Instruksi Mikroprosesor. Mode Pengalamatan-Jenis[1]

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Oleh. Junartho Halomoan (juned_new@yahoo.com)

Mikroprosesor dan Antarmuka

Instruksi Mikroprosesor

Mode Pengalamatan-Jenis[1]



_{Addressing Mode / Mode Penglamatan : adalah}

cara, bagaimana 8088 dapat mengakses

operand



_{Mode Pengalamatan pada 80x86:}

– (1) register

– (2) immediate

– (3) direct

– _{(4) register indirect/ register tidak langsung} – (5) based relative/ relatif berbasis

– (6) indexed relative/ relatif berindex

– (7) based indexed relative/ relatif berbasis dan index

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2

Mode Pengalamatan - Reg [2]

Menggunakan register untuk menyimpan data yang akan dimanipulasi

Pada mode ini tidak operasi pada memori Operasi relatif cepat

Contoh:

 MOV BX, DX ;copy isi DX ke BX

 MOV ES, AX ;copy isi AX ke ES

 _{ADD AL, BH ;jumlahkan isi BH dan AL, hasilnya di AL}

_{Register sumber dan tujuan mempunyai ukuran yang} sama

Mode Pengalamatan-Imm[3]

Operand (source) adalah konstanta, yang terletak setelah opcode

Operasinya sangat cepat

Immediate addressing mode dapat digunakan pada semua register, kecuali register segmen dan flag (?) Contoh:

 MOV AX,2550H ; bilangan 2550H dimasukkan ke AX

 MOV CX,625 ; bilangan 625d dimasukkan ke CX

 MOV BL, 40H ; bilangan 40H dimasukkan ke BL

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Ditandai dengan [ ]

Mode Pengalamatan-Direct[4]

 Operand dari instruksi ini merupakan alamat memori data yang akan diakses

 Alamat ini merupakan EA (Effective Address)  Contoh:

 MOV DL, [2400] ;copy isi memori dengan alamat DS:2400H ke DL

Mode Pengalamatan-Direct[5]

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4 Ditandai

dengan [ ]

Mode Pengalamatan-Indirect[6]

Alamat lokasi memori data yang akan diakses tersimpan dalam register

Register yang digunakan pada mode ini : SI, DI, dan BX Contoh:

 MOV AL,[BX]

PA (Physical Address) dan EA (Effective Address) ???

Mode Pengalamatan-Indirect[7]

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Mode Pengalamatan–Based.Re[8]

Menggunakan register BX and BP, untuk mendapatkan EA (effective address), ditambah dengan displacement Segment yang digunakan untuk mendapatkan physical

address (PA) adalah:

 DS untuk BX

 SS untuk BP Contoh :

 MOV CX,[BX]+10

Salinkan isi DS:BX+10 dan DS:BX+10+1 ke reg. CX (little endian convt.);

PA = ?

Mode Pengalamatan–Index.Re[7]

 Sama seperti based relative addressing mode, register yang digunakan adalah DI dan SI

 Contoh:

 MOV DX, [SI]+5 ;

 MOV CL, [DI]+20 ;

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6

Mode Pengalamatan–Base.Indx[8]

 Kombinasi based dan indexed addressing modes  Menggunakan satu base reg. dan satu index reg.  Contoh:  MOV CL, [BX][DI] + 8  MOV CH, [BX][SI]+20  MOV AH,[BP][DI]+12  MOV AH,[BP][SI]+29

Mode Pengalamatan–Base.Indx[9]

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- ITTELKOM

Jenis Instruksi.CLK - MOV [1]

MOV Clock Acc  mem 10 Mem  acc 10 R  R 2 Mem  R 12 + EA R  mem 13 + EA Immed  R 4 Immed  mem 10 + EA R  seg R 2 Mem  seg R 8 + EA Seg R  R 2 Seg R  mem 9 + EA

Jenis Instruksi. CLK-Aritmatika[2]

ADD/SUB Clock DIV Clock

R  R 3 8 bit reg 80 sd 90

Mem  R 9 + EA 16 bit reg 144 sd 162 R  mem 16 + EA 8 bit mem (86 sd 96)+EA Immed  R 4 16 bit mem (150 sd 168)+EA Immed  mem _{17 + EA}

MUL Clock Shift & Rotate Clock

8 bit reg 70 sd 77 Single bit reg 2 16 bit reg 118 sd 133 Var bit reg 8 + 4/bit 8 bit mem (76 sd 83)+EA Single bit mem 15 + EA 16 bit mem (124 sd 139)+EA Var bit mem 20+EA+4/bit

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8

JMP Clock Clock

short 15 JCXZ 6 (no branch) 18 (branch) Intrasegment direct 15 J condition 4 (no branch)

16 (branch) Intersegment direct 15

Intrasegment using reg

mode 11

Intrasegment indirect 18 + EA

Intersegment indirect 24 + EA

Jenis Instruksi.CLK-Kend.Prog[3]

Jenis Instruksi.CLK- lainnya [4]

Instruksi Clock Instruksi Clock INC/DEC reg8 3 MOV seg,reg 2 INC/DEC data 23+EA MOV reg,reg 2 INC/DEC reg16 3 MOV mem,reg 13+EA LOGIC reg,reg 3 MOV reg,mem 12+EA LOGIC mem,reg 24+EA MOV mem,imm 14+EA LOGIC reg,mem 13+EA MOV reg,imm 4 LOGIC reg,imm 4 MOV mem,acc 14 LOGIC mem,imm 23/25+E

A MOV acc,mem 14

LOGIC acc,imm 4 MOV seg,mem 12+EA MOV reg,seg 2 MOV mem,seg 13+EA

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Jenis Instruksi.CLK-Eff.Addr[5]

Addressing Mode Clock

Direct 6 Register indirect 5 Register relative 9 Based indexed (BP)+(DI) or (BX)+(SI) 7 (BP)+(SI) or (BX)+(DI) 8

Based indexed relative

(BP)+(DI)+disp or (BX)+(SI)+disp 11 (BP)+(SI)+disp or (BX)+(DI)+disp 12

Catatan: EA = Effective address

Jenis Instruksi.CLK-contoh [6]

Address Mnemonic Assembly Clock CS:0100 B8 34 12 MOV AX,1234 4 CS:0103 35 34 12 XOR AX,1234 4 CS:0106 74 02 JZ 010A 16 (branch) CS:0108 B3 12 MOV BL,12 - (skiped) CS:010A 8A 0E 34 12 MOV CL,[1234] 12 + 6(EA) CS:010E 88 16 34 12 MOV [1234],DL 13+ 6(EA)

Total 61

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10



Terdiri dari



_{MOV AX,BX  89 (opcode) D8 (operand)}

Format bhs mesin 8088[1]

bahasa mesin op-code 8 bit operand (data,register,dll) + mnemonic

Format bhs mesin 8088[2]



Sesungguhnya bhs mesin diciptakan untuk

kemudahan programmer (manusia)



_{Control Unit di CPU hanya mengerti pola bit}

perintah



_{MOV AX,BX}



_{8B C3 (versi EMU}

8086/ MASM), 89 D8 (versi debug)



_{MOV AL,[2400] }

_A0

_{00 24}



_{ADD AX,BX}



_{01 D8}

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Format bhs mesin 8088-MOV[3]

1 0 0 0 1 0 D W

(5 bit) mode pengalamatan Pemilihan register

Data byte/word ; 0=byte, 1=word

Arah transfer data, dari/ke register ; 0=dari, 1=ke kode operasi (operation code)

op code mod reg R/M low displacement high displacement or

direct address

low byte direct addresshigh byte byte 1 byte 2 byte 3 byte 4

Format bhs mesin 8088[4]

code 10 code 01 11 00 SS Seg. Reg. CS DS ES 101 110 W=0 W=1 001 010 100 111 BP SI Register 000 011 SP DI AX BX CX DX AH BH CH DH AL BL CL DL Pola Register

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- ITTELKOM

12

Format bhs mesin 8088[5]

d8 : 8 - bit displacement, d16 : 16 - bit displacement

[BX] + [DI] [BX] + [DI] + d8 [BX] + [DI] + d16

00 01

MOD R/M

[BX] + [SI] [BX] + [SI] + d8 [BX] + [SI] + d16 000

10

[BP] + [SI] [BP] + [SI] + d8 [BP] + [SI] + d16 [BP] + [DI] [BP] + [DI] + d8 [BP] + [DI] + d16

[BP] + d16 [BX] [BX] + d8 [BX] + d16 [SI] [SI] + d8 [SI] + d16 [DI] [DI] + d8 [DI] + d16

DH 101 110 111 001 010 011 100 d16 direct address [BP] + d8 SI DI W=0 W=1 AL CL DL BL AH CH 11

MEMORY MODE REGISTER MODE BH AX CX DX BX SP BP

Pola MOD dan R/M

Format bhs mesin 8088-contoh[6]

Contoh: coding MOV AL,BL

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Format bhs mesin 8088[7]



Contoh u/ berbagai Ad. Mode



_{mov SP,BX; register A. M.}



_{mov CX,[4372H]; direct A.M.}



_{mov CL,[BX]; register indirect A.M.}



mov [SI + 43H],DH; indexed relative A.M.



_{mov AL,9CH; immediate A.M. (?)}



_{mov CS:[BX],DL; segment ovverides (?)}

ES CS SS DS 4 3 2 1 CONTROL SYSTEM AH AL BH BL CH CL DH DL SP BP SI DI OPERAND FLAGS C-BUS IP ALU INSTRUCTION STREAM BYTE QUEUE A- BUS BIU EU  mov SP,BX; register A. M.  mov CX,[4372H]; direct A.M.  mov CL,[BX]; register indirect A.M.  mov [SI + 43H],DH;

indexed relative A.M.

 mov AL,9CH; immediate

A.M. (?)  mov CS:[BX],DL; segment ovverides (?)

Format bhs mesin 8088[8]

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- ITTELKOM

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Format bhs mesin 8088[9]

1 0 1 1 w

0 0 1 1 1 0 1 0 0 0 1 0 D W Immediate Addressing Mode

reg data - low byte data - high byte (w=1)

byte 1 byte 2 byte 3

reg

Segment ovveride prefix

byte 1 byte 2 op code mod reg R/M

Kode Operasi Immediate A.M. dan Segment Override

Format bhs mesin 8088[10]

1 0 1 0 0 0 d wbyte 1 Low byte addressbyte 2 High byte addressbyte 3

Op. Code Acc. ke/dari memori

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Format bhs mesin 8088-Lat[11]

Segment overides[1]

CPU 80x86 memungkinkan program untuk

mengganti register segmen yang seharusnya

dengan register segmen yang lain.



_{Contoh :}

MOV AL,[BX] ; penunjuk alamat fisik adalah

DS:BX



_{Bandingkan dengan :}

MOV AL,ES:[BX]

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16 SS ES DS CS Segment Register SP,BP SI, DI, BX SI, DI, BX IP Offset Register

Segment overides[2]

Offset Register untuk berbagai Segment

Segment overides-contoh[3]

Instruksi Segmen yg

digunakan Segmenseharusnya MOV AX,CS:[BP] CS:BP SS:BP

MOV DX,SS:[SI] SS:SI DS:SI MOV AX,DS:[BP] DS:BP SS:BP MOV CS,ES:[BX]+12 ES:BX+12 DS:BX+12 MOV SS:[BX][DI]+32,AX SS:BX+DI+32 DS:BX+DI+32

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Latihan Soal [1]

If DS=7FA2H and the offset is 438EH, determine: a) The physical address

b) The lower range of the data segment c) The upper range of the data segment d) Show the logical address

a) The Physical address is; 7FA20+438E= 83DAE b) The lower range: 7FA20(7FA20+0000)

c) The upper range: 8FA1F(7FA20+FFFF) d) The logical address is; 7FA2:438E

Latihan Soal [2]

Ex: SP=1236, AX=24B6, DI=85C2, and DX=5F93, show the contents of the stack as each nstruction is executed.PUSH AX, PUSH DI, PUSH DX

SP? SS:SP??

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Latihan Soal [3]

Ex: assume that the stack is shown below, and SP=18FA, show the contents of the stack and registers as each of the following instructions is executed.POP CX,POP DX,POP BX

SP? SS:SP??

Latihan Soal [4]

Ex: Show how the flag register is affected by the addition of 38H and 2FH. CF,PF,AF,ZF,SF?

MOV BH,38H ; ADD BH,2FH ;

CF = 0 since there is no carry beyond d7

PF = 0 since there is odd number of 1`s in the result

AF = 1 since there is a carry from d3 to d4 ZF = 0 since the result is not zero

SF = 0 since d7 of the result is zero

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Latihan Soal [5]

MOV AX,34F5H SUB AX,95EBH CF,PF,AF,ZF,SF?

CF = 1 since there is carry beyond d15

PF = 0 since there is odd number of 1s in the lower byte

AF = 1 since there is a carry from d3 to d4 ZF = 0 since the result is not zero

SF = 1 since d15 of the result is 1

Thank You!

Please study this subject at home

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