(王爽版)汇编实验7 寻址方式在结构化数据访问中的应用

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题目如下:

汇编代码如下:

assume cs:codesg 

;数据段
data segment

  db '1975','1976','1977','1978','1979','1980','1981','1982','1983'
  db '1984','1985','1986','1987','1988','1989','1990','1991','1992'
  db '1993','1994','1995'

  dd 16,22,382,1356,2390,8000,16000,24486,50065,97479,140417,197514
  dd 345980,590827,803530,1183000,1843000,2759000,3753000,4649000,5937000

  dw 3,7,9,13,28,38,130,220,476,778,1001,1442,2258,2793,4037,5635,8226
  dw 11542,14430,15257,17800

data ends


;表段
table segment
  
  db 21 dup('year summ ne ?? ')
            
table ends





;代码段
codesg segment


start:

 mov ax,data
 mov ds,ax

 mov ax,table
 mov es,ax



 mov di,0
 mov bx,0
 mov si,0
 mov cx,21

 s:


 

 ;复制年份 year

 mov ax,[si]
 mov es:[bx],ax

 mov ax,[si+2]
 mov es:[bx+2],ax

 

 ;复制总收入 summ

 mov ax,[si+84]
 mov es:[bx+5],ax

 mov ax,[si+84+2]
 mov es:[bx+5+2],ax

 ;复制雇员人数 ne

 mov ax,[di+168]
 mov es:[bx+10],ax

;被除数的低二位

 mov ax,es:[bx+5]

;被除数的高二位

 mov dx,es:[bx+5+2]

 ;取整数

 div word ptr es:[bx+10]

 ;商,余数在dx中
 mov es:[bx+13],ax

  add bx,16
  add si,4
  add di,2
 loop s 
  
 mov ax,4c00h
 int 21h
codesg ends


end start

运行截图

 

之后课上我又了解了一种更加简便的解法

assume cs:codesg ;es:table ds:data ss:stack

;数据段
data segment

  db '1975','1976','1977','1978','1979','1980','1981','1982','1983'
  db '1984','1985','1986','1987','1988','1989','1990','1991','1992'
  db '1993','1994','1995'

  dd 16,22,382,1356,2390,8000,16000,24486,50065,97479,140417,197514
  dd 345980,590827,803530,1183000,1843000,2759000,3753000,4649000,5937000

  dw 3,7,9,13,28,38,130,220,476,778,1001,1442,2258,2793,4037,5635,8226
  dw 11542,14430,15257,17800

data ends


;表段
table segment
  
  db 21 dup('year summ ne ?? ')
            
table ends





;代码段
codesg segment


start:

 mov ax,data
 mov ds,ax

 mov ax,table
 mov es,ax



 mov di,0
 mov bx,0
 mov si,0
 mov cx,21

 s:


 

 ;复制年份 year

 mov ax,[si]
 mov es:[bx],ax

 mov ax,[si+2]
 mov es:[bx+2],ax

 

 ;复制总收入 summ

 mov ax,[si+84]
 mov es:[bx+5],ax

 mov dx,[si+84+2];高位直接存在dx
 mov es:[bx+5+2],dx

 ;复制雇员人数 ne

 mov bp,[di+168]
 mov es:[bx+10],bp

 div word ptr es:[bx+10]

 ;商,余数在dx中
 mov es:[bx+13],ax

  add bx,16
  add si,4
  add di,2
 loop s 
  
 mov ax,4c00h
 int 21h
codesg ends


end start

 

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转载自blog.csdn.net/iostream992/article/details/82980989