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1、单片机原理及应用技术苏家建课后习题参考答案第三章MCS-51指令系统3-1MCS-51指令系统有哪几种寻址方式？按功能分类有哪几种指令？3-2设A=0FH,R0=30H,片内RAM的（30H）=0AH,（31H）=0BH,（32H）=0CH,下列程序段运行后的结果？MOVA,@R0;A=0AHMOV@R0,32H;(30H)=0CHMOV32H,A;(32H)=0AHMOVR0,#31H;R0=31HMOVA,@R0;A=(31H)=0BH3-3（1）R0的内容传送到R1MOVA,R0MOVR1,A(2)内部RAM20H单元的内容传送到AMOV
2、A,20H(3)外部RAM30H单元的内容传送到R0MOVR1,#30HMOVXA,@R1MOVR0,A(4)外部RAM30H单元的内容传送到内部RAM20H单元MOVR1,#30HMOVXA,@R1MOV20H,A(5)外部RAM1000H单元的内容传送到内部RAM20H单元MOVDPTR,#1000HMOVXA,@DPTRMOV20H,A(6)程序存储器ROM2000H单元的内容传送到R1MOVDPTR,#2000HCLRAMOVCA,@A+DPTRMOVR1,A(7)RAM2000H单元的内容传送到内部RAM20H单元MOVDPTR,#2
3、000HMOVXA,@DPTRMOV20H,A(8)RAM2000H单元的内容传送到外部RAM30H单元MOVDPTR,#2000HMOVXA,@DPTRMOVR0,#30HMOVX@R0,A(9)RAM2000H单元的内容传送到外部RAM1000H单元MOVDPTR,#2000HMOVXA,@DPTRMOVDPTR,#1000HMOVX@DPTR,A3-4设内部RAM30H=5AH,5AH=40H,40H=00H,P1=7FH,下面程序执行后的结果如何？MOVR0,#30H;R0=30HMOVA,@R0;A=(30H)=5AHMOVR1,A;
4、R1=5AHMOV@R1,P1;(5AH)=7FHMOVA,P1;A=7FHMOV40H,#20H;(40H)=20HMOV30H,40H;(30H)=00H3-5编程，将内部RAM20H-2FH共16个连续单元清零ORG0100HMOVR0,#20HMOVR2,#16LOOP:MOV@R0,#00HINCR0DJNZR2,LOOPEND3-6编程，求出内部RAM20H单元中的数据含“1”的个数，并将结果存入21H单元中。ORG0100HMOV21H,#0MOVR2,#8MOVA,20HCLRCLOOP:RLCAJCCOUNTDJNZR2,LO
5、OPSJMPSTOPCOUNT:INC21HDJNZR2,LOOPSTOP:SJMP$END3-7编程，找出内部RAM30H-50H单元中出现FFH的次数，并将结果存入51H单元中。ORG0100HMOV51H,#0MOVR0,#30HMOVR2,#20HLOOP:MOVA,@R0CJNEA,#0FFH,AGAININC51HAGAIN:INCR0DJNZR2,LOOPSJMP$END3-8编程，计算，并将结果存入内部的30H单元ORG0100HMOV30H,#0MOV31H,#0;i的初值MOVR0,#30HMOVR2,#11MOVA,31H
6、LOOP:INC31HADDA,@R0MOV@R0,AMOVA,31HRLADJNZR2,LOOPSJMP$END3-9已知片内RAM(30H)=38H,(38H)=40H,(40H)=48H,(48H)=90H.说明下列指令源操作数的寻址方式，及指令执行的结果MOVA,40H;A=48H直接寻址MOVR0,A；R0=48H寄存器寻址MOVP1,#0F0H;P1=0F0H直接寻址MOV@R0,30H;(48H)=38H寄存器间接寻址MOVDPTR,#3848H;DPTR=3848H直接寻址MOV40H,38H;(40H)=(38H)=40H直接
7、寻址MOVR0,30H;R0=(30H)=38H寄存器寻址MOVP0,R0;P0=38H直接寻址MOV18H,#30H;(18H)=30H直接寻址MOVA,@R0;A=(38H)=40H直接寻址MOVP2,P1;P2=0F0H直接寻址3-10设R0=32H,A=48H,片内RAM(32H)=80H,(40H)=08H.求指令运行的结果MOVA,@R0;A=(32H)=80HMOV@R0,40H;(32H)=（40H）=08HMOV40H,A;(40H)=80HMOVR0,#35H;R0=35H3-11设A=83H,R0=17H,片内RAM(17
8、H)=34H,求指令运行的结果ANLA,#17H;A=03HORLA,17H;(17H)=（37HXRLA,@R0;A=37HCPLA;A=C8H3-
