汇编语言编程题目资料答案

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mov dl,0AHmov ah,02Hint 21Hj mp 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】：DB 4 dup(?).stack 100H.codeMAIN PROC FARstart:push dsand ax,00Hpush axmov ax,1234Hmov cx,04Hmov si,00Hlop:push cxmov cl,04Hrol ax,clpush axand al,0FHmov BYTE PTR x[si],alpop axpop cxadd si,2Loop lopmov al,[X]mov bl,[X+2]mov cl,[X+4]mov dl,[X+6]print:mov ah,02Hint 21Hmov dl,almov ah,02Hint 21Hmov dl,blmov ah,02Hint 21Hmov dl,clmov ah,02Hint 21Hexit:retMAIN ENDPEND start5.3 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】：data segmentstring1 db "111111111",13,10,"$"string2 db "222222222",13,10,"$"inf1 db "match!",13,10,"$"inf2 db "no match!",13,10,"$"data endscode segmentmain proc farassume ds:data,cs:code,es:codestart: push dsmov ax,0push axmov ax,datamov ds,axmov es,axlea si,string1lea di,string2mov cx,9repz cmpsbjz matchlea dx,inf2jmp dispmatch:lea dx,inf1disp :mov ah,09int 21hretmain endpcode endsend start5.4 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)。

【解】：main proc farstart:push dssub ax,axpush axinput:mov ah,01Hint 21Hcmp al,20Hje exitcmp al,30Hjle inputcmp al,3AHjge inputsub al,30Hmov cx,axand cx,0FFHlop:mov dl,07Hmov ah,2Hint 21Hloop lopjmp inputexit :retmain endpend start5.5 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 SEGMENTBUF DB -32, 25, 36, -18, -64, 0, -3COUNT EQU $ - BUFPLUS DB ? ; 存放正数MINUS DB ? ; 存放负数DATA ENDSCODE SEGMENTASSUME CS:CODE,DS:DATASTART: MOV AX, DATAMOV DS, AXMOV 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 ; 小于0， BL加1JMP NEXT1 ; 转 NEXT1NEXT0: INC DL ;大于等于0， DL加1NEXT1: INC SI ; 指针增1DEC CX ; 循环次数减1JNZ 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 ENDSEND START5.6 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 4char 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,'$'data endscode segmentassume cs:code,ss:stacks,ds:datastart: mov ax,datamov ds,axlea dx,inputmov ah,9int 21hlea dx, charmov ah,0ahint 21hlea dx,entermov ah,9int 21hmov si,2mov cx,4l0: mov bl,char[si]cmp bl,30hjb startcmp bl,39hjbe l00cmp bl,61hjb startcmp bl,66hja startsub bl,57hjmp l333l00: sub bl,30hl333:mov char[si],bl inc sidec cxjnz l0lea dx,outputmov ah,9int 21hmov si,2l000:mov bl,char[si]mov ch,4mov cl,4shl bl,clll00:test bl,80hjnz l1mov dl,'0'mov ah,2int 21hjmp l2l1: mov dl,'1'mov ah,2int 21hl2: shl bl,1dec chjnz ll00inc sicmp si,6jne l000mov ah,4chint 21hcode endsend start5.7 设有一段英文，其字符变量名为ENG，并以$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,"$"jz toQuitcmp al,"S"jnz lp1lodsbcmp al,"U"jnz lp2lodsbcmp al,"N"jnz lp2inc Zjmp lp1toQuit:5.8 有一个首地址为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】：stack 100H.datamem dw 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, 5,1,2,0,2,4,0,2,54,0,12,0,0,0,0,0,1,2,3,6,4,7,8,2,1,0,0,54,5,0,2,4,7,8,0,5,6,2,1,4 ,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.codeMAIN PROC FARstart:push dsand ax,0push axmov ax,@datamov ds,ax;----------------------mov ax,0Hmov bx,64Hmov cx,64Hmov si,0FFFEHrepeat:ADD si,2Hcmp MEM[si],0HJE callsloop repeatcalls:INC axcall sortcmp ax,1HJE lastValueDEC cxjmp repeatexit :retlastValue:mov mem[bx],0HDEC cxjmp repeatMAIN ENDPsort PROC NEARpush cxpush sisub si,2Hs:add si,2Hmov dx,mem[si]mov mem[si+2],dxloop sreturn:pop sipop cxretsort ENDPEND start5.9 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,OFFSET STRINGMOV ECX,99START_LOOP:LODSBCMP AL,'0'JC LOOP_NEXTCMP AL,'9'JNC LOOP_NEXTOR BL,10HLOOP_NEXT:LOOP START_LOOPMOV 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】：data segmentTable dw 0,1,2,3,4,5,6,7,8,9,0,1,2,3,4,5,6,7,8,9dw 0,1,2,3,4,5,6,7,8,9,0,1,2,3,4,5,6,7,8,9dw 0,1,2,3,4,5,6,7,8,9,0,1,2,3,4,5,6,7,8,9dw 0,1,2,3,4,5,6,7,8,9,0,1,2,3,4,5,6,7,8,9dw 0,1,2,3,4,5,6,7,8,9,0,1,2,3,4,5,6,7,8,9dw 0,1,2,3,4,5,6,7,8,9,0,1,2,3,4,5,6,7,8,9dw 0,1,2,3,4,5,6,7,8,9,0,1,2,3,4,5,6,7,8,9dw 0,1,2,3,4,5,6,7,8,9,0,1,2,3,4,5,6,7,8,9dw 0,1,2,3,4,5,6,7,8,9,0,1,2,3,4,5,6,7,8,9dw 5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,-1,-1Result dw 0,0,0,0data endscode segmentassume cs:code,ds:datastart:mov ax, datamov ds, ax;-------------------------------mov si, offset Tablemov 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 siinc simov dx, [si]cmp dx, axjg @C002cmp dx, bxjl @C003jmp @C004@C002:mov ax, dxjmp @C004@C003:mov bx, dx@C004:loop @C001; 保存结果mov Result, axmov Result+2, bx;-------------------------------mov si, offset Tablemov cx, 200@C005:mov ax, [si]xor dx, dx;------------------------------- push cxmov di, offset Tablemov cx, 200@C006:mov bx, [di]cmp ax, bxjnz @C007inc dx@C007:inc diinc diloop @C006pop cx;-------------------------------cmp dx, Result+6jle @C008mov Result+4, axmov Result+6, dx@C008:inc siinc siloop @C005;-------------------------------mov ax,4c00hint 21hcode endsend start; 最后结果 Result 9, -1, 5, 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。