單片機程序分析題練習(xí)B2

1、1/ 86已知(R0)=20H, (20H ) =10H, (P0) =30H, (R2) =20H,執(zhí)行如下程序段后( 40H ) =15HMOV R0 , #11H (20H)=11HMOV A , R2 A=20H1程序如下：MOV SP，#13HMOV 33H，#7FHMOV 44H，#0ABHPUSH 33HPUSH 44HPOP 33HPOP 44H程序執(zhí)行后， (33H)=0ABh，（44H）=7FhA中的結(jié)果是什么？設(shè)(60H)=35H, (A)=19H, (RO)=30H, (30H)=OFH。1)MOVA，#48H；尋址方式：(A)= 48H2)ADDA，6OH；尋址方式：

2、(A)= 4DH3)ANLA，RO；尋址方式：(A)= 1OH2下列各條指令其源操作數(shù)的尋址方式是什么？各條指令單獨執(zhí)行后，3閱讀下列程序段，寫出每條指令執(zhí)行后的結(jié)果，并說明此程序段完成什么功能？MOVR1，#3OH；(R1)=3OHMOVA，#64H；(A)= 64HADDA，#47H；(A)=ABH，(CY)= O(AC)= ODAA；(A)=31H，(CY)=1(AC)=MOVR1，A；(R1)= 3OH，(3OH)= 31H此程序段完成的功能4設(shè)(A)=38H,R0=28H,(28H)=18H,執(zhí)行下列程序后， (A)=0 ORL A，#27HANL A，28HXCHD A，R0CPL

3、 A5設(shè)(A)=38H,(20H)=49H,PSW=00H,(B)=02H,填寫以下中間結(jié)果。SETB CADDC A, 20H (A)= 82H(CY)= 0RLC A(A)=04H (CY)=1MUL AB (A)=08H(B)=00H(AC)= 1(P)= 1(OV)=02/ 8ADDA , 20HA=40HMOVPSW , #80HSUBBA , P0A=10HXRLA , #45HA=15HMOV40H , A(40H)=15H7.已知(RO )=20H, (20H )=36H, (21H) =17H, (36H) =34H,執(zhí)行過程如下:MOVA , R0A=36HMOVR0 ,

4、AR0=36HMOVA , R0A=34HADDA , 21HA=4BHORLA , #21HA=21HRLAA=42HMOVR2 , AR2=42HRET則執(zhí)行結(jié)束(R0)=36H(R2)= 42H8設(shè)在31H單元存有#23H,執(zhí)行下面程序:MOVA, 31HA=23HANLA, #0FHA=03HMOV41H, A(41H)=03HMOVA, 31HA=23HANLA, #0F0HA=20HSWAPAA=02HMOV42H, A(42H)=02H則(41H)= 03H(42H)=02H 9.(R0)=4BH,(A)=84H,片內(nèi)RAM(4BH)=7FH,(40)=20HMOVA,R0A=7

5、FHMOVR0,40H(4BH)=20HMOV40H,A(40H)=7FH;MOVR0,#35HR0=35H問執(zhí)行程序后,R0=35H A= 7FH 4BH= 20H40H= 7FH10若PSW=00,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,#10HADD A,#7FH A=7AH CY=1; AC=1;OV=1; P=1;11.程序存儲器空間表格如下：地址2000H2001H2002H2003H.內(nèi)容3FH06H5EH4FH .已知：片內(nèi)RAM的2 0H中為01H,執(zhí)行下列程序后（30H）=A3H3/ 8MOV A,20H

6、A=O1H4/ 8INC AMOV DPTR,#2000HDPRT=2000HMOVC A,A + DPTRA=5BHCPLAA=A3HMOV30H,A(30H) =A3HSJMP$END執(zhí)行程序后,(30H)= A3H12.分析程序段：CLR CMOV A,#91HA寄存器與地址(61h)的值相減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62h13.設(shè)(R0)=7EH,DPTR=10FEH，片內(nèi)RAM7EH和7FH兩單元的內(nèi)容分別是FFH和38H，請寫出下列程序段的每條指令的執(zhí)行結(jié)果。INCR0(7E)=00HINCR0R0=7FHINCR0(7FH)=39

7、HINCDPTRDPTR=10FFHINCDPTRDPTR=1100HINCDPTRDPTR=1101H14.設(shè)R0=20H,R1=25H, (20H)=80H ,(21H)=90H ,(22H)=A0H ,(25H)=A0H , (26H)=6 FH ,(27H)=76H，下列程序執(zhí)行程序后，結(jié)果如何？ADDA,61HDAA；CLRCC=0;MOVR2,#3R2=3;LOOP: MOVA,R0A=80HA=90HADDCA,R1A=20HMOVR0,A(20H)=20H;(21H)=00H;INCROR0=21H;INCR1R仁26HDJNZR2,LOOPJNCNEXT1MOVR0,#01H

8、(23H)=01HSJMP$NEXT: DECR0SJMP$(22H)=16HA=02H清零cy指令 給A立即數(shù)(1)程序執(zhí)行何種操作?對91H做BDC碼調(diào)整;(2)若已知初值:(60H)=24H,(6伯)=72H,則運行后，(62H)=91HSUBB A,61HMOV 6 2H,A5/ 8(20H)=20H、(21H)= OOH、(22H)=16HSUBRET:NOP(23H)= O1H、CY=1A=16H、R0=23H、R仁27HMOVDPTR,#2000HMOVA,#80HMOVXDPTR,AINCDPTRMOVA,#90HMOVXDPTR,AMOVDPTR,#2000HMOVXA,DP

9、TRMOVB,AINCDPTRMOVXA,DPTR(DPTR)= 2000H(A) = 80H(2000H)=80H(DPTR)= 2001(A)=90H(2001H)= 90H(DPTR)= 2000H(A)= 80H(B)=80H(DPTR)= 2001H16.若(10H)=0FFH,(1伯)=00H,(12H)=0FFH,(13H)=00H,寫出每一步指令運行結(jié)果MOVA,10H(A)=FFH,ANLA,#01H(A)= _01H,MOVA,11H(A)=00H,ORLA,#01H(A)=01H,MOVA,12H(A)=,FFH,XRLA,#01H(A)=FEH,MOVA,13H(A)=

10、,00H,XRLA,#0AA(A)=:AAH,(10H)=FFH,(11H)= 00H, (12H)= FFH , (13H)=00H17.閱讀程序并回答下列問題:X DATA 50HORG00HJMPSTARTORG100HMOVSP,#60HMOVX,#10(50H)=10MOVY,#10(51H)=10SUB:LCALL SUBNOPINCX(50H)=11DECY(51H)=9MOVA,YA=9JZSUBRETLCALLSUBX+10=20Y=0;6/ 8RET7/ 8計算X和Y的值是多少？(X)=20;(Y)= _0明該段子程序的功能并寫出該程序的函數(shù)關(guān)系式。(7分)如果X=0,Y=

11、0;如果X128,Y=FEH;19.執(zhí)行下列程序段后，(P1)=_9AH_MOVP1,#5DHCPLP1.1CPLP1.2CLRP1.6SETBP1.720.執(zhí)行下列程序段后,MOV A,#C5HRL A21.下列程序段執(zhí)行后，(R0)=_7EH,(7EH)=_41H_,(7FH)=_01H_MOVR0,#7EHR0=7EH18.設(shè)自變量為X，存放在ARE單元，應(yīng)變量Y存放在BUF單兀。給程序標注注釋，說START:MOVDPTR,#AREMOVXA,DPTRJZSULJBACC.7,NEGMOVA,#02HSUL:MOVDPTR,#BUFMOVXDPTR,ARETNEG:MOV A,#0FE

12、HSJMPSUL;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(A)=_8BH_,(CY)=_0_。22.已知(SP)=60H，子程序SUBTRN雙字節(jié)指令后,SUBTRN的首地址為0345H，現(xiàn)執(zhí)行位于(PC)=,(61H)=_,(62H)=_0123H的ACALL8/ 8MOV7EH,#0FFH(7EH)=FFHMOV7FH,#40H(7EH)=40H1NCR0(7EH)=41H1NCR0R0=7FH;1NC R0(7FH)=1;23.閱讀下列程序，說明其功能。MOV R0,#dataMOV A,R0RL AMOV R1，ARL ARL AADD A，R1MOV

13、R0,ARET24.程序填空1)數(shù)據(jù)塊傳送，將RAM從30H開始的連續(xù)32個單元的內(nèi)容傳遞給片內(nèi)RAM從60H開 始的連續(xù)32個單元。ORG 1000HMOV R7，_MOV R0, #30HMOV R1,#60HLOOP：MOV A,R0MOV_,AINC R0INC R1DJNZ R7,_SJMP $END2)RAM中40H單元內(nèi)存有一個十六進制數(shù)，把這個數(shù)轉(zhuǎn)換為BCD碼的十進制數(shù)，BCD碼的十位和個位放在累加器A中，百位放在R2中。ORG 2200HMOV A ，_MOV B，#64HDIV ABMOV R2，AMOV A ，_XCH A，BDIV ABSWAP_ORL A，BSJMP

14、$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 R0，_9/ 8LOOP:MOV A,R0ANL A,_MOV R0,AINC RODJNZ R7,LOOPSJMP $END25.程序如下26.解讀下列程序，然后填寫有關(guān)寄存器內(nèi)容(1)MOVR1,#48HMOV48H,#51HCJNER1,#51H,00HJNCNEXT1MOVA,#0FFHSJMPNEXT2NEXT:MOVA,#0AAHNEXT2:SJMPNEXT2累加器A=()(2)MOVA,#0FBHMOVPSW,#10HADDA,#7FH若PSW=00,當(dāng)執(zhí)行完上述程序段后，將PSW各位狀態(tài)填入下表PSWCYACF0RS1RS0OVF1P2506H2509H250CH250FH2511HMOV10H,#0FH;MOV11H,#0BH;ACALLXHD;MOV20H,11HM5A:SJMPM5AXHD:PUSH10HPUSH11HPOP10HPOP11H問：（1）執(zhí)行POP10H后堆棧內(nèi)容?(2)執(zhí)行M5A:SJMPM5A后，(SP )=?( 20H )=?M5:MOVSP,#58H;