微机原理编程题

1、微机原理编程题分支程序设计 .编写一段程序，已知BUF1单元中有一带符号字节数据X,BUF2中有一带符号字节数据Y,根据以下函数关系编写程序求Z的值，结果存入RESULT单元。Z=|X-Y|；DATASEGMENTBUF1DB05HBUF2DB02HRESULTDB?DATAENDSCODESEGMENTASSUMECS:CODE,DS:DATASTART:MOVAX,DATAMOVDS,AXMOVAL,BUF1MOVBL,BUF2SUBAL,BLJNCGREATNEGALGREAT:MOVRESULT,ALEXIT:MOVAH,4CHINT21HCODEENDSENDSTART .编写一段程

2、序，已知BUF单元中有一无符号字节数据据X,假设为8,根以下函数关系编写程序求Y的值，结果存入RESULT单元。Z=5XX<10,X-5X>=10;DATASEGMENTBUFDB8RESULTDB?DATAENDSCODESEGMENTASSUMECS:CODE,DS:DATASTART:MOVAX,DATAMOVDS,AXMOVAL,BUFCMPAL,10JAEGREMOVBL,ALADDAL,ALADDAL,ALADDAL,BLJMPDONEGRE:SUBAL,5DONE:MOVRESULT,ALMOVAH,4CHINT21HCODEENDSENDSTART在内存单元BUF中

3、存放一个带符号字节数据X,假定为-2,试根据以下函数关系编写程序求丫的值，结果存入RESULT单元。Y=1X>0,0X=0,-1X<0;DATASEGMENTBUFDB-2RESULTDB?DATAENDSCODESEGMENTASSUMECS:CODE,DS:DATASTART:MOVAX,DATAMOVDS,AXMOVAL,BUFCMPAL,0JGELIMOVAL,-1JMPL3L1:JZL2MOVAL,1JMPL3L2:MOVAL,0L3:MOVRESULT,ALMOVAH,4CHINT21HCODEENDSENDSTARTN个数中求最大值、最小值假定N=10,已知原始数据存

4、放在BUF开始的内存单元中，将结果存入MAX,MIN内存单元中。DATASEGMENTBUFDB3,5,7,8,5,3,9,7,13,1COUNTEQU$-BUFMAXDB?MINDB?DATAENDSCODESEGMENTASSUMECS:CODE,DS:DATASTART:MOVAX,DATAMOVDS,AXMOVCX,COUNT-1MOVAL,BUFMOVMAX,ALMOVMIN,ALLEASI,BUFAGAIN:MOVAL,SI+1CMPAL,MINJGESIMOVMIN,AL51: CMPAL,MAXJLES2MOVMAX,AL52: INCSILOOPAGAINMOVAH,4CHI

5、NT21HCODEENDSENDSTART编写程序求1+2+3+N<10时最大的N值，将N值送NUM单元中，同时将1+2+3+N勺和送到SUM单元中。DATASEGMENTNUMDBSUMDB?DATAENDSCODESEGMENTASSUMECS:CODE,DS:DATASTART:MOVAX,DATAMOVDS,AXMOVAL,0MOVBL,0AGAIN:INCBLADDAL,BLCMPAL,100JBAGAINSUBAL,BLDECBLMOVNUM,BLMOVSUM,ALMOVAH,4CHINT21HCODEENDSENDSTART统计字符，正数，负数，零的个数。9/11COUNT

6、单元.若自STRING单元开始存放一个字符串，以飞结尾，试编写程序,统计字符串的长度，并将结果存入DATASEGMENTSTRINGDB'abcdefg$，COUNTDW?DATAENDSCODESEGMENTASSUMECS:CODE,DS:DATASTART:MOVAX,DATAMOVDS,AXLEASI,STRINGMOVCX,0NEXT:MOVAL,SICMPAL,卬JZDONEINCCXINCSIJMPNEXTDONE:MOVCOUNT,CXMOVAH,4CHINT21HCODEENDSENDSTART.自BUF单元开始存放10个带符号字节数据，将其中的正数、负数和零选出来，

7、并统计其个数，分别存入PLUSMINUS和ZERO单元中。DATASEGMENTBUFDB5,-2,0,-1,4,0,-4,8,0,-7PLUSDB?MINUSDB?ZERODB?DATAENDSCODESEGMENTASSUMECS:DATA,DS:DATASTART:MOVAX,DATAMOVDS,AXLEASI,BUFMOVCX,10AGAIN:MOVAL,BXINCBXANDAL,ALJSMlJZM2INCPLUSn /JMPNEXTMl:INCMINUSJMPNEXTM2:INCZEROJMPNEXTNEXT:LOOPAGAINMOVAH,4CHINT21HCODEENDSENDST

8、ART求压缩BCD码的和。已知从BUF1和BUF2单元开始，存放两个各为字节10个的BCD数，求这两个数的和，并将结果存入BUF3单元中。DATASEGMENTBUF1DB01H,01H,01H,01H,01H,01H,01H,01H,01H,01HBUF2DB03H,03H,03H,03H,03H,03H,03H,03H,03H,03HBUF3DB11DUP(7)DATAENDSCODESEGMENTASSUMECS:CODE,DS:DATASTART:MOVAX,DATAMOVDS,AXMOVCX,10LEASI,BUF1LEADI,BUF2LEABX,BUF3AGAIN:MOVAL,SIADCAL,DIDAAMOVBX,ALINCSIINCDIINCBXLOOPAGAINADCAH,0MOVBX,AHMOVAH,4CHINT21HCODEENDSENDSTART6.8255编程题：（课后实验）8255工作于方式0,此时PA、PBPC均为可独立输入/输出的并行口。8255的各寄存器对应的口地址为：PA：200H,PB：202H,PC:204H,口设置为输控制:206Ho要求8255工作于方式0,PA口设置为输入,出。DATASEGMENTC8255EQU206HP8255AEQU200HP8255CEQU204HDATAENDSCODESEGMENTASS