微机原理与接口技术第3章习题课件.ppt

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1、3-3写出执行下列程序段的中间结果和结果1.MOVAX，0809HMULAH；AX=AAM；AX=0048H0702H2.MOVAX，0809HMOVDL，5AAD；AX=DIVDL；AX=MOVDL，AHAAM；AX=DL=0059H0411H0107H04H3.MOVAX，0809HADDAL，AHMOVAH，0；AX=AAA；AX=0011H0107H4.MOVAX，0809HMOVDL，10XCHGAH，DLMULAH；AX=AAM；AX=ADDAL，DL；AX=005AH0900H0908H5.MOVAL,98HMOVAH,ALMOVCL,4SHRAH,CLANDAL,0F

2、HAAD；AL=H626.MOVCL,248XORAX，AXMOVCH,8AG：SHLCL，1ADCAL，ALDAAADCAH，AHDECCHJNZAG；AX=H02483-4编写程序，将字节变量BVAR中的压缩BCD数转换为二进制数，并存入原变量中。BVARDB99HMOVAL,BVARMOVCL,4SHRAL,CLMOVAH,10MULAHANDBVAR,0FHADDBVAR,AL本题还可以先将2位压缩BCD数转换为非压缩BCD数，然后用AAD指令将非压缩BCD数转换为二进制数。实现转换的程序段如下：MOVAL,BVARMOVAH,ALMOVCL,4SHRAH,CLANDAL,0

3、FHAADMOVBVAR,AL3-6编写求两个4位非压缩BCD数之和，将和送显示器显示的程序。D1DD06070809HD2DD04050607HOBFDB6DUP(0)MOVSI，0MOVDI,OFFSETOBF+5MOVBYTEPTR[DI],$MOVCX,4XORAH,AHAGAIN:MOVAL,BYTEPTRD1[SI]ADDAL,BYTEPTRD2[SI]ADDAL,AHMOVAH，0AAAADDAL,30HDECDIMOV[DI],ALINCSILOOPAGAINANDAH,AHJZNWW；无万位，转移DECDIMOVBYTEPTR[DI],1NWW:MOVDX,

4、DIMOVAH,9INT21H3-7编写求两个4位压缩BCD数之和，将和送显示器显示的程序。W1DW8931HW2DW5678HOBFDB6DUP(0)MOVAL,BYTEPTRW1ADDAL,BYTEPTRW2DAAMOVAH,ALMOVAL,BYTEPTRW1+1ADCAL,BYTEPTRW2+1DAAPUSHFMOVDIOFFSETOBF+5MOVBYTEPTR[DI],$MOVCL,4XCHGAH,ALAGAIN:MOVDL,0FHANDDL,ALADDDL,30HDECDIMOV[DI],DLSHRAX,CLANDAX,AXJNZAGAINPOPFJNCOPUTDECD

5、IMOVBYTEPTR[DI],1OPUT:MOVDX,DIMOVAH,9INT21H3-8编写程序，将字节变量BVAR中的二进制数（0~FFH）转换为BCD数，在屏幕上以十进制数形式显示结果。BVARDB99HOBUFDB4DUP(0)MOVAL,BVARMOVAH,0MOVDL,100DIVDLADDAL,30HMOVOBUF,ALMOVAL，AHAAMADDAX，3030HMOVOBUF+1，AHMOVOBUF+2，ALMOVOBUF+3，$MOVDX，OFFSETOBUFMOVAH，9INT21H3-30源程序如下，阅读后做如下试题：1.在分号后给指令或（向下）给程序

6、段加上注释（实质是做什么？例如，第1个注释若注为将2送BX，则视为非实质注释，不给分）。2.列举实例，说明该程序的功能（键入什么？显示什么？）。3.画出实例的数据存储图。IBUFDB255,0,255DUP(0)ABCDDB0AH,ABCD:DB255DUP(0)MNOPDB0AH,0DH,MNOP:DB255DUP(0)MOVDX,OFFSETIBUFMOVAH,10INT21HMOVBX,2；MOVSI,OFFSETABCD+7MOVDI,OFFSETMNOPQ+8AG:CMPIBUF[BX-1],0DH；JEEXITCMPIBUF[BX],'-'；JNEP1CALLMP

7、JMPAGP1:XCHGSI,DI；CALLMPXCHGSI,DIJMPAGEXIT:MOVBYTEPTR[SI-1],'$'MOVBYTEPTR[DI-1],'$'MOVAH,9MOVDX,OFFSETABCDINT21HMOVDX,OFFSETMNOPQINT21HMPPROC；MOVAL,IBUF[BX]MOV[DI],ALINCDIINCBXCMPIBUF[BX-1],0DHJEBACKCMPIBUF[BX-1],','JNEMPBACK:RETM