汇编语言编程题目资料答案.doc

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1、5.1编写程序，从键盘接收一个小写字母，然后找出它的前导字符和后续字符，再按顺序用大写字母显示这三个字符。【解】：MAIN PROC FARstart:push dssub ax,axpush axinput:mov ah,7Hint 21Hcmpe:cmp al,20HJE exitcmp al,61HJL inputcmp al,7AHJG inputprint:sub al,01Hmov dl,almov ah,02Hint 21Hadd al,01Hmov dl,almov ah,02Hint 21Hadd al,01Hmov dl,almov ah,02Hint 21H;输出换行回车

2、mov dl,0AHmov ah,02Hint 21Hjmp inputexit :retMAIN ENDPEND start5.2将AX寄存器中的16位数分成4组，每组4位，然后把这4组数分别放在AL、BL、CL和DL中。【解】：DB4dup(?).stack100H.codeMAINPROCFARstart:pushdsandax,00Hpushaxmovax,1234Hmovcx,04Hmovsi,00Hlop:pushcxmovcl,04Hrolax,clpushaxandal,0FHmovBYTEPTRxsi,alpopaxpopcxaddsi,2Looplopmoval,Xmovb

3、l,X+2movcl,X+4movdl,X+6print:movah,02Hint21Hmovdl,almovah,02Hint21Hmovdl,blmovah,02Hint21Hmovdl,clmovah,02Hint21Hexit:retMAINENDPENDstart5.3试编写一程序，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NO MATCH。【解】：datasegmentstring1db111111111,13,10,$string2db222222222,13,10,$inf1dbmatch!,13,10,$inf2d

4、bnomatch!,13,10,$dataendscodesegmentmainprocfarassumeds:data,cs:code,es:codestart:pushdsmovax,0pushaxmovax,datamovds,axmoves,axleasi,string1leadi,string2movcx,9repzcmpsbjzmatchleadx,inf2jmpdispmatch:leadx,inf1disp:movah,09int21hretmainendpcodeendsendstart5.4试编写一程序，要求能从键盘接收一个个位数N，然后响铃N次(响铃的ASCII码为07H

5、)。【解】：mainprocfarstart:pushdssubax,axpushaxinput:movah,01Hint21Hcmpal,20Hjeexitcmpal,30Hjleinputcmpal,3AHjgeinputsubal,30Hmovcx,axandcx,0FFHlop:movdl,07Hmovah,2Hint21Hlooplopjmpinputexit:retmainendpendstart5.5编写程序，将一个包含有20个数据的数组M分成两个数组：正数数组P和负数数组N，并分别把这两个数组中数据的个数显示出来。【解】：DATA SEGMENT BUF DB -32, 25,

6、 36, -18, -64, 0, -3 COUNT EQU $ - BUF PLUS DB ? ; 存放正数 MINUS DB ? ; 存放负数 DATA ENDS CODE SEGMENT ASSUME CS:CODE,DS:DATA START: MOV AX, DATA MOV DS, AX MOV BL, 0 ; 负数个数 MOV DL, 0 ; 正数个数 MOV SI, OFFSET BUF ; 首地址 MOV CX, COUNT ; 循环次数 LOP1: MOV AL, SI ; 取第一个数 CMP AL, 0 ; 和0比较 JGE NEXT0 ; 大于等于0，转 INC BL

7、; 小于0， BL加1 JMP NEXT1 ; 转 NEXT1 NEXT0: INC DL ;大于等于0， DL加1 NEXT1: INC SI ; 指针增1 DEC CX ; 循环次数减1 JNZ LOP1 ; CX不等于0，转回 MOV MINUS, BL ; 存储负数个数 MOV PLUS, DL ; 存储正数个数 CODE ENDS END START 5.6试编写一汇编语言程序，要求从键盘接收一个4位的十六进制数，并在终端上显示与它等值的二进制数。【解】：stacks segment stacks db 20 dup(0)stacks endsdata segmentn equ 4c

8、har db n+1 ;输入内存字符db 0db n+1 dup(0)input db Input the number:,0dh,0ah,$enter db 0dh,0ah,$output db Result is:,0dh,0ah,$data endscode segmentassume cs:code,ss:stacks,ds:datastart: mov ax,datamov ds,axlea dx,inputmov ah,9int 21hlea dx, charmov ah,0ahint 21hlea dx,entermov ah,9int 21hmov si,2mov cx,4l0

9、: mov bl,charsicmp bl,30hjb startcmp bl,39hjbe l00cmp bl,61hjb startcmp bl,66hja startsub bl,57hjmp l333l00: sub bl,30hl333:mov charsi,blinc sidec cxjnz l0lea dx,outputmov ah,9int 21hmov si,2l000:mov bl,charsimov ch,4mov cl,4shl bl,clll00:test bl,80hjnz l1mov dl,0mov ah,2int 21hjmp l2l1: mov dl,1mov

10、 ah,2int 21hl2: shl bl,1dec chjnz ll00inc sicmp si,6jne l000mov ah,4chint 21hcode endsend start5.7设有一段英文，其字符变量名为ENG，并以$字符结束。试编写一程序，查对单词SUN在该文中的出现次数，并以格式“SUNXXXX”显示出次数。【解】：mov Z,0lea esi,ENGlp1:lodsblp2:cmp al,$jztoQuitcmp al,Sjnz lp1lodsbcmp al,Ujnz lp2lodsbcmp al,Njnz lp2inc Zjmp lp1toQuit:5.8有一个首地

11、址为MEM的100D字数组，试编制程序删除数组中所有为零的项，并将后续项向前压缩，最后将数组的剩余部分补上零。【解】：stack100H.datamemdw12,0,0,0,0,0,1,2,3,6,4,7,8,2,1,0,0,54,5,0,2,4,7,8,0,5,6,2,1,4,8,5,1,45,7,5,1,2,0,2,4,0,2,54,0,12,0,0,0,0,0,1,2,3,6,4,7,8,2,1,0,0,54,5,0,2,4,7,8,0,5,6,2,1,4,8,5,1,45,7,5,1,2,0,2,4,0,2,54,0,45,7,5,1,2,0,2,4,0,2.codeMAINPROCF

12、ARstart:pushdsandax,0pushaxmovax,datamovds,ax;-movax,0Hmovbx,64Hmovcx,64Hmovsi,0FFFEHrepeat:ADDsi,2HcmpMEMsi,0HJEcallslooprepeatcalls:INCaxcallsortcmpax,1HJElastValueDECcxjmprepeatexit:retlastValue:movmembx,0HDECcxjmprepeatMAINENDPsortPROCNEARpushcxpushsisubsi,2Hs:addsi,2Hmovdx,memsimovmemsi+2,dxloo

13、psreturn:popsipopcxretsortENDPENDstart5.9在STRING到STRING+99单元中存放着一个字符串，试编制一程序测试该字符串中是否存在数字。如有，则把CL的第5位置1，否则将该位置0。【解】：XOR EBX,EBXMOV ESI,OFFSETSTRING MOV ECX,99START_LOOP: LODSB CMP AL,0 JC LOOP_NEXT CMP AL,9 JNC LOOP_NEXT OR BL,10HLOOP_NEXT: LOOP START_LOOP MOV CL,BL5.10在首地址为TABLE的数组中按递增次序存放着100H个16位

14、补码数。试编写一个程序把出现次数最多的数及其出现次数分别存放于AX和CX中。【解】：datasegment Table dw 0,1,2,3,4,5,6,7,8,9,0,1,2,3,4,5,6,7,8,9 dw 0,1,2,3,4,5,6,7,8,9,0,1,2,3,4,5,6,7,8,9 dw 0,1,2,3,4,5,6,7,8,9,0,1,2,3,4,5,6,7,8,9 dw 0,1,2,3,4,5,6,7,8,9,0,1,2,3,4,5,6,7,8,9 dw 0,1,2,3,4,5,6,7,8,9,0,1,2,3,4,5,6,7,8,9 dw 0,1,2,3,4,5,6,7,8,9,0,

15、1,2,3,4,5,6,7,8,9 dw 0,1,2,3,4,5,6,7,8,9,0,1,2,3,4,5,6,7,8,9 dw 0,1,2,3,4,5,6,7,8,9,0,1,2,3,4,5,6,7,8,9 dw 0,1,2,3,4,5,6,7,8,9,0,1,2,3,4,5,6,7,8,9 dw 5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,-1,-1 Result dw 0,0,0,0 data ends code segment assume cs:code,ds:data start: mov ax, data mov ds, ax ;- mov si, of

16、fset Table mov ax, si ; ax存放最大数 mov bx, si ; bx存放最小数 mov cx, 199 ; 循环199次 C001: inc si inc si mov dx, si cmp dx, ax jg C002 cmp dx, bx jl C003 jmp C004 C002: mov ax, dx jmp C004 C003: mov bx, dx C004: loop C001 ; 保存结果 mov Result, ax mov Result+2, bx ;- mov si, offset Table mov cx, 200 C005: mov ax,

17、si xor dx, dx ;- push cx mov di, offset Table mov cx, 200 C006: mov bx, di cmp ax, bx jnz C007 inc dx C007: inc di inc di loop C006 pop cx ;- cmp dx, Result+6 jle C008 mov Result+4, ax mov Result+6, dx C008: inc si inc si loop C005 ;- mov ax,4c00h int 21h code ends end start ; 最后结果 Result 9, -1, 5,

18、365.11试编制一个程序，把AX中的十六进制数转换为ASCII码，并将对应的ASCII码依次存放到MEM数组中的4个字节中。例如，当(AX)2A49H时，程序执行完后，MEM中的4个字节内容分别为39H，34H，41H和32H。【解】：datasegmentxdbY,$ydbN,$dataendscodesegmentassumecs:code,ds,datastart:movax,datamovds,axmovah,01int21hcmpal,ajaabcabc:cmpal,zjbabc1retabc1:movdl,offsetxmovah,9int21hretcmpal,Ajaabab:

19、cmpal,Zjbab1retab1:movdl,offsetxmovah,9int21hretcmpal,zjaaccmpal,Ajbacac:movdl,offsetymovah,9int21hretmovah,4chint21hcodeendsendstart5.12已知数组A包含15个互不相等的整数，数组B包含20个互不相等的整数。试编制一程序，把既在A中又在B中出现的整数存放于数组C中。【解】：DATASEGMENTADB 1,3,4,6,10,12,15,21,33,2,40,42,46,48,55BDB 2,3,6,21,33,41,55,88,8,66,21,10,22,23,

20、56,48,50,51,89,100ZDB 15 DUP (?)DATAENDSCODESEGMENTASSUME DS:DATA,CS:CODESTART:MOVAX,DATAMOVDS,AXLEASI,ALEABX,ZMOVCX,15CLDLOP1:LODSBPUSHCX(保存外环CX)MOVCX,20（内环次数）LEADI,B(每次内环完，重新赋值)LOP2:CMPAL,DIJELOP3（相等，退出内环）INCDILOOPLOP2JMPLOP4（内环完毕而没有相等的就直接到外环）LOP3:MOVBX,ALINCBXLOP4:POPCX（外环出栈）LOOPLOP1MOVAH,4CHINT2

21、1HCODEENDSENDSTART5.13从键盘输入一系列字符(以回车符结束)，并按字母、数字及其他字符分类计数，最后显示出这三类的计数结果。【解】：datarea segment count dw 0,0,0 ;分别存放字母、字母和其他字符的个数 datarea ends code segment main proc far assume cs:code,ds:datarea start:push ds sub ax,ax push ax mov ax,datarea mov dx,ax next: mov ah,01 ;判断是否为字母 int 21h cmp al,0dh jz exit

22、 cmp al,5ah jg other cmp al,41h jl shuzi add count0,1 shuzi:cmp al,39h ;判断是否为数字 jg other cmp al,31h jl other add count1,1 other:add count2,1 ;其他字符的个数加1 jmp next mov si,0 exit: mov bx,countsi ;打印输出count的内容，为16进制 mov ch,4 rotate: mov cl,4 rol bx,cl mov al,bl and al,0fh add al,30h cmp al,3ah jl printit add al,7h printit: mov dl,al mov ah,2 int 21h dec ch jnz rotate add si,1 cmp si,2 jnz exit ret main endp code ends end start