1、實(shí)用文檔1程序如下：MOVSP , #13HMOV 33H , #7FHMOV 44H , #0ABHPUSH 33HPUSH 44HPOP 33HPOP44H,(44H) =7Fh程序執(zhí)行后(33H) =0ABh2下列各條指令其源操作數(shù)的尋址方式是什么？各條指令單獨(dú)執(zhí)行后，A中的結(jié)果是什么?設(shè)(60H ) =35H , (A) =19H , ( RO) =30H , (30H ) =OFH。(1) MOV A, #48H(A) =48H；尋址方式:(2) ADD A, 60H；尋址方式:(A) =4DH(3) ANLA, R0；尋址方式:(A) =10H3閱讀下列程序段，寫(xiě)出每條指令執(zhí)行后的

2、結(jié)果，并說(shuō)明此程序段完成什么功 能？MOVR1 ,#30H;(R1)=30HMOVA,#64H ; ( A)=64HADD(AC )=0A,#47H ; ( A)=ABH,(CY) =0DA(AC )=A;(A)=31H,(CY )=1MOVR1 ,A;(R1此程序段完成的功能:)=30H,(30H) = 31H4.(A)=38H,R0=28H,(28H)=18H5 執(zhí)行下列程序后'ORLA, #27HANLA , 28HXCHD A, R06.已知(RO)CPL A5.設(shè)(A)=38H,(20H)=49H,PSW=00H,(B)=02H,填寫(xiě)以下中間結(jié)果。SETBCADDCA,20H

3、(A)= 82H(CY)= 0(AC)= 1RLCA(A)=04H(CY)=1(P)=1MULAB(A)=08H(B)=00H(OV)=0=20H, (20H ) =10H, (P0) =30H, (R2) =20H,執(zhí)行如下程序段后(40H )標(biāo)準(zhǔn)實(shí)用文檔=15HMOVRO,#11HMOV A, R2(20H)=11HA=20H標(biāo)準(zhǔn)實(shí)用文檔ADDA,20HA=40HMOVPSW , #80HSUBBA , P0A=10HXRLA , #45HA=15HMOV40H, A(40H)=15H7.已知(RO )=20H, (20H )=36H,(21H) =17H, (36H) =34H,執(zhí)行過(guò)程

4、如下：MOVA , R0A=36HMOVRO , AR0=36HMOVA , R0A=34HADDA,21HA=4BHORLA ,#21HA=21HRLAA=42HMOVR2 , AR2=42HRET則執(zhí)行結(jié)束(RO) =36H(R2)= 42H8設(shè)在31H單元存有# 23H,執(zhí)行下面程序:MOV A,31HA=23HANL A, #0FH A=03HMOVMOV41H, A(41H)=03HA, 31HA=23HANL A, #0F0HA=20HSWAP AA=02HMOV 42H, A (42H)=02H則(41H)=03H(42H)=02H9. (R0)=4BH,(A)=84H,片內(nèi) R

5、AM(4BH)=7FH,(40)=20HMOV A,R0A=7FHMOV R0540H(4BH)=20HMOV 40H,A(40H)=7FH;MOV R05#35HR0=35H40H= 7FH問(wèn)執(zhí)行程序后,R0=35H A= 7FH 4BH= 20H10若PSW=OO,執(zhí)行下列程序后，PSW的CY,AC,OV,P各位狀態(tài)如何？MOV A,#0FBH A=FBHMOV PSW,#1 OHADD A,#7FH A=7AH CY=1; AC=1 ;OV=1; P=1;11 程序存儲(chǔ)器空間表格如下：地址 2000H2001H2002H2003H-內(nèi)容 3FH06H5EH4FH 已知：片內(nèi)RAM的2 0

6、H中為01H,執(zhí)行下列程序后(30H) =A3H標(biāo)準(zhǔn)實(shí)用文檔MOV A，20H A=O1H標(biāo)準(zhǔn)實(shí)用文檔INCAMOV DPTR,#2000HMOVC A,A + DPTRCPL AMOV30H,ASJMP$ENDA=02HDPRT=2000HA=5BHA=A3H(30H) =A3H執(zhí)行程序后,（30H）= A3H 12.分析程序段:CLRCMOV A,#91HA寄存器與地址(61h )的 值相減SUBB A,61HADD A,61HDA A;清零cy指令給A立即數(shù)A寄存器與地址（61h ）的值相加BDC碼調(diào)整指令把累加器A的值送給地址寄存器62hMOV 6 2H,A（1 ）程序執(zhí)行何種操對(duì)91

7、H做BDC碼調(diào)整；13.設(shè)(RO) =7EH,DPTR=10FEH，片內(nèi) RAM 7EH 和 7FH 兩單（2）若已知初值：（60H） =24H,（6伯）=72H,則運(yùn)行后，（62H）=91H元的內(nèi)容分別是FFH和38H，請(qǐng)寫(xiě)出下列程序段的每條指令的執(zhí)行結(jié)果。INC RO(7E)=00HINCROR0=7FHINC RO(7FH)=39HINCDPTRDPTR=1OFF1 1INCDPTRDPTR=11OOHINCDPTRDPTR=11O1H14.設(shè) R0=20H,R 1=25H, (20H)=80H ,(21H)=90H ,(22H)=A0H ,(25H)=A0H , (26H)=6FH ,

8、（27H）=76H,下列程序執(zhí)行程序后，結(jié)果如何？CLRCc=o；MOVR2,#3R2=3;LOOP: MOVA,R0A=80HA=90HADDCA,R1A=20HMOVR0,A(20H)=20H;(21H)=00H;(22H)=16HINCROR0=21H;INCR1R 仁 26HDJNZR2,LOOPJNCNEXT1MOVR0,#01H(23H)=01HSJMP$NEXT: DECROSJMP$標(biāo)準(zhǔn)實(shí)用文檔(20H) =20H(21H)=OOH(22H)=16HA=16H R0=23H(23H)=01H CY=1oMOVDPTR,#2000(DPTR) = 2000HMOVA,#80H(A

9、) = 80HMOVXDPTR,A(2000H) =80HINCDPTR(DPTR) = 2001MOVA,#90H(A) =90HMOVXDPTR,A(2001H) = 90HMOVDPTR,#2000H(DPTR) = 2000HMOVXA,DPTR(A) = 80HMOVB,A(B)=80HINCDPTR(DPTR) = 20011MOVXA,DPTR(A)=90H寫(xiě)出每一步指令運(yùn)行結(jié)果MOVA,10H(A)=FFH ,ANLA,#01H(A)=_01H,MOVA,11H(A)=OOH,ORLA,#01H(A)=01H,MOVA,12H(A)=,FFH,XRLA,#01H(A)=FEH,

10、MOVA,13H(A)=,OOH,XRLA,#0AA(A)=:AAH,(10H)=FFH,(11H)= OOH , (12H)= FFH,(13H)=16 若(1 OH)=OFFH,(11 H)=00H,(12H)=0FFH,(13H)=00H,OOH17閱讀程序并回答下列問(wèn)題:X DATA50HYDATA51HORG OOHJMP STARTORG 100HSTART: MOVSP,#60HMOV X5#10MOV Y,#10(50H)=10(51H)=10LCALLSUBSUB:INCX(50H)=11DECY(51H)=9MOVA,YA=9JZSUBRETLCALLSUBNOPX+10=

11、20Y=0;SUBRET:NOPRET標(biāo)準(zhǔn)實(shí)用文檔Y存放在BUF單兀。給程序標(biāo)注注釋，說(shuō)(7分)計(jì)算X和丫的值是多少？(X )二20 ； (Y )二 _018設(shè)自變量為X，存放在ARE單元，應(yīng)變量明該段子程序的功能并寫(xiě)岀該程序的函數(shù)尖系式。START：MOVDPTR,#AREMOVXA , DPTRJZSULJBACC.7 , NEGMOVA, #02HSUL :MOVDPTR,#BUFMOVXDPTR , ARETNEG :MOV A,#OFEHSJMPSUL如果 X=0,Y=0;如果 X<128,Y=02h;；注釋：如果A的值大于128跳轉(zhuǎn)NEG;注釋：把A的值傳送給BUF地址中女口

12、果 X>128,Y=FEH ；19執(zhí)行下列程序段后，(P1)= 9AHMOV P1, #5DHCPL P1.1CPL P1.2CLR P1.6SETB P1.7(A)=8BH,(C y)=_0_ o20 執(zhí)行下列程序段后,MOVA, #C5HRLA21 下列程序段執(zhí)行后，(R°)=7EH,(7EH)=_41 H_,(7FH)=_01H0123H 的 ACALL22已知(SP)=60H，子程序SUBTRN的首地址為0345H，現(xiàn)執(zhí)行位于SUBTRN 雙字節(jié)指令后，(PC)=,(61H)二 ,(62H)=MOVRo, #7EHR0=7EHMOV7EH ,#OFFH(7EH)=FFM

13、OV7FH , #40H(7EH)=40H1NCR0(7EH)=41H1NCROR0=7FH;1NCR0(7FH)=1;23.閱讀下列程序，說(shuō)明其功能。標(biāo)準(zhǔn)實(shí)用文檔MOV Ro,#dataMOV A,RoRL AMOV Ri, ARL ARL AADD A , RiMOV Ro,ARET24.程序填空1 ）數(shù)據(jù)塊傳送，將RAM從30H開(kāi)始的連續(xù)32個(gè)單元的內(nèi)容傳遞給片內(nèi)RAM從60H開(kāi)始的連續(xù) 32個(gè)單元。ORG 1000HMOV R7,MOV RO, #30HMOVR1, #60HLOOP： MOV A,R0MOV ,AINC ROINC R1DJNZ R7,SJMP$END2） RAM中4

14、0H單元內(nèi)存有一個(gè)十六進(jìn)制數(shù)，把這個(gè)數(shù)轉(zhuǎn)換為BCD碼的十進(jìn)制數(shù)，BCD碼的十位和個(gè)位放在累加器A中，百位放在R2中。ORG 2200HMOVA, MOV B，#64HDIV ABMOV R2，AMOVA, XCH A，BDIV ABSWAPORLA，BSJMP$END3 ）編程序?qū)⑵瑑?nèi)40H-46H單元內(nèi)容的高4位清零，保持低4位不變。ORG 1000HMOV R7，#07HMOV RO, 標(biāo)準(zhǔn)實(shí)用文檔LOOP : MOV A , R0ANL A,MOVRO, AINC RODJNZ R7 , LOOPSJMP$END25程序如下問(wèn)：（1）執(zhí)行POP10H后堆棧內(nèi)容？2506HM5:MOVSP

15、,#58H;2509 HMOV10H,#0FH;250CHMOV11H,#0BH;250FHACALLXHD;2511HMOV20H,11H2514HM5A:SJMPM5A 10HXHD:PUSH11HPUSH10HPOP11HPOPRET(20H )=?將PSW各位狀態(tài)填入下表26解讀下列程序，然后填與有矢寄存器內(nèi)容R1,#48H48H,#51H(1)MOVMOVCJNER1,#51H,00HJNCNEXT1MOVA,#0FFHSJMPNEXT2NEXT:MOVA,#0AAHNEXT2:SJMPNEXT2累加器A=()(2)MOVA,#0FBHMOVPSW,#1 OHADDA,#7FH執(zhí)行M5A:SJMP M5A 后,(SP )=?若 PSW=00,當(dāng)執(zhí)行完上述程序段后，PSWCYACF0RSIRSOOVFlP27-改正下列指令MOV R3,A標(biāo)準(zhǔn)