Mikroprosesor dan Antarmuka
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Instruksi Mikroprosesor
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Oleh. Junartho Halomoan (
[email protected])
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Mode Pengalamatan-Jenis[1]
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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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Mode Pengalamatan - Reg [2]
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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
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Mode Pengalamatan-Imm[3]
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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
2
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
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Ditandai dengan [ ]
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Mode Pengalamatan-Direct[5]
3
Mode Pengalamatan-Indirect[6]
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Alamat lokasi memori data yang akan diakses tersimpan dalam register Register yang digunakan pada mode ini : SI, DI, dan BX Ditandai Contoh: dengan [ ] MOV AL,[BX] PA (Physical Address) dan EA (Effective Address) ???
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Mode Pengalamatan-Indirect[7]
4
Mode Pengalamatan–Based.Re[8]
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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 = ?
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Mode Pengalamatan–Index.Re[7]
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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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Mode Pengalamatan–Base.Indx[8]
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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
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Mode Pengalamatan–Base.Indx[9]
6
MOV
Clock
Acc mem
10
Mem acc
10
RR
2
Mem R
12 + EA
R mem
13 + EA
Immed R
4
Immed mem
10 + EA 2 8 + EA
Seg R R
2
9 + EA
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Seg R mem
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R seg R Mem seg R
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Jenis Instruksi.CLK - MOV [1]
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Jenis Instruksi. CLK-Aritmatika[2] Clock
DIV
Clock
RR
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
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ADD/SUB
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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Jenis Instruksi.CLK-Kend.Prog[3] 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
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Intrasegment indirect
18 + EA
Intersegment indirect
24 + EA
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JMP
Jenis Instruksi.CLK- lainnya [4]
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Instruksi
Instruksi
Clock
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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INC/DEC reg8
Clock
8
Jenis Instruksi.CLK-Eff.Addr[5] Addressing Mode
Clock
Direct
6
Register indirect
5
Register relative
9
Based indexed 7
(BP)+(SI) or (BX)+(DI)
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(BP)+(DI) or (BX)+(SI)
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Based indexed relative
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(BP)+(SI)+disp or (BX)+(DI)+disp
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(BP)+(DI)+disp or (BX)+(SI)+disp
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Catatan: EA = Effective address
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Jenis Instruksi.CLK-contoh [6] Mnemonic
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Address
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
CS:0108
B3 12
MOV BL,12
CS:010A
8A 0E 34 12
MOV CL,[1234]
12 + 6(EA)
CS:010E
88 16 34 12
MOV [1234],DL
13+ 6(EA)
16 (branch)
- (skiped)
Total
61
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Format bhs mesin 8088[1] Terdiri dari op-code 8 bit + operand (data,register,dll)
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MOV AX,BX 89 (opcode) D8 (operand)
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mnemonic
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bahasa mesin
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Format bhs mesin 8088[2]
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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]
byte 1 1 0 0 0 1 0 op code D W mod
byte 2 reg
byte 3 R/M
byte 4
low displacement
high displacement or
direct address low byte
direct address high byte
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(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)
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Format bhs mesin 8088[4]
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Pola Register
Register W=0 W=1 AL AX BL BX CL CX DL DX AH SP BH DI CH BP DH SI
code 000 011 001 010 100 111 101 110
Seg. Reg. CS DS ES SS
code 01 11 00 10
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Format bhs mesin 8088[5] Pola MOD dan R/M
111
01
10
[BX] + [SI] [BX] + [DI] [BP] + [SI] [BP] + [DI] [SI] [DI] d16 direct address [BX]
[BX] + [SI] + d8 [BX] + [DI] + d8 [BP] + [SI] + d8 [BP] + [DI] + d8 [SI] + d8 [DI] + d8
[BX] + [SI] + d16 [BX] + [DI] + d16 [BP] + [SI] + d16 [BP] + [DI] + d16 [SI] + d16 [DI] + d16
W=0 AL CL DL BL AH CH
[BP] + d8
[BP] + d16
DH
SI
[BX] + d8
[BX] + d16
BH
DI
MEMORY MODE
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110
11
00
W=1 AX CX DX BX SP BP
REGISTER MODE
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MOD R/M 000 001 010 011 100 101
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d8 : 8 - bit displacement, d16 : 16 - bit displacement
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Format bhs mesin 8088-contoh[6]
Contoh: coding MOV AL,BL
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Format bhs mesin 8088[7]
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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 (?)
Format bhs mesin 8088[8] M.
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mov SP,BX; register A.
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mov CX,[4372H]; direct
BIU C-BUS
A.M. mov CL,[BX]; register indirect A.M. mov [SI + 43H],DH; indexed relative A.M. mov AL,9CH; immediate A.M. (?)
ES CS SS DS IP EU
mov CS:[BX],DL;
segment ovverides (?)
4 3 2 1
INSTRUCTION STREAM BYTE QUEUE
CONTROL SYSTEM A- BUS
AH BH CH DH
SP BP SI DI
AL BL CL DL
ALU OPERAND FLAGS
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Format bhs mesin 8088[9] Kode Operasi Immediate A.M. dan Segment Override
byte 1 1 0 1 1 w
byte 2 data - low byte
reg
byte 3 data - high byte (w=1)
Immediate Addressing Mode
byte 2
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byte 1 1 0 0 0 1 0 op code D W mod
reg
R/M
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0 0 1 reg 1 1 0 Segment ovveride prefix
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Format bhs mesin 8088[10]
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Op. Code Acc. ke/dari memori byte 1 1 0 1 0 0 0 d w
byte 2 Low byte address
byte 3 High byte address
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Format bhs mesin 8088-Lat[11]
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Segment overides[1]
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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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Segment overides[2] Offset Register untuk berbagai Segment CS
DS
ES
SS
Offset Register
IP
SI, DI, BX
SI, DI, BX
SP,BP
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Segment Register
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Segment overides-contoh[3] Segmen yg digunakan
Segmen seharusnya
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
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Instruksi
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
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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
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Latihan Soal [2]
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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]
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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??
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Latihan Soal [4]
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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?
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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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