04 Lap Trinh Voi Assembly

5/27/2018 04 Lap Trinh Voi Assembly

1/671Pham Quoc Thinh, ictu.edu.vn.

Lp trnh vi Assembly cho 8051

ICTU 2010

Chng 3 Vi iu khin 8051v lp trnh hp ng


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Department of Electronics & Telecommunications

Pham Quoc Thinh, ictu.edu.vn.

Cc lnh ton shc

Cc lnh logic

Cc lnh quay

Cc lnh so snh

Cc cha ch8051

Ni dung bi ging


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ADD v ADDC

SUBB

MUL

DIV

INC

DECDA

Cc lnh s hc

C 24 opcode s hc c phn thnh cc loi sau y:


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Flag: N l 1 bit thanh ghi, cho bit tnhtrng ca kt qutmt hot ng

Cc th1 trong hai trng thi vigi tr0hoc 1

Cc cshc cho bit trng thi cakt quca cc hot ng ton hc ( +,

, *, / )

Cc c s hc


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C 4 cshc trong 8051Cnh(C)Cnhph(AC)Ctrn(OV)

Cchn l(P)

Tt ccc ctrn y c lu trtrong ttrng thi chng trnh (PSW)

Cc c s hc (Cc c c iu kin)

CY AC -- RS1 RS0 0V -- PPSW.7 PSW.6 PSW.5 PSW.4 PSW.3 PSW.2 PSW.1 PSW.0


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CC theo di cc hot ng khng duCOV theo di cc hot ng c du

Cc c s hc (Cc c c iu kin)

CY PSW.7 C nh AC PSW.6 C nh ph -- PSW.5 Cho ngi s dng RS1 PSW.4 Bit 1 la chn bank thanh ghi RS0 PSW.3 Bit 0 la chn bank thanh ghi 0V PSW.2 C trn -- PSW.1 C ngi s dng nh ngha P PSW.0 C chn l


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Cc lnh nh hng ti c (1/2)

Lnh Mnemonic Cc cnh hngADD C AC OV

ADDC C AC OV

SUBB C AC OV

MUL C = 0 OV

DIV C = 0 OV

DA A CSETB C C = 1

MOV C, bit C


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Cc lnh nh hng ti c (2/2)

Lnh Mnemonic Cc cnh hngORL C, bit C

ANL C, bit C

RLC C

RRC C

CLR C C = 0

CPL C C = /C

CJNE C


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ADD A, source ; A = A + source

ADDC A, source ; A = A + source + C

Thanh ghi Aphi tham gia trong cc php cng

CCc t ln 1 nu c mt bit nhbit 7

CACc t ln 1 nu c mt bit nhbit 3

Cc lnh ADD v ADDC


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F5h 1111 0101

+ 0Bh + 0000 1011

100h 0000 0000

Sau ghi cng, thanh ghi A (ch) cha 00 v cc cnh

sau:CY = 1 c nhtD7

P = 0 v scc bit 1 l 0

AC = 1 v c c nhtD3 sang D4

V d 1

Chra cbnh hng bi cc lnh sau.MOV A, #0F5h ; A = F5hADD A, #0Bh ; A = F5 + 0B = 00


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V d 2Gisvtr RAM trong khong 40h42h c gi trcho trc.

Vit mt chng trnh tm tng ca cc gi tr vtr trongRAM . cui chng trnh, A cha byte thp v R7 chabyte cao

Vtr RAM: 40h = (7Dh), 41h = (EBh), 42h = (C5h)

Gii php:MOV A, 40h ; A = gi trRAM nh40h trong RAMMOV R7, #0 ; R7 = 0ADD A, 41h ; Cng A vi nh41h trong RAMJNC NEXT ;nu CY = 0, A khng btrnINC R7 ; Tng R7 nu CY=1

NEXT: ADD A, 42h ; Cng A vi nh42h trong RAMJNC NEXT1 ; nu CY = 0, A khng btrnINC R7 ; Tng R7 nu CY=1

NEXT1:END


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V d 3

Vit mt on chng trnh cng hai s16-bit. Hai sl 3CE7h v 3B8Dh. t kt qutrong R7 v R6; R6 chabyte thp.

CLR C ; xo C=0

MOV A, #0E7h ; A=E7h

ADD A, #8Dh ; cng vi byte thp A=74h v C=1

MOV R6, A ; R6 = A(byte thp)

MOV A, #3Ch ; da byte cao vo A

ADDC A, #3Bh ; cng byte cao v cCY

; 3B + 3C + 1 = 78 (hex)

MOV R7, A ; Ghi byte cao vo R7


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DA A Hnh ng Hiu chnh thp phn thanh ghi A Sdng sau khi cng hai sBCD

Lnh DA

V d4:MOV A, #47h ; A=47h ton hng BCD thnht

MOV B, #25h ; B=25h ton hng BCD th

hai

ADD A, B ; kt qu(A=6Ch)

DA A ; hiu chnh (A=72h)

ADDC ..DA A

ADD ..DA A


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V d 4 ca lnh DA

Hex BCD

47 0100 0111

+ 25 + 0010 01016C 0110 1100

+ 6 + 0110

72 0111 0010


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SUBB A, ngun

Khng mn: A = A ngunVi mn:A = A ngun c

Sau khi thc thi:CCset lnh 1 nu mn thm vo bit 7

CACset lnh 1 nu mn thm vo bit 3

Lnh SUBB SUBB A, #dataSUBB A, directSUBB A, @Ri , vi i =0 hoc 1SUBB A, Rn, vi n =0,1,,7


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MUL AB

Sdng thanh ghi A v B nhcc thanh ghi ch

v ngunNhn 2 strong A v B, sau phn byte cao dttrong B v phn byte thp t trong A

Ctrn OVt ln 1 nu kt qu>FFhCh cC = 0

Lnh MUL


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DIV AB

Sdng thanh ghi A v B nhcc thanh ghi chv ngun

Strong A chia cho strong B. Kt qut trongA, phn dt trong B

Ctrn OVt ln 1 nu Strong Bl 00h (li

chia cho 0)

CC lun bng 0

Lnh DIV


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tng (INC) hoc gim (DEC) vtr bnhcchnh bi cc ton hngKhng thay i cc cshc

V d: INC 7Fh ; Ni dung 7Fh tng 1DEC R1 ; Ni dung trong R1 gim 1

Cc lnh INC v DEC

INC A

INC direct

INC @Ri Vi i=0 hay 1

INC Rn Vi n=0,,7


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Cc hot ng logic

Cc hot ng quay v trao i

Cc hot ng so snh

Hot ng logic trong 8051


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ANDOR

XOR

NOT

Cc chc nng logic chung

C sn cc lnh cho 8051 thc hin ccchc nng logic


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ANL ch, ngunch = ch ANDngun

ORL ch, ngun

ch = ch ORngun

XRL ch, ngun

ch = ch XORngun

Thng thng, ch l thanh ghi A hay mta chtrc tip trong RAM ni

Cc lnh logicANL direct, A

ANL direct, #data

ANL A, #data

ANL A, direct

ANL A,@Ri vi i=0 hay 1ANL A, Rn vi n=0,,7


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Cc ton hng ngun c thmt trong 4 cha ch(tc thi/thanh ghi/ trc tip/gin tip)

ANLc thc sdng xo (0) cc bit nht nh

ORLc thc sdng set (1) cc bit nht nh

Cc lnh logic

Instruction ANL A,R0 ORL A,R0 XRL A,R0

A before: 10010111 10010111 10010111

R0 before: 11110010 11110010 11110010

A afterwards: 10010010 11110111 01100101

V d


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CLR A

Tt ccc bt trong thanh ghi Ac xo

CPL ATt ccc bt trong thanh ghi Abo

Ch cc lnh CLRv CPLchhot ng trn

thanh ghi A

Cc lnh CLR v CPL


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Thanh ghi A quay mt mt bit tri hoc phi (ch

hot ng vi A)

C thbao gm cnhtrong hot ng

quay

Cc lnh quay RL ARR A

RLC A

RRC A


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RL AQuay A mt bit sang tri

RLC A

Quay A v cnhmt bit sang tri

RR A

Quay A mt bit sang phi

RRC A

Quay A v cnhmt bit sang phi

Cc lnh quay


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01234567

RL A

01234567C

RLC ACarry Flag

01234567

RR A

Before: 10011100 After: 00111001

C01234567

Carry FlagRRC A

Before: 10011100 CY = 0 After: 00111000 CY = 1

Before: 10011100 After: 01001110

Before: 10011100 CY = 1 After: 11001110 CY = 0

Cc lnh quay


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Lnh hon i 4 bit phn thp v 4 bit phn caoca thanh ghi A.

017 6 5 4 3 2

High Nibble Low Nibble

SWAP A

Lnh trao i SWAP

Thanh ghi A = 5Eh (gi tr ban u) sau khi SWAP thanh ghi A =E5h


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CJNE ch, ngun, a chtng i

u tin so snh ton h

ng ngu

n v ch

Nhy ti a chtng i (chng trnh con)nu chng khng bng

Cc lnh so snh

C nh = 1, nu ch nh hn ngun,Nu khng, CY = 0.


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Hot ng so snhCJNE A, #data, relative

CJNE A, direct, relativeCJNE @Ri, #data, relative vi i=0 hay 1

CJNE Rn, #data, relative Vi i=0,,7

; v d !!

CJNE R7, #60H, NOT_EQ; hin ti R7 = 60H..

NOT_EQ:

JC REG_LOW

; hin ti C = 0, R7 > 60H..

REG_LOW:

; hin ti C = 1. R7 < 60H


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V d 5

Vit mt on chng trnh gim st cng P1 cho tikhi P1=63h.

Gii php:

MOV P1, #0FFh ; P1 l cng vo

HERE: MOV A, P1 ; c P1

CJNE A, #63h, HERE ; gim st cho ti khi; P1=63h


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Cha ch: Mt phng php

Chra ni ton hng(ngun v ch)

Cc ton hng c thc truy cp nhthno

Cc kiu nh a ch


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a chtc thi (v d: MOV A, #55H)

a chthanh ghi (v d: MOV A,R0)

a chtrc tip (v d: MOV A,30H)

a chgin tip thanh ghi (v d: MOV A, @R0)

a chchs(v d: MOVC A, @A+PC)

a chtuyt i (v d: ACALL address11)

a chdi (v d: LCALLaddr16 )

a chtng i (v d: SJMPrelative)

Cc ch a ch


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Ton hng ngun l mt hng s/k tdliu tcthi

Nu ton hng l mt s, chng tphi thm du #trc n

V d: ADD A, #56h ; cng 56(16)vi strong thanhghi A

V d: MOV R6, #81 ; np 81(10)vo R6

a ch tc thi


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Lnh Hot ng

MOV A, #0AFh Chuyn gi trAFh vo thanh ghi A

ANL 15h, #88h AND (bit vi bit) a ch15h vi gi tr88h

MOV DPTR, #0ABCDh Chuyn ABCDhv thanh ghi DPTR

MOV R3, #1Ch Chuyn 1Chvo R3

MOV R2, #A Chuyn k tASCII A (41h)v thanh ghi R2

V d ca a ch tc thi


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Thm #trc bt k gi trtc thi

Chc ton hng ngun c thl tc thi

Thm h sau sh16, bsau sh2;

Sdng bao cc k t

Cc sh16 bt u vi A-F th thm 0 ngtrc

Ch ca a ch tc thi

MOV A,#ABh


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Ngun/ch u c nh vtrong Cc thanh ghiCPU (tc l R0 R7; A; DPTR)

V d: MOV A, R5 ; Sao chp gi trtrong R5 v A

V d: MOV R3, A ; Sao chp gi trtrong A v R3V d: ADD A, R2 ; Thm R2 vo A

V d: MOV R7, DPL

MOV R6, DPH

a ch thanh ghi


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Vic gii quyt chhiu qunht:

Khng cn truy xut bnh

Cc lnh thc thi nhanh hn nhiu

Kt qu: tc (v thhiu qu) tng

C thchuyn dliu giaAccv Rn(n = 0 n 7)nhng khng thchuyn dliu gia cc thanh ghiRn

V d: MOV R4, R7 (Khng hp l)

Ch ca a ch thanh ghi


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Cc cu hi n tp

1. Ngi lp trnh c thto ra mt cha chmi?

2. Chra lnh chuyn gi tr1000 0000 (binary) vo R3.

3. Ti sao lnh sau y khng hp l? MOV R2, DPTR

4. ng hay sai. DPTR l mt thanh ghi 16 bit cng c thtruy xut vo nh ng byte thp byte cao.

5. PC (Bm chng trnh) cng c nh dng byte thpbyte cao?


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Ngun/ch u l trng a ch8-bit trong lnh

Sdng chny truy cp 128 byte RAMvcc thanh ghi SFR

Vtr ca ton hng l cnh khng ththay i khi chng trnh chy nhng ni dung cththay i c

Khng linh hot

a ch

cc thnh ph

n trong m

tbng dliu

a ch trc tip


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Lnh Hot ngMOV 80h, A hayMOV P0, A

Sao chp ni dung thanh ghi A vo a ch80h

MOV A, 80h hay

MOV A, P0

Sao chp ni dung ti a ch80h (P0) vothanh ghi A

ABC EQU 80h

MOV A, ABC

Sao ni dung ta chtrc tip vi nhn ABCvo thanh ghi A, c Port 0 vo A

MOV R0, 12h Sao chp ni dung ta ch12h trong RAMvo R0

MOV 0A8h, 77hhay

MOV IE, 77h

V d a ch trc tip

MOV direct,direct

D f El i & T l i i


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Vi d a ch trc tip

MOV R2, #5 ; R2 = 05

MOV A, 2 ; Sao vtr 02 (R2) ti A

MOV B, 2 ; Sao vtr 02 (R2) ti B

MOV 7, 2 ; Sao vtr 02 ti 07 (R2 ti R7)

MOV direct,direct

D t t f El t i & T l i ti


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V d 6

Gii php :

PUSH 05 ; t R5 vo ngn xp

PUSH 06 ; t R6 vo ngn xpPUSH 0E0h ; t ni dung A vo ngn xp

POP 0F0h ; Ly nh stack vo B

; now register B = register A

POP 02 ; Ly nh stack vo R2; now R2 = R6

POP 03 ; Ly nh stack vo R3

; now R3 = R5

Ch ra m t R5, R6 v A vo stack v sau ly chng vo

R2, R3 v B thanh ghi B = thanh ghi A, R2 = R6 v R3 = R5.

D t t f El t i & T l i ti


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Ngn xp v ch a ch trc tip

ChCha chtrc tip c cho phpt vo stack

PUSH A (Khng hp l)

PUSH 0E0h (Hp l

PUSH R3 (Khng hp l)

PUSH 03 (hp l)

POP R4 (khng hp l)

POP 04 (hp l)

PUSH direct

POP direct



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Gi tra chbgii hn 1 byte 00FFh (128-byteRAM v SFR)

Sdng MOV di chuyn dliu tchnh nvo chnh n c thdn n kt qukhng thon trc li

Ch ca a ch trc tip

MOV A, A



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Sdng mt thanh ghi gia chca ton hng; tcl sdng thanh ghi nhmt con tr

Ch R0 v R1c thc sdng khi dliu trong CPU

(Khong

a ch

t

00h7Fh) v d

. MOV A,@R1

R0 ,R1v DPTR c thc sdng khi a chtvtrbnhngoi V d. MOVX A,@R1 MOVX A,@DPTR

Phi t du @trc tn thanh ghi

a ch gin tip thanh ghi



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Sau

B nh

chng trnh

Trc

a ch

ACC

R0ADD A, @R0200

201

B nh dliu

1231

32

30

10

31

ACC

R0

31

22

a ch gin tip thanh ghi (ADD A,@R0)



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Lnh Hot ngMOV @R1, A Sao chp ni dung ca A ti a chc tr

bi R1

MOV A, @R0 Sao chp ni dung ca a chc trbi R0vo A

MOV @R1, #35h Chuyn gi tr35h vo a chc trbi R1

MOV @R0, 80h hay

MOV @R0, P0

Sao chp dliu tP0 v vtr c trbi R0

MOVX A, @R0 Sao chp ni dung a chdliu ngoi ctrbi R0 vo thanh ghi A

MOVX A, @DPTR Sao chp ni dung a chdliu ngoi ctrbi DPTR vo thanh ghi A

V d ca a ch gin tip

MOV @Ri,#data where i=0 or 1



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Vit mt on chng trnh sao chp gi tr55h vo vtr 40h n 44h trong RAM sdng:

(c) Cha chtrc tip;

(d) Cha chgin tip thanh ghi khng sdng vnglp;

(e) vi mt vng lp

V d 7



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MOV A, #55h ; np gi tr55H vo A

MOV 40h, A ; Sao A vo vtr 40h trong RAM

MOV 41h, A ; Sao A vo vtr 41h trong RAMMOV 42h, A ; Sao A vo vtr 42h trong RAM



Gii php ca v d 7(a) Ch a ch trc tip



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50

p


MOV A, #55h ; np gi trA=55h

MOV R0, #40h ; np con trR0 = 40h

MOV @R0, A ; Sao chp A vo trong RAM c trbi R0

INC R0 ; Tng con tr. R0 = 41hMOV @R0, A ; Sao chp A vo trong RAM c trbi R0

INC R0 ; Tng con tr. R0 = 42h


INC R0 ; Tng con tr. R0 = 43hMOV @R0, A ; Sao chp A vo trong RAM c trbi R0

INC R0 ; Tng con tr. R0 = 44h


Gii php ca v d 7 (b)Ch a ch gin tip thanh ghi khng c vng lp



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51

p


MOV A, #55h ; A = 55h

MOV R0, #40h ; np con tr. R0 = 40h, a chRAM

MOV R2, #05 ; Np bm, R2 = 5

AGAIN: MOV @R0, A ; Sao chp A vo trong RAM c trbi R0

INC R0 ; Tng con trR0

DJNZ R2, AGAIN ; Lp li cho ti khi R2=0

Gii phpV d C) Loop method

DJNZ : gim v nhy nu khc 0

DJNZ direct, relative

DJNZ Rn, relative vi n = 0,1,,,7

MOV R2, #05h ; v d

LP:;--------------------------------

; do 5 timesinside the loop

;--------------------------------

DJNZ R2, LP ; R2 as counter



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p


V d 8 (looping)Vit on chng trnh xo 15 vtr RAM btu ta ch60h.

CLR A ; A = 0

MOV R1, #60h ; Np con tr. R1 = 60h

MOV R7, #15 ; np bm, R7 = 15 (0F trong HEX)

AGAIN: MOV @R1, A ; Xo vtr RAM trbi R1

INC R1 ; Tng R1

DJNZ R7, AGAIN ; Lp li khi bm =0; xoa mt v RAM a ch60hCLR A

MOV R1,#60h

MOV @R1,ASetup a loop using DJNZ and register R7 as counter



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53Pham Quoc Thinh, ictu.edu.vn.

V d 9 (block transfer)Vit mt on chng trnh sao chp mt khi

10byte dliu tRAM bt u a ch35h ti RAMbt u ta ch60h

MOV R0, #35h ; con trngunMOV R1, #60h ; con trchMOV R3, #10 ; bm

BACK:MOV A, @R0 ; nhn mt byte tngun

MOV @R1, A ; sao n n chINC R0 ; tng con trngunINC R1 ; tng con trchDJNZ R3, BACK ; lm d10 byte

; COPY 1 TIMES

MOV R0,#35h

MOV R1,#60h

MOV A,@R0

MOV @R1,A



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Sdng con trtrong chng trnh cho phpxl Cu trc dliu ng mt lith

Dliu ng: Gi trdliu khng bcnh

Ch ca a ch gin tip



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Sdng thanh ghi nnv mt lch-offset to thnh mt a chc hiu qucho lnh JMPhay MOV

c sdng theo c php mt chmc camt mng hay mt bng look-up

Thng thng, DPTRl thanh ghi nn v thanhghi Al offset lch

Atng/gim theo c php trong danh sch

a ch ch sMOVC A, @A+DPTR

MOVC A, @A+PC

JMP @A+DPTR



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Sau

Program

memory

Trc

ACC

DPTR

MOVC A, @A + DPTR2000

2001

41

00 10

31

ACC

56

56

a ch ch s

MOVC A, @A + DPTR

V d: MOVC A,@A+DPTR



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Lnh Hot ng

MOVC A, @A + DPTR Sao chp byte m lnh tm thy a chROM c nh dng bi thanh ghi A vthah ghi DPTR ti A

MOVC A, @A + PC Sao chp byte m lnh tm thy a chROM c nh dng bi vic cng thm Av PC ti thanh ghi A

JMP @A + DPTR Nhy ti a chc nh dng bi victhm A v DPTR, y l mt lnh nhykhng c iu kin v slun thc hin.

V d ca a ch ch s



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V d 10 (Bng look-up)Vit mt chng trnh nhn gi trx tP1 v gix2ra P2, tip tc

ORG 0h

MOV DPTR, #300h ; np a chbng look-up sMOV A, #0FFh ; A = FF

MOV P1, A ; P1 l cng voBACK: MOV A, P1 ; c X

MOV A, @A+DPTR ; c x tbngMOV P2, A ;xut ra P2SJMP BACK ; lm li

ORG 300h

TABLE: DB 0, 1, 4, 9, 16, 25, 36, 49, 64, 81

END



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Sdng trong cc lnh nhy (JMP)

a chtng i: mt gi tr8 bit(-128 n+127)

Bn c thxl a chtng i nhlmt offset

Nhn chra ch nhy (ni dng li)Assembler pht hin ra a chtng ibng cch sdng nhn

a ch tng i SJMP tng iDJNZ trc tip, tng iDJNZ Rn, tng i vi n=0,1,,7



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a chtng i c cng vo PC

Tng l a chca lnh tip theo c thcthi

Nhn chra ch ca ch nhy(ni dng li).

a ch tng i



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B m chng trnh+ offset

= a ch hiu qu

= a ch ca lnh tip theo

+ Offset

Branch OpcodeOffset

Next Opcode

Next Instruction

B m chng trnh

a ch tng i



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Lnh Hot ng

SJMP NXT Nhy ti a chtngi vi nhn 'NXT'; y lmt lnh nhy khng c iu kin

DJNZ R1, DWN Gim thanh ghi R1 i 1 v nhy ti a chtng

i 'DWN' nu kt quR1 khc 0

V d v a ch tng i

0035



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Chsdng vi cc lnhACALLvAJMP

Bc "nhyln nht c ththc hin c l 2K

a ch tuyt i

Chsdng vi cc lnh LCALLv LJMP

Bc "nhyln nht c th thc hin c l 64K

a ch di

ACALL address11

AJMP address11

211= 2048=2K



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a chtuyt i: 11-bit a chlnh 2-byte

a chdi: 16-bit a chtrong lnh 3-byte

Phm vi ca nhy" ca di ln hn tuyt iTuy nhin, chtuyt i c code ngn hn(2 byte), do thc hin nhanh hn

Ch a ch di v tuyt i



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C rt nhiu phng php lm mt nhim vMt shiu quhn trong khi mt sth khng

Chn cha chng shon thnh nhim v

hiu quChng ta hy gii quyt mt trng hp ngin nhv d: Xoa vtr nht30h n 7Fh.

Chng ta c thsdng lnh MOV trong chachtrc tip

Nhng chng trnh sdi v tn khng gian m

Ti sao phi s dng nhiu ch ?



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MOV 30h, #00h ; Clear Array [0]

MOV 31h, #00h ; and Array [1]

. ; Keep on going

.

.

MOV 7Eh, #00h ; Clear Array [78]

MOV 7Fh, #00h ; Clear Array [79]

Ch a ch trc tip sdng n 240 byte

MOV R0, #30h ;t con tr bt u mng

clear_arr: MOV @R0, #00 ;Xoa v tr tr bi R0

INC R0 ; tng con tr 1CJNE R0, #80H, clear_arr ;

NEXT: . . . . . . . . ;

; lnh tip

S dng a ch gin tip ch cn 8 byte, tit kim232 byte !

Ti sao phi s dng nhiu ch a ch?

V vy, thsdng cha chgin tip nhsau:

80x3 = 240

3 byte cho lnh MOV direct, #data


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Lp trnh vi Assembly cho 8051

ICTU 2010

END of Lecture

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04 Lap Trinh Voi Assembly