3_25.asm

来自「华工电信系汇编习题的编程实现」· 汇编 代码 · 共 46 行

stack segment stack 'stack'
      dw 32 dup(?)
stack ends

data segment 
     BUFA dw 01211H,0BACDH,0H,03415H;,0H,5678H
     MYSIZE equ $ - BUFA
data ends

code segment 
begin proc far
      assume ss:stack,cs:code,ds:data
      push ds
      sub ax,ax
      push ax
      mov ax,data
      mov ds,ax

      mov SI,0
      mov DI,0
AGAIN:       
      mov ax,word ptr BUFA[SI]
      and ax,ax
      JZ ZERO
      mov word ptr BUFA[DI],ax
      add DI,2      
ZERO:             
      add SI,2
      mov bx,MYSIZE
      add bx,bx
      CMP SI,bx
      JZ NEXT
      JMP AGAIN

NEXT:
      CMP SI,DI
      JZ MYEND
      mov word ptr BUFA[DI],0
      add DI,2
      JMP NEXT
MYEND:      

      ret
begin   endp
code    ends
      end begin