微机原理整理后

2012微机原理14道题目1. 在XXX单元中存放一个数x(0WxW15)，请用查表法的方法计算x的平方，并将结果保存到YYY 单元中。课本197页2. 已知数组由100个字组成，存放在数据段中以偏移地址为ARY开始的存中，试编写程序求出这个 数组元数之和，结果存放在的数据段中以偏移地址为SUM开始的存中。课本184页3. 试编写一程序，将数据段中偏移地址为BUF1开始的100个数据传送到偏移地址为BUF2开始的连续存单元中去。STACK SEGMENTSTACKSTACKDW100 DUP(?)STACK ENDSDATA SEGMENTBUF1 DB100 DUP(?)BUF2 DB100 DUP(?)SRCADR DW?DSTADR DW?LEN DW?DATA ENDSCODE SEGMENTASSUMECS:CODE,DS:DATA,SS:STACK,ES:DATAMAIN PROCSTART: MOVAX,DATAMOVDS,AXMOVES,AXLEAAX,BUF1MOVSRCADR,AXMOVLEN,100CALLMVDATMOVAX,4CHINT21HMAIN ENDPMVDAT PROCMOVSI,SRCADRMOVDI,DSTADRMOVCX,LENCLDCMPSI,DIJADONESTDADDSI,CXDECSIADDDI,CXDECDIDONE:REPMOVSBRETMVDATENDPCODEENDSENDSTART4. 编写一个程序，实现SUM=a1+a2+a20。已知ala20依次存放在以BUF为首址的数据区， 每个数据占两个字节，SUM也是两个字节。DATASEGMENTDATDW2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2SUMDW?DATAENDSCODESEGMENTMAINPROCFARASSUMECS:CODE,DS:DATASTART:MOVAX,DATAMOVDS,AXMOVAX,0MOVSUM,AXMOVSI,OFFSET DATMOVCX,20LP1:ADDAX,SIINCSIINCSILOOPLP1MOVSUM,AXMOVAH,4CHINT21HMAINENDPCODEENDSENDSTART5. 试编写一程序，将AX中的各位去反，然后统计出AX中“0”的个数，将结果存到CL中。CODESEGMENTMAINPROCFARASSUMECS:CODESTART:MOVAX,0E001HMOVDL,0MOVCL,16NOTAXRETEST:ANDAX,AXJSSKIPINCDLSKIP:SHLAX,1LOOPRETESTMOVCL,DLEXIT:MOVAH,4CHINT21HMAINENDPCODEENDSENDSTART6. 在存数据段存储器中，从ADDR单元开始存有一个字符串，它以“$”符号作为结束标志，试编写 程序统计该字符串的长度（$字符不计入长度），并把长度值存入LETH字单元。DATASEGMENTADDRDBLIUBIN$LETHDW?DATAENDSSTACKSEGMENTPARA STACKSTACKDB100 DUP(?)STACKENDSCODESEGMENTASSUMECS:CODE,DS:DATA,SS:STACKSTART:MOVAX,DATAMOVDS,AXMOVBX,OFFSET ADDRMOVCX,0LP:MOVAL,BXCMPAL,$JEDONEINCCXINCBXJMPLPDONE:MOVADDR,CXMOVAH,4CHINT21HCODEENDSENDSTART7. 试编写一程序，要现将ASCII码表示的两位十进制数转换为一字节二进制数，其中高地址单元存 放十位数。课本200页8. 编程在显示器上输出你自己的汉语拼音的子程序。课本155页.9将数据0-63置入到存中以BUF为首址的连续64个字节单元中。课本198页10. 试编写一程序，把数组STRING中存放的20个8位二进制数分成正数数组和负数数 组，并统计正数、负数和零的个数，结果分别存放到P、M、Z三个单元。DATASEGMENTSTRINGDW2 DUP(3045H,0FD34H,0D3DH,9899H,0,3DF2H,0,0FFDEH,93FDH,0DE6CH)PDW20 DUP(?)MDW20 DUP(?)ZDW20 DUP(?)DATAENDSCODESEGMENTASSUMECS:CODE,DS:DATASTART:MOVAX,DATAMOVDS,AXXORBP,BPLEASI ,P;正数LEADI ,M；负数LEABP ,Z；零PUSHSIPUSHDIPUSHBPLEABX,STRINGMOVCX,20LL:MOVAX,BXPUSHAXADDAX,AXJZL1POPAXPUSHAXSALAX,1JCL2POPAXMOVADDJMPLA:ADDLOOPXORMOVPOPMOVSUBSHRMOVPOPMOVSUBSHRMOVPOPMOVSUBSHRMOVMOVINTL1:POPMOVADDJMPL2:POPMOVADDJMPSI ,AXSI,2LABX ,2LLCX,CXCX,2AXDX,BPDX,AXDX,CL BP ,DXAXDX,DIDX,AXDX,CL DI ,DXAXDX,SIDX,AXDX,CL SI ,DXAH,4CH21HAXBP,AXBP ,2LAAXDI ,AXDI,2LACOD ENDSEND START11、试编写一程序，完成10个一位十进制数累加，累加结果以分离式BCD码形式存放 于AH（高位）、AL（低位）寄存器中。DATASSEGMENTTABDB41H,42H,43H,44H,45H,46HBINDB1101101110011110BUFDB4 DUP(?)DATASENDSCODESSEGMENTASSUMECS:CODES,DS:DATASSTART:MOVAX,DATASMOVDS,AXLEABX,TABLEASI,BINLEADI,BUFMOVCX,4LP1:MOVAX,SI+2PUSHCXXORCX,CXMOVCL,8RORAX,CLSUBAX,3030HMOVDX,SIRORDX,CLSUBDX,3030HSHLAH,1ADDAH,ALMOVCL,2SHLDL,CLMOVCL,3SHLDH,CLADCDH,DLADCAH,DHCMPAH,0AHJBLP2SUBAH,0AHMOVAL,AHXLATLP3:MOV DI,ALADDDI,1ADDSI,4POPCXLOOPLP1MOVAH,4CHINT21HLP2:ADDAH,30HMOVAL,AHJMPLP3CODES ENDSENDSTART12、试编写一程序，将2个字节二进制数，变换成用ASCII码表示的四位十六进制数(用 四字节表示)。DATASSEGMENTTABDB41H,42H,43H,44H,45H,46HBINDB1101101110011110BUFDB4 DUP(?)DATASENDSCODESSEGMENTASSUMECS:CODES,DS:DATASSTART:MOVAX,DATASMOVDS,AXLEABX,TABLEASI,BINLEAMOVDI,BUFCX,4LP1:MOVAX,SI+2PUSHCXXORCX,CXMOVCL,8RORAX,CLSUBAX,3030HMOVDX,SIRORDX,CLSUBDX,3030HSHLAH,1ADDAH,ALMOVCL,2SHLDL,CLMOVCL,3SHLDH,CLADCDH,DLADCAH,DHCMPAH,0AHJBLP2SUBAH,0AHMOVAL,AHXLATLP3:MOVDI,ALADDDI,1ADDSI,4POPCXLOOPLP1MOVAH,4CHINT21HLP2:ADDAH,30HMOVAL,AHJMPLP3CODES ENDSENDSTART13,试用串操作指令SCAS，在10个字节的数据块BLOCK中，搜索与2EH相等的数，若 找到，则将该数地址存放于ADR中，并在SIGNAL单元中做标记0FFH ；否则，SIGNAL单元中为00H。DATASEGMENTBLOCKDBN DUP(0)SINGALDB00HADRDB00HDATAENDSCODESEGMENTASSUMEDS:DATA,CS:CODE,ES:DATASTART:MOVDI OFFSET BLOCKCLDMOVCX,10MOVSL,IEHREPNISCASBJNZAAAMOVSIGNAL,0FFHMOVADR,IEHAAA:MOVSIGNAL,00HCODEENDSENDSTART14.编写一程序，计算100个16位正整数之和，如果和不超过16位值的围，则保存其 和到SUM，如果超过则显示“ Overflow”。DATSEGMENTDAT1DW25 DUP (2D4EH,5611H,1234H,7891H)BUFDW?STR1DBoverflow!,0DH,0AH,$DATENDSCODSEGMENTASSUMECS:COD,DS:DATSTART:MOVAX,DATMOVDS,AXMOVSI,OFFSET DAT1MOVBX,OFFSET BUFMOVCX,100XORAX,AXL1:ADDAX,SIJCDISPLOOPL1MOVBX,AXMOVAH,4CHINT21HDISP PROCNEARMOVDX,OFFSET STR1MOVAH,9INT21HMOVAH,4CHINT21HDISP ENDPCOD ENDSENDSTART