单片机程序分析题练习

1、1.程序如下：MOV SP #13HMOV 33H #7FHMOV 44H #0ABHPUSH 33HPUSH44HPOP 33HPOP 44H,(44H) =7Fh程序执行后,(33H) =0ABh2.以下各条指令其源操作数的寻址方式是什么？各条指令单独执行后,A中的结果是什么?设(60H) =35H, (A) =19H, (R0) =30H, (303 =0FH.(1) MOVA #48H;寻址方式：(A)= 48H(2) ADDA 60H;寻址方式：(A)= 4DH(3) ANLA R0;寻址方式：(A)= 10H3.阅读以下程序段,写出每条指令执行后的结果,并说明此程序段完成什么功能?

2、MOV R1 #30H；(R1) =30HMOV A #64H ;(A)= 64HADD A #47H ;(A)=ABH, (CYO =0,(AC) = 0DAA;(A) =31H,(CYO =1(AC)=MOV R 1 A ;(R1)=30H, ( 30H)= 31H此程序段完成的功能4. 设(A)=38H,R0=28H,(28H)=18H,执行以下程序后,(A) =0ORL A, #27HANL A , 28HXCHD A, R0CPL A5. 设(A)=38H,(20H)=49H,PSW=00H,(B)=02H,填写以下中间结果.SETB CADDC A, 20H (A)= 82H (C

3、Y)= 0 (AC)= 1RLC A(A)=04H(CY)=1(P)= 1MUL AB(A)=08H(B)=00H(OV)=06. (R0) =20H, (20H ) =10H, (P0) =30H, (R2) =20H,执行如下程序段后(40H ) =15HMOV R0 , #11H(20H)=11HMOV A , R2A=20HADD A , 20HA=40HMOV PSW , #80HSUBB A , P0A=10HXRL A , #45HA=15HMOV 40H , A(403 =15H7. (R0 )=20H, (20H )=36H, (21H) =17H, (36H) =34H,执

4、行过程如下：MOV A , R0A=36HMOV R0 , AR0=36HMOV A , R0A=34HADD A , 21HA=4BHORL A , #21HA=21HRL AA=42HMOV R2 , AR2=42HRET那么执行结束R0) =36H(R2)= 42H8.设在31H单元存有# 23H,执行下面程序：MOV A, 31H A=23HANL A, #0FHA=03HMOV 41H, A (41H)=03HMOV A, 31H A=23HANL A, #0F0HA=20HSWAP AA=02HMOV 42H, A(42H)=02H那么(41H)= 03H(42H)= 02H9.

5、(R0)=4BH,(A)=84H,片内 RAM(4BH)=7FH,(40)=20HMOV A,R0A=7FHMOV R0,40H (4BH)=20HMOV 40H,A(40H)=7FH;MOV R0,#35H R0=35H问执行程序后,R0=35H A= 7FH 4BH= 20H 40H=7FH10. 假设PSW=00,执行以下程序后,PSW的CY,AC,OV,P各位状态如何？MOV A,#0FBH A=FBHMOV PSW,#10HADD A,#7FH A=7AH CY=1; AC=1;OV=1; P=1;11. 程序存储器空间表格如下：地址 2000H2001H2002H2003H.内容

6、3FH06H 5BH 4FH .：片内RAM的2 0 H中为0 1 H,执行以下程序后(3 0 H) = A3HMOV A, 2 0 H A=O1HINC A A=02HMOV DPTR, #2000H DPRT=2000HMOVC A, A + DPTR A=5BHC P L AA=A3HMOV 30H,A(30H) =A3HSJMP $END执行程序后,(30H)= A3H12. 分析程序段：C L R C清零cy指令MOV A, #9 1H给A立即数SUBB A, 6 1H ; A存放器与地址(61h)的值相减ADD A, 6 1H ; A存放器与地址(61h)的值相加D A AMOV

7、6 2 H, A(1)程序执行何种操作？(2)假设初值：(60H)BDg调整指令把累加器A的值送给地址存放器 62h对91H做BDg调整；=24H,(61H)=72H,那么运行后,(62H) = 91H7 EH和7 F H两单元的13.设(RO) = 7EH, DPTR=1 0 F EH,片内 RAM内容分别是F FH和3 8 HINC R 0INCROINC R 0请写出以下程序段的每条指令的执行结果.(7E)=00HR0=7FH(7FH)=39HINCDPTRINCDPTRINCDPTRDPTR=10FFHDPTR=1100HDPTR=1101H14. 设 R0=20H,R1=25H,(2

8、0H)=80H , (21H)=90H , (22H)=A0H, (25H)=A0H, (26H)=6 FH ,(27H)=76H,以下程序执行程序后,结果如何？CLR CC=0;MOV R2,#3R2=3;LOOP: MOVA,R0A=80HA=90HADDCA,R1A=20HMOV R0,A (20H)=20H;(21H)=00H; (22H)=16HINCR0R0=21H;INCR1R1=26HDJNZ R2,LOOP JNC NEXT1MOV R0,#01H (23H)=01HSJMP $NEXT: DEC R0 SJMP $(20H) =20H、(21H)= 00H 、(22H)=1

9、6H 、(23H)= 01H 、CY=1A=16H、R0=23H、R1=27H.(DPTR = 2000H(A) = 80H(2000H) = 80H(DPTR = 2001(A) = 90H (2001R = 90H(DPTR = 2000H(A) = 80H(B) =80H(DPTR = 2001H(A) = 90H15. MOV DPTR,#2000HMOV A,#80H MOVX DPTR,A INC DPTR MOV A,#90H MOVX DPTR,AMOV DPTR,#2000H MOVX A,DPTR MOV B,A INC DPTR MOVX A,DPTR16. 假设(10H

10、)=0FFH,(11H)=00H,(12H)=0FFH,(13H)=00H,写出每一步指令运行结果MOVA,10H(A)=FFHANLA,#01H(A)=01H _,MOVA,11H(A)=00H,ORLA,#01H(A)=01H _,MOVA,12H(A)=FFH ,XRL A,#01H(A)= FEH ,MOV A,13H(A)= 00H,XRL A,#0AA(A)= AAH ,(10H)= FFH , (11H)= 00H , (12H)= FFH , (13H)= 00H17. 阅读程序并答复以下问题：XDATA50HYDATA51HORG 00HJMP STARTORG 100HST

11、ART: MOV SP,#60HMOV X,#10(503 =10MOV Y,#10 (51H)=10LCALL SUBSUB:NOPINCX(50H)=11X+10=20DEC Y (51H)=9Y=0;MOVA,YA=9JZSUBRETLCALLSUBSUBRET:NOPRET计算X和Y的值是多少？(X) = 20;(Y) = 0 ;18. 设自变量为X,存放在ARE单元,应变量Y存放在BUF单元.给程序标注注释,说明该段子程序的功能并写出该程序的函数关系式.(7分)START MOV DPTR,#AREMOVX A DPTRJZ SULJB ACC.7 , NEG ;注释：如果 A的值大

12、于128跳转NEGMOVA, #02HSUL MOV DPTR,#BUFMOVX DPTRA;注释：把 A的值传送给 BUF地址中RETNEG MOV A #0FEHSJMP SUL如果 X=0,Y=0;如果 X128,Y=FEH;19. 执行以下程序段后,(P1)= 9AH .MOV P1 , #5DHCPL P1.1CPL P1.2CLR P1.6SETB P1.7 20.执行以下程序段后,(A)=8BH,(Cy)=_0_.MOV A , #C5H RL A21.以下程序段执行后, MOV R 0, #7EH MOV 7EH , #0FFH MOV 7FH , #40H 1NC R0 1N

13、C R0 1NC R0(Ro)=_7EH,(7EH)=_41H_,(7FH)=01H R0=7EH(7EH)=FFH(7EH)=40H(7EH)=41HR0=7FH;(7FH)=1;22. (SP)=60H,子程序 SUBTRN勺首地址为 0345H,现执行位于 0123H的ACALLSUBTRN双字节指令后,(PC)=,(61H)=,(62H)=.23. 阅读以下程序,说明其功能.MOV R 0,#dataMOV A,R0RL AMOV R 1, ARL ARL AADD A , RiMOV R0,A RET24. 程序填空1) 数据块传送,将 RAMa 30H开始的连续32个单元的内容传递

14、给片内RA耿60H开始的 连续32个单元.ORG 1000HMOV R7 , MOV R0, #30HMOV R1,#60HLOOP MOV A,R0MOV ,AINC R0INC R1DJNZ R7,SJMP $END2) RAM中40H单元内存有一个十六进制数,把这个数转换为BC咽的十进制数,BCD马的十位和个位放在累加器 A中,百位放在 R2中.ORG 2200HMOV A MOV B #64HDIV ABMOV R2 AMOV A XCH A, B DIV AB SWAP ORL A, BSJMP $END3)编程序将片内40H-46H单元内容的高4位清零,保持低 4位不变.ORG 1

15、000HMOV R7 , #07HMOV R0, LOOP MOV A, R0ANL A , MOV R0 , AINC R0DJNZ R7 , LOOP SJMP $END25.程序如下2506HM5: MOV SP,#58H;2509 HMOV 10H,#0FH;250CHMOV11H,#0BH;250FHACALL XHD;2511HMOV20H,11H2514HM5A: SJMP M5AXHD:PUSH10HPUSHPOPPOP11H10H11HRET问：(1 )执行POP 10H后堆栈内容？(2)执行 M5A: SJMP M5A 后,(SP )=?( 20H )=?26.解读以下程序,然后填写有关存放器内容MOV CJNE JNC MOV SJMPMOV R1,#48H48H,#51HR1,#51H,00HNEXT1A,#0FFHNEXT: MOVNEXT2: SJMPA,#0AAHNEXT2A=()MOV A,#0FBHMOVADDPSW,#10HA,#7FHNEXT2假设P SW= 0 0 ,当执行完上述程序段后,将P SW各位状态填入下表:P S WC YACF 0R S 1R S 0O VF