我再次调试并从头到尾看了下代码,将S3循环部分更改,并用更直观结构的方式表示寻址,详细代码如下,调试通过,哈哈。。。。。。- assume cs:codesg,ds:data,es:table
- 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
- table segment
- db 21 dup('year summ ne ?? ')
- table ends
- stack segment
- dw 0,0,0,0,0,0,0,0
- stack ends
- codesg segment
- start: mov ax,stack
- mov ss,ax
- mov sp,10h
- mov ax,data
- mov ds,ax
- mov ax,table
- mov es,ax
- mov dx,0
- mov si,0
- mov cx,21
-
- s0: mov ax,[bx+0]
- mov es:[si+0],ax
- mov ax,[bx+2]
- mov es:[si+2],ax
- add si,16
- add bx,4
- loop s0
- mov si,0
- mov cx,21
- ;以上将年份数据放入table
- s1: mov ax,[bx+0]
- mov es:[si+5],ax
- mov ax,[bx+2]
- mov es:[si+7],ax
- add si,16
- add bx,4
- loop s1
- mov si,0
- mov cx,21
- ;以上将收入数据放入table
-
- s2: mov ax,[bx+0]
- mov es:[si+10],ax
- add si,16
- add bx,2
- loop s2
- mov si,0
- mov cx,21
- ;以上将雇员数放入table
- s3: mov ax,es:[si+5] ;总收入,被除数,低16位放ax
- mov dx,es:[si+7] ;总收入,被除数,高16位放dx
- div word ptr es:[si+10];做除法运算
- mov es:[si+13],ax ;将商写到table段
- add si,16
- loop s3
- ;计算人均收入
-
- mov ax,4c00h
- int 21h
- codesg ends
- end start
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