汇编第五章课后答题案.doc

1.从屏幕上输入小写字母，转化为大写字母输出(解法1)?DATA SEGMENT?? MESSAGE?? DB ENTER A STRING:,0AH,0DH,$?? MAXLENGTH? DB 50,?,50 DUP(?)? ;每次最多可以输入49个字符DATA ENDS??CODE SEGMENT??? ASSUME DS:DATA,CS:CODESTART:?? MOV AX,DATA???? MOV DS,AX?? LEA DX,MESSAGE ;输出ENTER A STRING?? MOV AH,09H?? INT 21H?????????? ?? LEA DX,MAXLENGTH? ;输入字符串?? MOV AH,0AH????? ;键盘输入到缓冲区,DS:DX=缓冲区首址?? INT 21H??? ;(DS:DX)=缓冲区最大字符数,(DS:DX+1)=实际输入的字符数????? MOV AH,02H???? ;输出回车换行?? MOV DL,0AH??? INT 21H?? MOV AH,02H?? MOV DL,0DH????? INT 21H ?? MOV CL,MAXLENGTH+1;把字符的实际长度放入寄存器CL????? MOV CH,0?? MOV BH,02H?? LEA SI,MAXLENGTH+2;取字符串的基地址放入SI?XUN:???? MOV AL,[SI]?? CMP AL,Z?? JBE S1??? ;小于等于Z转移??????? JMP S3???S1:CMP AL,A???? JAE DA?? ;大于等于A转移?? JMP OUTPUTDA:ADD AL,20H??? JMP OUTPUTS3:CMP AL,z? ;小于等于小Z转移?? JBE S4S4:CMP AL,a? ;大于等于小a转移?? JAE XIAO?? JMP OUTPUTXIAO:?? SUB AL,32?? JMP OUTPUTOUTPUT:?? MOV DL,AL?? MOV AH,02H? ;显示输出?? INT 21H?? INC SILOOP XUN?? MOV AH,4CH???? INT 21HCODE ENDS?? END START(解法2);将输入的小写字母转化为大写字母输出，输入回车结束CODE SEGMENT?? ASSUME CS:CODEBEGIN:?? MOV AH,01H?? INT 21H?CMP AL,0DH? ;ASCII OF CARRIAGE RETURN?JE EXIT?? CMP AL,61H?;ASCII OF a?? JB? STOP?? CMP AL,7AH?;ASCII OF z?? JA? STOP?SUB AL,20HSTOP:?? MOV DL,AL?? MOV AH,2?? INT 21H?? JMP BEGINEXIT:????MOV AH,4CH?INT 21H?? RETCODE ENDS?? END BEGIN ?2.输入一个字符,找出它的前导字符和后续字符,并按顺序显示这三个字符.?(解法1)CODE SEGMENTMAIN PROC FAR?ASSUME CS:CODESTART:PUSH DSXOR AX,AXPUSH AXMOV AH,01HINT 21HCMP AL,61HJB EXITCMP AL,7AHJA EXITMOV CL,ALDEC ALMOV DL,ALMOV AH,02HINT 21HMOV DL,CLMOV AH,02HINT 21HAND AL,01HINC CLMOV DL,CLMOV AH,02HINT 21H EXIT:MOV AX,4C00HINT 21HRETMAIN ENDPCODE ENDS?END START (解法2).MODEL SMALL.STACK 200H.DATAimsg db Input:$ ;输入提示信息?omsg db 0Dh,0Ah,Output:$ ;输出提示信息?string db 3 dup(0),$ ;存放三个字符?.CODE?START:?mov ax,@data?mov ds,ax?lea dx,imsg?mov ah,9?int 21h?mov ah,1?int 21h?lea di,string?dec al?mov cx,3?s:?mov [di],al?inc di?inc al?loo