汇编程序参考

程序1：试编写一个程序，要求比较两个字符串string1和string2所含字符是否相同，若相同则显示“MATCH”，若不相同则显示”NOMATCH”。datasegmentstr1db "hello$"mess1db"match$"mess2db"nomatch$"dataendsextrasegmentstr2db"heloo$"extraendscodesegmentassumecs:code,ds:data,es:extrastart:movax,datamovds,axmovax,extramoves,axleasi,str1leadi,str2movcx,6repecmpsbjematchleadx,mess2jmpexitmatch:leadx,mess1exit:movah,9int21hmovah,4chint21hcodeendsendstart程序2：从键盘接受字符串，以$符结束，并将该字符串存入附加段中的str中。extrasegmentstrdb10dup(?)extraendscodesegmentassumecs:code,es:extrastart:pushdssubax,axpushaxmovax,extramoves,axleasi,strbegin:movah,1int21hmoves:[si],alincsicmpal,'$'jnebeginmovah,4chint21hcodeendsendstart程序3：在屏幕上输出：*************************CODESEGMENT ASSUMECS:CODESTART: movbx,9rotate: movdl,'*' movcx,bxprint: movah,02h int21h loopprint movdl,0ah movah,02h int21h movdl,0dh movah,02h int21h subbx,2 jnsrotate MOVAH,4CH INT21HCODEENDS ENDSTART 程序4：将BX的内容以十六进制的形式显示在屏幕上。CODESEGMENT ASSUMECS:CODESTART: MOVBX,6AC4H MOVCH,4ROTATE: MOVCL,4 ROLBX,CL MOVAL,BL ANDAL,0FH ADDAL,30H CMPAL,3AH JLPRINTIT ADDAL,7 PRINTIT:MOVDL,AL MOVAH,2 INT21H DECCH JNZROTATE MOVAH,4CH INT21HCODEENDS ENDSTART 程序5：首地址为BUF的内存单元存有10个字节数，统计其中单元内容为0的单元个数，并将结果显示出来。DATASSEGMENTBUF DB 12,0,11,0,51,56,0,0,44,-74DATASENDSCODESSEGMENT ASSUMECS:CODES,DS:DATASSTART: MOVAX,DATAS MOVDS,AX MOVCX,10 MOVSI,0 MOVDL,0ROTATE:CMPBUF[SI],0 JNENEXT INCDLNEXT: INCSI LOOPROTATE ADDDL,30H MOVAH,02H INT21H MOVAH,4CH INT21HCODESENDSENDSTART程序6：统计一个有10人的班中，男、女生的人数各为多少。将统计结果的男生人数存入变量MAN中，女生人数存入变量WOMAN中。数据段定义如下：DATASSEGMENTBUF DB 1,1,0,0,1,0,1,1,0,0;男生为1，女生为0mandb ?woman db ?DATASENDS源代码：DATASSEGMENTBUF DB 1,1,0,0,1,0,1,1,0,0;男生为1，女生为0mandb ?woman db ?DATASENDSCODESSEGMENT ASSUMECS:CODES,DS:DATASSTART: MOVAX,DATAS MOVDS,AX MOVCX,10 MOVSI,0 MOVDL,0;存放男生数量 MOVBL,0;存放女生数量ROTATE:CMPBUF[SI],0 JNENEXT INCDL JMPNEXT2NEXT: INCBLNEXT2: INCSI LOOPROTATE MOVMAN,DL MOVWOMAN,BL MOVAH,4CH INT21HCODESENDSENDSTART程序7：编程序段计算SUM=a1+a2+……+a20，已知a1……a20依次存放在BUF为首地址的数据区，每个数据占两个字节。和数SUM也为2个字节。DATASSEGMENTBUF Dw 20dup(3)sumdw ?DATASENDSCODESSEGMENT ASSUMECS:CODES,DS:DATASSTART: MOVAX,DATAS MOVDS,AX MOVCX,20 LEASI,BUF movax,0;存放和数ROTATE:ADDAX,[SI] ADDSI,2 LOOPROTATE MOVSUM,AX MOVAH,4CH INT21HCODESENDSENDSTART程序8：在BUF1和BUF2两个数据区中，各定义有10个带符号字数据，试编写一完整程序，求它们对应项的绝对值之和，并将和数存入以SUM为首地址的数据区中。DATASSEGMENT BUF1 DW -56,24,54,-1,89,-8,43,23,-65,4 BUF2 DW 45,-23,124,345,-265,-6,44,23,-21,43 SUM DW 10DUP(0)DATASENDSCODESSEGMENT ASSUMECS:CODES,DS:DATASSTART: MOVAX,DATAS MOVDS,AX MOVCX,10 MOVSI,0ROTATE:MOVAX,BUF1[SI] CMPAX,0 JNSNEXT1 NEGAXNEXT1: MOVBX,BUF2[SI] CMPBX,0 JNSNEXT2 NEGBXNEXT2: ADDAX,BX MOVSUM[SI],AX ADDSI,2 LOOPROTATE MOVAH,4CH INT21HCODESENDSENDSTART字符串传送DATASEGMENTSTR1DB'HELLO'DATAENDS

EXTRASEGMENTSTR2DB5DUP(?)EXTRAENDS

CODESEGMENTASSUMECS:CODE,DS:DATA,ES:EXTRASTART:MOVAX,DATAMOVDS,AX

MOVAX,EXTRAMOVES,AX

LEASI,STR1LEADI,STR2CLDMOVCX,5REPMOVSB

MOVAH,4CHINT21HCODEENDSENDSTART编程序求W=（V-（X*Y+Z-540））/XDATASSEGMENTxdd034a0020hydd3hzdd4hwdd?DATASENDSCODESSEGMENTASSUMECS:CODES,DS:DATASSTART:MOVAX,DATASMOVDS,AX;此处输入代码段代码movax,wordptrxmovdx,wordptrx+2