微机原理：顺序、分支、循环、子程序设计

LOOP DO NEXT0:

SUB BL, 48 MOV AL, BL MOV BL, 2 DIV BL CMP AH, 1 JNZ NEXT1 JZ NEXT2 NEXT1:

LEA DX, HEX1 MOV AH, 09H INT 21H HLT NEXT2:

LEA DX, HEX2 MOV AH, 09H INT 21H ends end start

若输入的为偶数

若输入的为奇数

3.4 experiment 2-4 name \

data segment

HEX DB 16 DUP(0) ends

code segment start:

ASSUME DS:DATA MOV AX, DATA MOV DS, AX MOV CX, 16 ;read from keyboard read:

CMP CX, 1 JZ STOP

MOV AH, 01H INT 21H CMP AL, 65 JZ count

CMP AL, 13

JZ STOP LOOP read ;count 'A' characters count:

INC BL LOOP read

STOP:

MOV AH, 02H MOV DL, 0AH

INT 21H; 打印换行 MOV DL, 0DH INT 21H;打印回车 ADD BL, 48 MOV DL, BL INT 21H ends end

start

3.5 experiment 2-5

①冒泡排序，没有在BUF3中除去重复的字符 此题设两个字符串为 BUF1 = ”ACEGIK” BUF2 = ”BDFJLHMN”

name \

;此程序不需要压入堆栈保护的数据 data segment

BUF1 DB \ BUF2 DB \ BUF3 DB 14 DUP(0) ends

stack segment

DB 256 DUP(0) ends

code segment start:

ASSUME DS:DATA, CS:CODEL, SS:STACK MOV AX, DATA MOV DS, AX LEA BX, BUF1 LEA SI, BUF2 LEA DI, BUF3 CALL remove

CALL reorder;段内转移，若段间转移，测试一下格式 ;print on the screen LEA DX, BUF3 MOV AH, 09H INT 21H HLT

remove proc near MOV CX, 6 DO1:

MOV AX, [BX] MOV [DI], AX INC BX INC DI LOOP DO1 MOV CX, 8 DO2:

MOV AX, [SI] MOV [DI], AX INC SI INC DI