微机原理实验报告程序文档格式.docx

1、拆字ORG 1000HNUM DW 1234HORG 4000HA DB ？B DB ? MOV AX,NUM MOV A,AH MOV B,AL实验33INFOR1 DB 0AH,0DH, Please Press any key to input a letter:$ ;0AH为换行，0DH为回车INFOR2 DB 0AH,0DH, You Input a lowercase letter! $INFOR3 DB 0AH,0DH, You input a uppercase letter!INFOR4 DB 0AH,0DH, You input a digit!INFOR5 DB 0ah,

2、0dh, You input other letter! CODE SEGMENT ASSUME CS: MOV AX,DATA MOV DX,OFFSET INFOR1 MOV AH,09H ;输出字符串INFOR1MOV AH,1 MOV AH,01H ;从键盘输入一个字符 CMP AL, 0 JB OTHER ;小于0，为其他字符9 JBE DIGIT ;大于等于0，小于等于9，为数字A JB OTHERZ JBE UPPERaz JBE LOWER JMP PENDLOWER: MOV DX,OFFSET INFOR2 MOV AH,09HUPPER: MOV DX,OFFSET IN

3、FOR3DIGIT: MOV DX,OFFSET INFOR4OTHER: MOV DX,OFFSET INFOR5 PEND: CODE ENDS END START实验31 STRING DB I am a student LEN EQU $-STRING MOV BX,OFFSET STRING MOV CL,LEN NEXT:MOV AL,BX CMP AL, JB LOP JA LOP SUB AL,20H MOV BX,AL LOP:INC BX DEC CL JNZ NEXT实验41A DB 1234567890N EQU $-AB DB N DUP（?）ASSUME CS:MO

4、V DS,AXMOV ES,AXLEA SI,ALEA DI,BADD DI,NMOV CX,NMOVE:MOV AL,SI MOV ES:DI,AL INC SI DEC DI LOOP MOVE实验42 DB 100 DUP（ORG 1050HBDATA,ES: MOV SI,1000H MOV DI,1050H MOV CX,100AGAIN: MOV AL,SI MOV DI,AL INC DI LOOP AGAIN实验43 BUF DW 1,2,3,4,5,6,7,8,9,0 MAX DW 0 MOV SI,OFFSET BUF MOV CX,10 MOV AX,SI CMP AX,

5、MAX JB NEXT MOV MAX,AXNEXT:实验61DATA SEGMENT STR DB Please input a character:STR1 DB 0ah,0dhSTR3 DB ?,H$CODE SEGMENT MOV DX,OFFSET STR MOV AH,01H MOV STR1,AL MOV DX,OFFSET STR1实验62STR1 DB 0dh,0ah,STR2 DB 0dh,0ah,You input a number!STR3 DB 0dh,0ah,The character you input is not a number! CMP AL,30H JB

6、 NEXT1 CMP AL,39H JA NEXT1 MOV DX,OFFSET STR2 JMP NEXT2NEXT1:MOV DX,OFFSET STR3NEXT2:MOV AH,4CH实验63info1 db 0dh,0ah, input string:info2 db 0dh,0ah, output string:STR DB 20 DUP（? ADD SI,10 MOV SI,BYTE PTR MOV DX,OFFSET info1MOV AH,01H DEC SI MOV SI,AL输入字符后回车 CODE START:MOV AX,CODE MOV DL,0AH PUSH AX

7、AND AX,00FFH MOV CL,4 SHR AL,CL ADD AL,30H MOV DL,AL MOV AH,02H POP AX AND AL,0FH CMP AL,0AH JNC NEXT JMP finalADD AL,37H final:MOV DL,AL MOV DL,H DATA SEGMENTinfo1 DB 0dh,0ah,info2 DB 0dh,0ah, STR DB 0Bh,0,20 dup（? n EQU $-STR DB DATA ENDS ASSUME cs:code,ds:datamov ax,DATA mov DS,ax lea DX,info1 mo

8、v AH,09H lea DX,STR mov AH,0AH MOV CX,0 lea SI,STR mov DI,SI add DI,n-1number:MOV BL,SI CMP BL,0DH JZ NEXT INC CX JMP numberlea SI,STRagain:mov bl,SI mov DI,bl loop again lea DX,info2 mov DX,DI mov AH,4ch INT 21hCODE ends end startcode segment assume cs:code start: mov ah,1 int 21h cmp al,1bhjz tjmov bl,al mov ch,2 rotate:mov cl,4 rol bl,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 dec ch jnz rotate jmp start tj:mov ax,4c00h code ends end start