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楼主 |
发表于 2012-11-15 14:39:59
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我也来个。- assume cs:codesg,ds:datasg,es:table,ss:stack
- table segment
- db '1975','1976','1977','1978','1979','1980','1981','1982','1983'
- db '1984','1985','1986','1987','1988','1989','1990','1991','1991'
- db '1993','1994','1995'
- ;以上是表示21年的21个字符串
-
- dd 16,22,382,1356,2390,8000,16000,24486,50065,97479,140417,197514
- dd 345980,590827,803530,1183000,1843000,2795000,3753000,4649000,5937000
- ;以上是表示21年公司总收入的21个dword型数据
-
- dw 3,7,9,13,28,38,130,220,476,778,1001,1442,2258,2793,4037,5635,8226
- dw 11542,14430,15257,17800
- ;以上是表示21年公司雇员人数的21个word型数据
-
- dw 5,3,42,104,85,210,123,111,105,125,140,136,153,211,199,209,224,239
- dw 260,304,333
- ;以上是表示21年公司人均收入
-
- dd 10 dup (0)
- table ends
- datasg segment
-
- db 1000 dup (0)
-
- datasg ends
- stack segment
- db 100 dup (0)
-
- stack ends
- codesg segment
- start:
-
- mov ax,datasg
- mov ds,ax
-
- mov ax,table
- mov es,ax
- mov si,0
- mov bp,0
-
- mov ax,stack
- mov ss,ax
- mov sp,100
-
- mov cx,21
- s7: call dtoc
- add bp,16 ;bp加16换行
- add di,4 ;si+4
- loop s7
-
- mov ax,4c00h
- int 21h
-
-
- dtoc:
- push cx
- push di
- push si
- push bp
- ;cx、si入栈
- s4:
- mov ax,es:[di+54h]
- mov dx,es:[di+56h]
-
-
- mov bx,10
-
- s1: mov bp,ax ;转移到L位
-
- mov ax,dx ;dx值等于ax
-
- xor dx,dx ;-----------H转AX D清零
-
- div bx ;H除以bx
-
- mov di,ax ;H结果=di
-
- mov ax,bp ;ax植出栈,L重赋给ax
-
- div bx ;-----------REG+L/10 DX=余娄 H结果=DI
- ;=========== L结果AX
-
- mov cx,bp ;商为零cx
-
- jcxz d_wk ;-----------余数为空jmp 到出栈
- add dx,30h
- push dx
- inc si
- xor dx,dx
- ;------------余数+30H=ASCII
- mov dx,di
- ;------------H重赋给DX
- jmp s1
- ;------------判断H为0,不为0循环
- ;============ 为0做L位除法
-
- d_wk: mov cx,si
- ;------------设定循环数
- ;------------CX为0 jmpECS
-
- jcxz s3 ;cx为零转移到s2
-
-
- mov si,0
- mov di,0 ;di等于0
-
- s3: pop ax ;dx植出栈放入ax
-
- mov es:[di+100h],al ;[bp+di+05h]=al
- add di,1 ;di加1
-
- loop s3 ;loop 循环
-
- pop bp
- mov cx,di
-
- s2: mov al,es:[si+100h]
- mov ds:[si+bp+05h],al
-
- add si,1
-
- loop s2
-
- okay: pop si
- pop di
- pop cx
-
- ret
- codesg ends
- end start
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