练习一：统计复数的个数
DTSEG SEGMENT MES1 DB 'The result is:$' NUMB DB 12H,88H,82H,89H,33H,90H,01H,10H,0BDH,01H DTSEG ENDS CDSEG SEGMENT ASSUME CS:CDSEG,DS:DTSEG START:MOV AX,DTSEG MOV DS,AX MOV DX,OFFSET MES1 MOV AH,09H INT 21H MOV BL,0H MOV CX,0AH MOV SI,OFFSET NUMB NEXT: MOV AX,[SI] AND AX,80H ；判断是否是负数 CMP AX,0H JG MIN JMP CON MIN: INC BL CON: INC SI LOOP NEXT CALL SHOW MOV AH,4CH INT 21H SHOW PROC NEAR PUSH AX PUSH DX MOV AL,BL AND AL,0F0H ;取高4位 SHR AL,4 CMP AL,0AH ;是否是A以上的数 JB C2 ADD AL,07H C2: ADD AL,30H MOV DL,AL ;show character MOV AH,02H INT 21H MOV AL,BL AND AL,0FH ;取低4位 CMP AL,0AH JB C3 ADD AL,07H C3: ADD AL,30H MOV DL,AL ;show character MOV AH,02H INT 21H POP DX POP AX RET ENDP CDSEG ENDS END START
联系2：找出最大最小值
; You may customize this and other start-up templates; ; The location of this template is c:\emu8086\inc\0_com_template.txt STACK1 SEGMENT STACK DW 256 DUP(?) STACK1 ENDS DDATA SEGMENT MES1 DB 'The least number is:$' MES2 DB 0AH,0DH,'The largest number is:$' NUMB DB 0D9H,07H,8BH,0C5H,0EBH,04H,9DH,0F9H DDATA ENDS CODE SEGMENT ASSUME CS:CODE,DS:DDATA START: MOV AX,DDATA MOV DS,AX MOV SI,OFFSET NUMB MOV CX,0008H ;JCXZ A4 MOV BH,[SI] MOV BL,BH ；BL记录最小值，BH记录最大值 A1: LODSB ;AL=DS:[SI],SI=SI+1 取串指令 CMP AL,BH JBE A2 MOV BH,AL JMP A3 A2: CMP AL,BL JAE A3 MOV BL,AL A3: LOOP A1 A4: MOV DX,OFFSET MES1 ;show mes1 MOV AH,09H INT 21H MOV AL,BL ;show the least number AND AL,0F0H ;get the highest 4 bits SHR AL,4 CMP AL,0AH JB C2 ADD AL,07H C2: ADD AL,30H MOV DL,AL ;show character MOV AH,02H INT 21H MOV AL,BL AND AL,0FH ;get the lowest 4 bits CMP AL,0AH JB C3 ADD AL,07H C3: ADD AL,30H MOV DL,AL ;show character MOV AH,02H INT 21H MOV DX,OFFSET MES2 ;show mes2 MOV AH,09H INT 21H MOV AL,BH ;show the largest number AND AL,0F0H ;get the highest 4 bits SHR AL,4 CMP AL,0AH JB C22 ADD AL,07H C22: ADD AL,30H MOV DL,AL ;show character MOV AH,02H INT 21H MOV AL,BH AND AL,0FH ;get the lowest 4 bits CMP AL,0AH JB C33 ADD AL,07H C33: ADD AL,30H MOV DL,AL ;show character MOV AH,02H INT 21H MOV AX,4C00H ;go back to dos INT 21H CODE ENDS END START
练习3：双精度加法计算
要求计算X+Y=Z，将结果Z输出到屏幕上，其中X=001565A0H，Y=0021B79EH。实验利用累加器AX,先求低十六位和，并存入地址存储单元，后求高16位和，再存入高址存储单元。由于地位可能向高位有进位，因而高位相加语句需用ADC指令，则地位相加有进位时，CF=1，高位字相加时，同时加上CF中的1。在80386以上微机中可以直接使用32位寄存器和32位加法指令完成。 ;二进制双精度加法运算 STACK1 SEGMENT STACK DW 256 DUP(?) STACK1 ENDS DATA SEGMENT MES1 DB 'The result is:$' XL DW 65A0H XH DW 0015H YL DW 0B79EH YH DW 0021H DATA ENDS CODE SEGMENT ASSUME CS:CODE,DS:DATA START: MOV AX,DATA MOV DS,AX MOV DX,OFFSET MES1 ;将MES1偏移地址给DX MOV AH,09H ;将MES1中的内容输出到屏幕 INT 21H MOV AX,XL ADD AX,YL MOV BX,AX MOV AX,XH ADC AX,YH PUSH BX ;入栈保存BX CALL SHWORD ;先去执行SHWORD POP BX ;返回继续执行BX出栈处理 MOV AX,BX ;此时AX中是低16位之和 CALL SHWORD ;再次调用SHWORD MOV AX,4C00H INT 21H SHWORD PROC NEAR MOV BL,AH ;保持高16位的高8位 CALL SHOW ;显示结果 MOV BL,AL ;保持高16位的低8位 CALL SHOW ;显示结果 RET ENDP SHOW PROC NEAR PUSH AX ;保持AX PUSH DX ;保存DX MOV AL,BL AND AL,0F0H ;AL与0F0H相与取高4位 SHR AL,4 CMP AL,0AH ;是否是A以上的数 JB C2 ADD AL,07H ;A以上的数加07H折显示的是字母 C2: ADD AL,30H MOV DL,AL ;show character MOV AH,02H INT 21H MOV AL,BL AND AL,0FH ;AL与0FH相与取低4位 CMP AL,0AH JB C3 ADD AL,07H C3: ADD AL,30H MOV DL,AL ;show character MOV AH,02H INT 21H POP DX POP AX RET ENDP CODE ENDS END START
文章来源: 汇编练习