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主题 : :  [求解]16.3中一个程序的一个步骤,想了许久,未解,请牛人点拨  [待解决] 回复[ 2次 ]   点击[ 483次 ]  
kamjia
[帖 主]   [ 发表时间:2010-11-24 17:16 ]   [引用]   [回复]   [ top ] 
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注册日期:2009-05-25 18:12
assume cs:code

code segment

start:                                                        mov ax,60     ;测试数据
                                                        call showsin
                                                        
                                                        mov ax,4c00h
                                                        int 21h


showsin:                jmp short show
                                                        
table                        dw ag0,ag30,ag60,ag90,ag120,ag150,ag180
                        ag0                db '0',0
                        ag30                db '0.5',0
                        ag60                db '0.866',0
                        ag90                db '1',0
                        ag120                db '0.866',0
                        ag150                db '0.5',0
                        ag180                db '0',0
                        
show:                                                        push bx
                                                        push es
                                                        push si
                                                        
                                                        mov bx,0b800h
                                                        mov es,bx
                                                        
                                                        cmp ax,0
                                                        jb showret
                                                        cmp ax,180
                                                        ja showret
                                                         
                                                        ;mov ah,0       ;???疑问在这里???
                                                        mov bl,30
                                                        div bl
                                                        mov bl,al
                                                        mov bh,0
                                                        add bx,bx
                                                        mov bx,table[bx]
                                                        
                                                        mov si,160*12+40*2
shows:                                                        mov ah,cs:[bx]
                                                        cmp ah,0
                                                        je showret
                                                        mov es:[si],ah
                                                        inc bx
                                                        add si,2
                                                        jmp short shows
                                                        
showret:                                                pop si
                                                        pop es
                                                        pop bx
                                                        ret
                                                        
code ends

end start

-------------------------------------------------------------------------
疑问:为什么要加这条指令,不知道意义何在,我将之注释掉,对结果并无影响
kamjia
[第1楼]   [ 回复时间:2010-11-24 17:37 ]   [引用]   [回复]   [ top ] 
荣誉值:1
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注册日期:2009-05-25 18:12
我觉得有一种可能是:防止除尘的溢出
kamjia
[第2楼]   [ 回复时间:2010-11-24 17:37 ]   [引用]   [回复]   [ top ] 
荣誉值:1
信誉值:2
注册日期:2009-05-25 18:12
我觉得有一种可能是:防止除尘的溢出
------------------
回复:是除法的溢出,笔误
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