[SI] – ES:[DI]After string opr, if DF = 0, then SI and DI is incremented by 1and viceversa

Problem statement block transfer

Transfer contents of 52500 to 62500 and 52501 to 62501

Block transferMOV AX,5000 HMOV DS,AXMOV SI,2500HMOV AX,6000HMOV DI,2500HCLDMOV CX,0002HREP MOVSB DS:[SI] =>ES:[DI] and increment DI and SI (rep this CX times)

(CX=0002h) (DS=5000h)(SI=2500h)(DI=2500h)(ES=6000h)

• 1st time REP MOVSB DS:[SI] =>ES:[DI] i.e. [52500]=>[62500]

SI =2501 and DI = 2501CX=0001h

• 2ND time REP MOVSBDS:[SI] =>ES:[DI] i.e. [52501]=>[62501]

SI =2502 and DI = 2502CX=0000h

• 3rd Time will not happen cos CX = 0000h

8086 sample programs

When both the numbers are positive(+03)h x (+02)h = (+06)hMOV AL,03HMOV BL,02HIMUL BLResult: (+0006)h => AX

8086 sample programsWhen both the numbers are not positive(+03)h x (-02)h = (-06)h2’s complement of 02 = 1’s complement of 00000010 + 1=11111110 = FEhMOV AL,03HMOV BL,FEHIMUL BLResult: (FFFA)h => AX(FFFA)h is actually –(0006)h

(+09)h / (-02)h MOV AX,0009HMOV BL,FEHIDIV BLResult: (01FC)h => AX01 =>AH is the remainderFC =>AL is the quotient (-04)h

(-09)h / (+02)h MOV AX,FFF7HMOV BL,02HIDIV BLResult: (FFFC)h => AXFF =>AH is the remainder (-01)hFC =>AL is the quotient (-04)h

MOV AX,5000 H MOV DS,AXMOV SI,2500HMOV AX,6000HMOV ES,AXMOV DI,2500HCLDMOV CX,0004HREPNE CMPSB DS:[SI] - ES:[DI] and increment DI and SI (REPNE) HLT

MEMORY SOURCE

CONTENT MEMORYDEST

CONTENT

52500 42 62500 98

52501 45 62501 34

52502 34 62502 34

52503 23 62503 23

(CX=0004h) (DS=5000h)(SI=2500h)(DI=2500h)(ES=6000h)• 1st time REPNE CMPSB

DS:[SI] - ES:[DI] i.e. [52500]-[62500] i.e. 42-98(NE) SI =2501 and DI = 2501

CX=0003h(this is cos of REP prefix)• 2ND time REP CMPSB DS:[SI] - ES:[DI] i.e. [52501]-[62501] i.e. 45-34(NE) SI =2502 and DI = 2502

CX=0002h(this is cos of REP prefix)• 3rd Time REPNE CMPSBDS:[SI] - ES:[DI] i.e. [52502]-[62502] i.e. 34-34(E) SI =2503 and DI = 2503

CX=0001h(this is cos of REP prefix)• 4TH Time REPNE CMPSB will not happen cos locations are no

longer NE

8086 segmentation and addressing modes

Documents

University of Mumbai - library.spit.ac.inlibrary.spit.ac.in/syllabus/TE5-6-COMP-2009.pdf · 8086 microprocessor in details, Addressing modes of 8086/88, Programming the 8086 in assembly

1327 Addressing Modes Of 8086

8086 Addressing Modes - Rajdhani College is in register and Instruction Specifies the perticular register ... 8086 has Base register and Index register

80x86 ADDRESSING MODES - EduTechLearnersedutechlearners.com/download/MP/Addressing Modes.pdf · 3. Data Memory Addressing Mode. 80x86 Memory Addressing Modes. The 8086 provides 17

8086 Addressing Modes - ShiningStudy · 8086 Addressing Modes. 8086 Addressing Modes. 1.Register Addressing Mode 7 ... 8086 has Base register and Index register

College of Arts, Commerce and Science, Parbhani€¦ · 8086 Microprocessor-EU and BIU, Features of 8086, Pin diagram of 8086, Addressing modes, Instruction set classification, Assembly

2.5Addressing modes in 8086

Addressing Modes of 8086 Processor Ammar Anwar Khan Electrical Engineer King Saud University Riyadh Saudi Arabia

8memory Segmentation of 8086

EC6504-MICROPROCESSOR AND MICROCONTROLLER UNIT-I THE 8086 … · COURSE OBJECTIVE: Know about the Architecture, addressing modes and instruction sets of 8086 Microprocessor. PART-A

Module 2 Programming 8086 - swatitcet.files.wordpress.com · Module 2 Programming 8086 2.1 Motivation Motivation of this course is to provide the basic knowledge of addressing modes

UNIT II Addressing Modes, Instruction Set and Programming ... · Addressing Modes, Instruction Set and Programming of 8086 Addressing Modes of 8086: Addressing mode indicates a way

INTERRUPT INTERFACE OF THE 8088 AND 8086 ...un.uobasrah.edu.iq/lectures/3108.pdf8088 or 8086 microprocessor to initiate variations in the basic operating modes defined by the ICW commands

8086/8088 Addressing Modes, Instruction Set ... - XMU …mirel.xmu.edu.cn/mirel/public/Study/PMI/Chapter2.pdf · Summary of the addressing modes Addressing Mode Register Immediate

8086 Addressing Modes - · PDF file8086 Programmers Model 2. 8086 Programmers Model. 8086 Addressing Modes 4. 1. Register Addressing Mode 2. Immediate Addressing Mode 3

8086 ppt-address modes-instruction-set

P9 8086 Addressing Modes

Memory Segmentation of Intel 8086

MELJUN CORTES Memory segmentation, stacks and addressing modes

8086/8088 Addressing Modes, Instruction Set & Machine Codes