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K THUT VI X LCu hi 1 loi im:

1.1 Nu cc thnh phn chnh ca mt h vi x l ? Chc nng ca cc thnh phn? V s khi tng qut h vi x l?

1.2 Cu trc ca b vi x l 8086/8088 v chc nng ca cc thnh phn? V hnh?

1.3 Trnh by cc thanh ghi trong m hnh lp trnh ca b vi x l 8086?

1.4 Mt nh trong h VXL 8086 c a ch vt l l 7A45CH, hy tnh a ch lch (offset ) ca n nu a ch on l 75FAH? Khi CPU thc hin np lnh ti a ch ny th CS=?, IP=?.

achvtl=Thanh_ghi_an16+Thanh_ghi_lch

a ch lch (offset) l: 7A45CH-75FA0H = 44BCH.

CS = 75FAH

IP = 44BCH(Ch : Php nhn vi 16 (hay 10H) c c bng cch thm 4 bt 0 (hay 0H) vo bn phi)

1.5 Mt nh trong h VXL 8086 c a ch vt l l 9B4C6H, hy tnh a ch on ca n nu a ch lch (offset) l 31A6H? Khi CPU thc hin np lnh ti a ch ny th CS=?, IP=?.achvtl=Thanh_ghi_an16+Thanh_ghi_lch

9B4C6H - 31A6H = 98320H => a ch on l 9832H

CS = 9832H

IP = 31A6H

1.6 Cu trc ca dng lnh hp ng? Ly v d mt dng lnh cho VXL 8086?

1.7 Cu trc ca chng trnh hp ng?

1.8 Hy vit cc nh ngha d liu cho cc bin v hng sau:

a) Mt bin kiu Word tn W1, c khi to vi gi tr 57BH.

W1 DW 57BH

b) Mt bin kiu Word tn W2, khng c khi to.

W2 DB?

c) Mt bin kiu Byte tn X1, c khi to vi gi tr 120.

X1 DB 120

d) Mt mng kiu Byte tn MB, c khi to vi 5 gi tr u: 5D, 32H, 1FH, C, 0011B.

MB DB 5D, 32H, 1FH, C, 0011B

1.9 Hy vit cc nh ngha d liu cho cc bin v hng sau:

a) Mt hng kiu xu k t tn MSG, vi gi tr c gn l Xau KT.

MSG DB Xau KT

b) Mt bin kiu k t tn KT, c khi to vi gi tr A.

KT DB A

c) Mt bin kiu Word tn X1, khng c khi to.

X1 DW?

d) Mt mng kiu Byte tn MB, c khi to vi 5 gi tr u: 12, A, F, 123, 1AH.

MB DB 12, A, F, 123, 1AH

1.10 Khung ca chng trnh hp ng dch ra dng *.EXE?

. Model small

. Stack 100

. Data

; cc nh ngha cho bin v hng ti y

. Code

MAIN Proc

; Khi u cho DS

MOV AX, @Data

MOV DS, AX

; Cc lnh ca chng trnh chnh ti y

; Tr v DOS dng hm 4CH ca INT 21H

MOV AH, 4CH

INT 21 H

MAIN Endp

; cc chng trnh con (nu c) ti y

END MAIN

1.11 Khung ca chng trnh hp ng dch ra dng *.COM?

. Model Tiny

. Code

ORG 100h

START: JMP CONTINUE

; cc nh ngha cho bin v hng ti y

CONTINUE:

MAIN Proc

; cc lnh ca chng trnh chnh ti y

INT 20H ; Tr v DOS

MAIN Endp

; cc chng trnh con (nu c) ti y

END START

1.12 Trnh by cch trao i d liu vi ngoi vi theo phng php hi vng polling ?

1.13 Trnh by cch trao i d liu vi ngoi vi theo phng php ngt ?

1.14 Trnh by cch trao i d liu vi ngoi vi theo phng php DMA ?

1.15 Phn tch u nhc im ca 3 cch trao i d liu vi ngoi vi theo phng php DMA (trao i tng khi d liu, trao i tng byte v phng php tn dng chu k bus ri)?

1.16 Trnh by phng php nh a ch segment:offset ? Ti sao ni phng php nh a ch ny c tnh cht a tr ? Cho v d minh ha ?

Ni dung cc thanh ghi on (16bit) xc nh a ch on (a ch nh u on).

Ni dung thanh ghi lch (16bit) xc nh a ch lch ( lch) ca nh c th so vi a ch on. Cp a ch segment: offset l a ch logic.

a ch vt l nh c c bng cng thc:


Phng php ny c tnh a tr v t mt a ch vt l ta c th to ra cc gi tr khc nhau ca thanh ghi on v thanh ghi lch, nn s c v s a ch logic ty vo cc cch la chn on.

V d: a ch vt l 12345H c th c to ra t cc gi tr:

Thanh ghi on Thanh ghi lch

1000H

2345H

1200H

0345H

1004H

2305H

Cu hi loi 2 im.

Cu hi 2.1:Trnh by s khi ca h vi x l tiu biu v gii thch ngn gn vai tr ca cc khi chc nng chnh?

Cc khi chc nng chnh ca h vi x l tiu biu l:

CPU (Central Processing Unit): B x l trung tm c nhim v tnh ton v iu khin

ALU (Arithmetic and Logic Unit): khi tnh ton

CU (Control Unit): khi iu khin

B nh (Memory): lu tr d liu (data) v lnh (instruction) cho CPU x l. B nh ROM: lu lnh v d liu ca h thng

B nh RAM: lu lnh v d liu ca h thng v ca ngi dng

Khi phi ghp vo/ra:

Cc thit b vo (Inputs):Tip nhn d liu v thng tin iu khin, chuyn cho CPU x l

Cc thit b ra (Oututs):Kt xut thng tin ra (mn hnh, giy, ...);Lu tr thng tin lu di (a t, a quang, ....)

Cc BUS truyn thng tin: Gm Bus a ch, Bus d liu v Bus iu khin. C nhim v lin kt cc khi chc nngtrong h thng my tnh.Cu hi 2.2:Trnh by s chc nng vi x l v gii thch vai tr ca cc khi chc nng trong qu trnh thc hin chng trnh?Vi x l chnh l n v x l trung tm CPU ca my vi tnh. Vi xl c thc chia thnh cc n vchc nng chnh nh sau:

Cc thanh ghi:

Thanh ghi lnh: vi x l lu m lnh trong thanh ghi lnh sau khi np m lnh t b nh.Gi trtrong thanh ghi ny lun c vi xl gii m xc nh lnh.

Bm chng trnh: cha a chca lnh hay m thc thi (op-code).Thng thng,thanh ghi ny cha a chca cu lnh k tip theo tun t chng trnh.

Thanh ghi a chbnh:cha a chca dliu.Vi xl sdng cc a ch ny nh l cc con tr trc tip ti b nh. Gi tr ca cc a ch ny chnh l d liu ang c trao i v x l . Thanh ghi dng chung:cn c gi l thanh ghitch ly(accumulator), dng thlu hu ht cc kt qutnh ton ca n vxl s hc v l-gc ALU v cn dng trao i d liu vi cc thit b vo/ra.

n v x l s hc logic (ALU): ALU thc hin tt ccc thao tc xl dliu bn trong vi xl nh l cc php ton l-gc,shc, c th l cc php ton: Cng nh phn v cc php l-gc

Tnh s b mt ca d liu

Dch hoc quay tri phi cc thanh ghi dng chung.

n v iu khin (CU): C chc nng l c v gii m cc lnh tbnh chng trnh.thc hin lnh,CU kch hot khi ph hp trong ALU cn cvo mlnh (op-code) trong thanh ghi lnh. M lnh xc nh thao tc CU thc thi. Qu trnh thc hin chng trnh:

Np (Fetch): Vi x l np (c) lnh t b nh chnh vo thanh ghi lnh.

Gii m (Decode):Vi xl gii m hay dch lnh nhn viu khin CU. CU nhp ni dung ca thanh ghi lnh v gii m xc nh kiu lnh.

Thc hin (Execute): Vi x l thc hin lnh nh CU. hon thnh nhim v, CU sinh ra mt chui cc tn hiu iu khin tng ng vi lnh.

Cu hi 2.5: Phn bit ch a ch trc tip v ch a ch gin tip qua thanh ghi? Cho v d?

V d:

Ch a ch trc tip: MOV AL, (1234H);chuyn nh DS:1234 vo AL.

MOV (4320H), CX ;chuyn CX vo 2 nh lin tip DS:4320 v DS:4321

Ch a ch gin tip qua thanh ghi:MOV AL, (BX); chuyn nh c a ch DS:BX vo AL.

MOV (SI), CL; chuyn CL vo nh c a ch DS:SI.

MOV (DI), AX; chuyn AX vo 2 nh lin tip ti DS:DI v DS: (DI + 1).

Phn bit: Ging nhau: Cho phpCPU tm thy ton hng chocc lnh ca n khi hot ng. Khc nhau: Ch a ch trc tip:Mt ton hng l mt hng biu din a ch lch (offset) ca nh. Ton hng cn li c th l thanh ghi (khng c l nh). Ch a ch gin tip qua thanh ghi:Mt ton hng l mt thanh ghi cha a ch lch ca nh. Ton hng cn li c th l thanh ghi hoc hng (ko c l nh) Ch a ch gin tip qua thanh ghi chm hn ch a ch trc tip nhng mm do hn.

Cu hi 2.6: Phn bit ch a ch thanh ghi v ch a ch tc th? Cho v d?

V d:

Ch a ch thanh ghi:

MOV BX, DX ; chuyn ni dung DX vo BX.

Ch a ch tc th:

MOV CL, 100 ; chuyn 100 vo CL.

Phn bit:

Ging nhau: Cho phpCPU tm thy ton hng chocc lnh ca n khi hot ng. Khc nhau:

Ch a ch thanh ghi: S dng cc thanh ghi bn trong cpu nh l cc ton hng cha d liu cn thao tc.C ton hng gc v ch u l cc thanh ghi. Ch a ch tc th: Ton hng ch l mt thanh ghi hay mt nh. Ton hng gc l mt hng s.Cu hi 2.7:Phn bit ch a ch gin tip qua thanh ghi v cc ch a ch tng i c s? Cho v d?

V d:

Ch a ch gin tip qua thanh ghi:

MOV AL, (BX); chuyn nh c a ch DS:BX vo AL.

MOV (SI), CL; chuyn CL vo nh c a ch DS:SI.

MOV (DI), AX; chuyn AX vo 2 nh lin tip ti DS:DI v DS: (DI + 1).

Ch a ch tng i c s:

MOV AL, [BX+10]; chuyn nh lin tip c a ch [DS: BX + 10] vo CX.

Phn bit:

Ging nhau: Cho phpCPU tm thy ton hng chocc lnh ca n khi hot ng. Khc nhau: Ch a ch gin tip qua thanh ghi:Mt ton hng l mt thanh ghi cha a ch lch ca nh. Ton hng cn li c th l thanh ghi hoc hng (ko c l nh)

Ch a ch tng i c s: Mt ton hng l /c ca nh./c ca nh c to bi vic s dng thanh ghi c s nh BX (on DS) hoc BP (on SS) v mt hng s. Hng s trong a ch tng i c s biu din cc gi tr dch chuyn (displacement) c dng tnh a ch hiu dng ca cc ton hng trong cc vng nh DS v SS. Ton hng cn li c th l thanh ghi (ko c l nh).Cu hi 2.8: Ti sao phi ngt n v x l trung tm? Cc loi ngt trong h 8086?

S cn thit phi ngt CPU:

Ngt l vic tm dng vic chng trnh ang chy CPU c th chy mt chng trnh khc nhm x l mt yu cu do bn ngoi a ti CPU nh yu cu vo/ra hoc do chnh yu cu ca bn trong CPU nh li trong khi tnh ton. Trong cch tchc trao i dliu thng qua vic thm d trng thi sn sng ca thit b ngoi vi, trc khi tin hnh bt k mt cuc trao i d liu no CPU phi dnh ton bthi gian vo vic xc nh trng thi sn sng lm vic ca thit bngoi vi.

Ngt gip tn dng khnng lm vic ca CPU,chkhi no c yu cu trao i dliu th mi yu cuCPU tm dng cng vic hin ti phc v vic trao i d liu. Sau khi hon thnh vic trao i d liu th CPU li phi quay v lm tip cng vic hin ang b gin on.

Cc loi ngt trong h 8086:

Nhm cc ngt cng: l cc yu cu ngt CPU do cc tn hiu n tcc chn INTR v NMI. Ngt cng INTR lyu cu ngt che c.Cc lnh CLI v STI cnh hng trc tip ti trng thi ca cIF trong bvi xl ,tc l nh hng ti vic CPU c nhn bit yu cu ngt ti chn ny hay khng. Yu cu ngt ti chn INTR c thc kiu ngt N nm trong khong0-FFH.Kiu ngt ny phi c a vo butdliu CPU c thc c khi c xung trong chu ktrli chp nhn ngt.

Nhm cc ngt mm: khi CPU thc hin cc lnh ngt dng INT N, trong N l s hiu (kiu) ngt nm trong khong 00-FFH (0-255).

Nhm cc hin tng ngoi l: l cc ngt do cc li ny sinh trong qu trnh hot ng ca CPU nh php chia cho 0, xy ra trn khi tnh ton.

Cu hi 2.9: Trnh by p ng ca CPU khi c yu cu ngt ? Khi c nhiu yu cu ngt th CPU phi x l th no ?

p ng ca CPU khi c yu cu ngt:Khi c ngt xy ra th chng trnh chnh tm dng vic thc hin v lu cc thanh ghi cn thit nh thanh ghi c.Sau con trlnh ca CPU sctrti on m cachng trnh con phc vngt.Khi chng trnh con phc v ngt kt thc, CPU khi phc li trng thi cc thanh ghi ca chng trnh chnh v t con trlnh vvtr bngng khi phc vngt. C th, CPU thc hin cc cng vic sau:

1.SP ( SP-2, [SP] ( FR, trong [SP] l nh do SP ch ra. (chra nh mi ca ngn xp,ctthanh ghi cvo nh ngn xp)

2.IF ( 0, TF ( 0. (cm cc ngt khc tc ng vo CPU,cho CPU chychbnh thng)

3.SP ( SP-2, [SP] ( CS.(chra nh mi ca ngn xp,ct phn a chon ca a chtr v vo nh ngn xp)

4.SP ( SP-2, [SP] ( IP (chra nh mi ca ngn xp,ct phn a chlch ca a chtr v vo nh ngn xp)

5.[N*4] ( IP, [N*4+2] ( CS (ly lnh ti a ch mi ca chng trnh con phc vngt kiu N tng ng trong bng vect ngt)

6.Ti cui chng trnh phc v ngt, khi gp lnh IRET

[SP] ( IP, SP ( SP+2

[SP] ( CS, SP ( SP+2

[SP] ( FR, SP ( SP+2

(bvi xl quay li chng trnh chnh ti a chtrvv vi gi tr c ca thanh ghi c c ly ra t ngn xp).

Khi c nhiu yu cu ngt, CPU phc v ngt c mc u tin cao nht.Nu mt hay nhiu yu cu ngt xut hin khi CPUang phc vchng trnh ngt khc th ngt mic u tin cao hn s c phc v.

Cc ngt c chia thnh cc mc u tin theo th t t cao n thp nh sau: Ngt trong: INT 0 (php chia cho 0), INT N, INTO (Mc u tin cao nht). Ngt khng che c NMI

Ngt che c INTR

Ngt chy tng lng INT 1 (Mc u tin thp nht).Cu hi 2.10: V lu v gii thch ngha ca on m sau:

CMP AX, 0

JL TIEP

MOV CL, l

JMP THOAT

TIEP:XOR CL, CL

THOAT:

Lu :

ngha: on m dng kim tra gi tr trong AX. Kt qu kim tra lu trong thanh ghi CL, CL = 0 nn AX < 0, v CL = 1 nu AX 0.

Cu hi 2.11:V lu v gii thch ngha ca on m sau:

MOV CX, 80

XOR AX,AX

XULY:ADD AX,CX

LOOPXULY

RA:

Lu :

ngha: on m ny dng tnh tng cc s nguyn t 1 n 80, lu kt qu vo thanh ghi AX. Kt qu trng trnh, AX = 00001100 10101000 (= 3240), CX=0.

Cu hi 2.14: Phn loi b nh? Trnh by cc tn hiu cn bn ca vi mch nh khi

qut ? Phn loi b nh: Cc b nh bn thng dng vi b vi x l bao gm:

BnhcnhROM(ReadOnlyMemory)haybnhchcra. Thngtinghitrongmchkhngbmtkhimtnguninnuicho mch.

Bnhbn cnh EPROM (ErasableProgrammable ROM)l bnh ROM c th lp trnh c bng xung in v xo c bng tia cc tm.

Bnhtruy nhp ngu nhinRAM(RandomAccessMemory)thng tin ghi trong mch bmt khi mt ngun in nui cho mch.Trong cc b nhRAMcnphnbitraloiRAMtnh(StaticRAM),trongmi phn tnhl mt mch lt hay trng thin nh) v loi RAM ng (Dynamic RAM)trong mi phn tnhl mt tin rt nhc ch to bng cng ngh MOS.

Cc tn hiu cn bn ca vi mch nh:

Nhm tn hiu a ch (A0-Am): Cc tn hiu a chc tc dng chn ra mt nhca vi mch nh.Cc nhc di khc nhau(cn gi ltnh)tutheo nh sn xut:1,4,8, bt.Sng tn hiu a chc lin quan n dung lng ca mch nh.

Nhm tn hiu d liu (D0-Dn):Cctn hiu dliu thng l u ra i vi mch ROM hoc u vo/ra dliu chung (hai chiu) i vi mch RAM.Ngoi ra cloimch nhRAM vi u ra v u vo d liu ring bit. Cc mch nh thng c ura dliu kiu 3 trng thi.Sng dy dliu quyt nh di tnhca mch nh.V dmch nhdung lng 1 Kx8 (tc l 1KB) hoc 16Kx4. . .

Tn hiu chn vi mch ( hoc ):Cctnhiuchnvimchl (chipselect)vi mch RAM hoc (chipenable) vi mch ROM thng c dng chn ra vi mch nh c th ghi/c.

Nhm tn hiu iu khin:Cc mch nhROM thng c mt u vo iu khin (output enable) cho php dliu c a ra bus.MtmchnhRAMnuchcmttnhiuiukhinth thng l iu khin qu trnh ghi/c. Nu mch nhRAM c hai tn hiu iu khin thng l (write enable) iu khin ghi viu khin c.

Cc tn hiu chn vi mch v tn hiu iu khin u hot ng mc tch cc thp.

Cu hi 2.18: So snh truyn thng ni tip ng b v d b?

Ging nhau: D liu di dng song song u pht c chuyn thnh d liu dng ni tip, tn hiu ni tip sau c truyn i lin tip tng bt trn mt ng dy. u thutnhiu ni tip sc biningclititohiudngsongsongthchhpchovicxltiptheo. Khc nhau: Khc cn bn cu trc khung truyn. Truyn thng ni tip ng b: d liu c truyn theo tng khi vi mt tc xc nh. Khi d liu trc khi c truyn i s c b sung thm cc phn t c bitu vcuito thnh khung.Cc phn tny dng nh du im bt u ca khi dliu hay cc thng tin gip pht hin li trongqu trnh truyn.y thc cht lcchiu khin hngktvccktcbitcdngnhduccphnkhcnhautrong khung.

Truyn thng ni tip d b: dliu c truyn i theo tng k tring bit. di k t c th thay i t 5 n 8 bt. K tcn truyn i c gn thm 1 bt nh du u bo bt u k t (Start bit) v mt hoc hai bt bo kt thc k t (Stop bit), v mt bt kim tra tnh ton vn dliu (Parity bit).V mi k tc nhn dng ring bit nn n c thc truyn i vo bt k lc no.Gia cc k ttruyn i c thc cc khong cch vthi gian.

Cu hi 2.19: Trnh by phng php vo/ra thm d ? So snh phng php ny viphng php vo/ra s dng ngt?

Trnh by phng pho vo/ra thm d: Yu cu d liu xut pht t pha CPU.

CPU qun l danh sch cc thit b vo ra km theo a ch cc cng giao tip;Mi bt ca thanh ghi trng thi c gn cho mt thit b. Cc thit b vo ra nh k cp nht trng thi sn sng lm vic ca mnh ln cc bt c trng thi vo ra ca mnh;

CPU nh k ln lt qut cc thit b vo ra c cc bit c trng thi vo ra;

Nu gp mt thit b sn sng lm vic, 2 bn tin hnh trao i d liu;

Trao i d liu xong, CPU tip tc qut thit b khc.

C 2 kiu vo ra bng thm d:

Thm d c u tin:CPU c th qut tt c cc thit b trong mt chu trnh thm d nhng ch trao i d liu vi ti a 1 thit b trong mt chu trnh thm d v bt u chu trnh thm d mi ngay sau khi trao i xong vi mt thit b; Cc thit b c mc u tin cao lun c thm trc; Thm d khng u tin:CPU qut tt c cc thit b trong mt chu trnh thm d, CPU c th trao i d liu vi nhiu hn 1 thit b trong mt chu trnh thm d. Cc thit b c thm ln lt, khng ph thuc vo thit b ng trc chu trnh.CPU bt u 1 chu trnh thm d mi sau khi qut qua tt c cc thit b. So snh phng php vo/ra thm d vi phng php vo/ra s dng ngt:

Ging nhau:

Mc ch cho php trao i d liu gia my tnh v cc thit b ngoi vi.

u c CPU can thip vo qu trnh trao i d liu.

Khc nhau:

Vo/ra bng thm dVo/ra bng ngt

CPU l bn ch ng trong qu trnh trao i d liu. CPU phi tin hnh thm d tng thit b ngoi vi, nn hiu qu thp v khng kh thi khi c nhiu thit b trong danh sch thm d. n gin, d ci t, c th ci t bng phn mm. Thit b vo ra ch ng yu cu trao i d liu.

Hiu qu hn vo/ra bng thm d do CPU khng phi thm d tng thit b.

Phc tp hn, v cn thm mch phn cng iu khin ngt.

Cu hi 2.20: Trnh by phng php vo/ra trc tip b nh ? So snh vi phng php vo/ra s dng ngt? Trnh by phng php vo/ra trc tip b nh (vo/ra bng DMA):

DRQ (DMA Request): Yu cu DMA.

DACK (DMA Acknowledgment): Xc nhn DMA.

HLDA (Hold Acknowledgment): Xc nhn treo.

HOLD: Treo.

Thit b vo ra c yu cu trao i d liu gi yu cu DRQ n CPU thng qua B iu khin DMA (DMAC); DMAC chuyn yu cu DRQ thnh HRQ v gi n chn tn hiu HOLD ca CPU; Nhn c yu cu s dng bus HRQ, CPU: Gi cc tham s iu khin trao i d liu v tn hiu xc nhn yu cu s dng bus HACK cho DMAC qua chn tn hiu HLDA; CPU tm thi treo BUS (khng s dng BUS na). T tch ra khi bus h thng (100% cc tn hiu ca bus A v D v mt s tn hiu ca bus C). Nhn c HACK, DMAC chim quyn iu khin bus h thng v gi tn hiu xc nhn DACK cho thit b vo ra; DMAC iu khin qu trnh trao i d liu trc tip gia thit b vo ra v b nh; Kt thc qu trnh DMA, DMAC tr quyn iu khin bus cho CPU. So snh vo/ra bng DMA vi vo/ra s dng ngt:

Ging nhau:

Mc ch cho php trao i d liu gia my tnh v cc thit b ngoi vi.

Yu cu trao i d liu u xut pht t pha thit b ngoi vi.

Khc nhau:

Vo/ra bng ngtVo/ra bng DMA

CPU can thip qu trnh trao i d liu. Khng nht thit cn thm phn cng. DMA can thip qu trnh trao i d liu, khng thng qua CPU nn hiu sut cao hn v tc cao hn nhiu ln so vi vo/ra bng ngt.

Phc tp hn vo/ra bng ngt do bt buc cn thm phn cng iu khin qu trnh DMA.

Cu hi loi 2 im (Ngn hng in T):2.1 Ch a ch (addressing mode) ca vi x l l g? M t ch a ch gin tip qua thanh ghi ca b vi x l ntel 8086/8088. Cho v d minh ho.

2.2 Cc ch a ch trong lp trnh hp ng ca vi x l 8088/8086? Cho v d minh ha i vi tng ch ?

2.3 Gi s DS = 2344, BX = 56. Hi nh c gn ni dung ca thanh ghi AL trong lnh sau c a ch vt l bng bao nhiu?

MOV 15[BX], AL; ( tc l MOV [BX]+15,AL)

DS = 2344 = 0928H

BX = 56 = 0038H

a ch logic nh: DS: (BX+15) = 0928H: 47Hachvtl=Thanh_ghi_an16+Thanh_ghi_lch

a ch vt l nh l: 09280H + 47H = 092C7 H2.4 Gi s DS = 4423, SI = 23. Hi nh c ni dung gn cho thanh ghi AL trong lnh sau c a ch vt l bng bao nhiu?

MOV AL,[SI] +16

DS = 4423 = 1147H

SI =23 = 17H

a ch logic nh l DS: (SI + 16) = 1147H: 27H


a ch vt l nh l: 11470H+ 27H = 11497H

2.5 Gi s DS = 4498H, DI = 6AH. Hi nh c gn ni dung ca thanh ghi CL trong lnh sau c a ch vt l bng bao nhiu?

MOV [DI +35 ], CL

a ch logic nh: DS: (DI + 35) = 4498H: 8DH

achvtl=Thanh_ghi_an16 + Thanh_ghi_lch

a ch vt l ca nh l: 44980H + 8DH = 44A0DH

2.6 Gi s DS = 44F8H, BX = 15H, SI = 10H. Hi nh c ni dung gn cho thanh ghi AL trong lnh sau c a ch vt l bng bao nhiu?

MOV AL, [BX][SI]+10H

V d:

MOV [BX][SI]+10H,CX ;DS+SI+BX+15

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Đáp án KTVXL PTit