Instruksi Mikroprosesor. Mode Pengalamatan-Jenis[1]

Transkripsi

1 Mikroprosesor dan Antarmuka Instruksi Mikroprosesor Oleh. Junartho Halomoan Mode Pengalamatan-Jenis[1] LOGO 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 1

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 2

3 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 Ditandai dengan [ ] Mode Pengalamatan-Direct[5] 3

4 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: Ditandai dengan [ ] MOV AL,[BX] PA (Physical Address) dan EA (Effective Address)??? Mode Pengalamatan-Indirect[7] 4

5 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 ; 5

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] 6

7 Jenis Instruksi.CLK - MOV [1] MOV Clock Acc mem 10 Mem acc 10 R R 2 Mem R R mem 12 + EA 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 7

8 Jenis Instruksi.CLK-Kend.Prog[3] 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 Intrasegment indirect Intersegment indirect EA 24 + EA 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 8

9 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 B MOV AX, CS: XOR AX, CS: JZ 010A 16 (branch) CS:0108 B3 12 MOV BL,12 - (skiped) CS:010A 8A 0E MOV CL,[1234] (EA) CS:010E MOV [1234],DL 13+ 6(EA) Total 61 9

10 Format bhs mesin 8088[1] Terdiri dari op-code 8 bit + operand (data,register,dll) MOV AX,BX 89 (opcode) D8 (operand) mnemonic bahasa mesin 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] A ADD AX,BX 01 D8 10

11 Format bhs mesin 8088-MOV[3] byte 1 byte 2 byte 3 byte op code D W mod reg R/M low displacement high displacement or direct address direct address low byte high byte (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) Format bhs mesin 8088[4] Pola Register Register W=0 W=1 AL BL CL DL AH BH CH DH AX BX CX DX SP DI BP SI code Seg. Reg. CS DS ES SS code

12 Format bhs mesin 8088[5] Pola MOD dan R/M MOD R/M [BX] + [SI] [BX] + [SI] + d8 [BX] + [SI] + d16 [BX] + [DI] [BX] + [DI] + d8 [BX] + [DI] + d16 [BP] + [SI] [BP] + [SI] + d8 [BP] + [SI] + d16 [BP] + [DI] [BP] + [DI] + d8 [BP] + [DI] + d16 [SI] [SI] + d8 [SI] + d16 [DI] [DI] + d8 [DI] + d16 d16 direct address [BP] + d8 [BP] + d16 [BX] [BX] + d8 [BX] + d16 MEMORY MODE d8 : 8 - bit displacement, d16 : 16 - bit displacement W=0 W=1 AL AX CL CX DL DX BL BX AH SP CH BP DH BH SI DI REGISTER MODE Format bhs mesin 8088-contoh[6] Contoh: coding MOV AL,BL 12

13 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 (?) Format bhs mesin 8088[8] 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. (?) BIU ES CS SS DS IP C-BUS INSTRUCTION STREAM BYTE QUEUE CONTROL SYSTEM mov CS:[BX],DL; segment ovverides (?) EU AH BH CH DH SP BP SI DI AL BL CL DL A- BUS ALU OPERAND FLAGS 13

14 Format bhs mesin 8088[9] Kode Operasi Immediate A.M. dan Segment Override byte 1 byte 2 byte w reg data - low byte data - high byte (w=1) Immediate Addressing Mode byte 1 byte reg Segment ovveride prefix op code D W mod reg R/M Format bhs mesin 8088[10] Op. Code Acc. ke/dari memori byte 1 byte 2 byte d w Low byte address High byte address 14

15 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] 15

16 Segment overides[2] Offset Register untuk berbagai Segment Segment Register CS DS ES SS Offset Register IP SI, DI, BX SI, DI, BX SP,BP Segment overides-contoh[3] Instruksi 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 MOV SS:[BX][DI]+32,AX SS:BX+DI+32 DS:BX+DI+32 16

17 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(7FA ) 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?? 17

18 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 18

19 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 LOGO 19