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主题 : : 实验10-3: source code! [待解决] |
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[帖 主]
[ 发表时间：2011-07-04 16:00 ]
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注册日期：2011-05-30 14:33 |
assume cs:code
data segment
db 10 dup(0)
data ends
code segment
start: mov ax,12666
mov bx,data
mov ds,bx
mov si,0
call dtoc
mov dh,8
mov dl,3
mov cl,2
mov ax,data
mov ds,ax
mov si,0
call show_str
mov ax,4c00h
int 21h
dtoc: push ax
push bx
push cx
push dx
push si
mov bx,10
to: mov dx,0
div bx
mov cx,ax
add dl,30h
push dx
jcxz good
inc si
jmp short to
good: inc si
mov cx,si
mov si,0
circle: pop [si]
inc si
loop circle
pop si
pop dx
pop cx
pop bx
pop ax
ret
show_str: push ax
push bx
push cx
push dx
push si
push di
push es
address: mov ax,0b800h
mov es,ax
mov ax,160
mul dh
mov bx,ax
dec dl
add dl,dl
mov dh,0
add bx,dx
mov di,0
mov ah,cl
mov ch,0
show: mov cl,[si]
jcxz ok
mov al,[si]
mov es:[bx+di],ax
inc si
add di,2
jmp short show
ok: pop es
pop di
pop si
pop dx
pop cx
pop bx
pop ax
ret
code ends
end start | | |
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[第1楼]
[ 回复时间：2011-08-11 23:36 ]
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注册日期：2011-08-11 23:23 |
assume cs:code,ds:data
data segment
db 10 dup (0)
data ends
code segment
start: mov ax,0ffffh
mov dx,0ffffh
mov bx,data
mov ds,bx
mov si,0 ;Index of the string
call dtoc
mov dh,8 ;Line no
mov dl,3 ;Column no
mov cl,00000010b ;Color, green
call show_str
mov ax,4c00h ;Terminate procerss
int 21h ;Call dos interupt
dtoc: push dx
push cx
push di
push si
mov di,0 ;Record number of character
s0: mov cx,10 ;Divisor is 10
call divdw
add cx,30h ;Covert to ASCII character
push cx ;Push character into stack
inc di ;Count +1
; mov [si],cl ;Save the character
; mov byte ptr [si+1],0 ;Save 0 in the end
; inc si ;Next
mov cx,ax
jcxz eof0 ;Return
jmp short s0
eof0: mov byte ptr [di],0 ;Set the end of flag
mov cx,di ;Set the loop times
l: pop bx ;Get the character
mov [si],bl ;Set the character
inc si
loop l
pop si
pop di
pop cx
pop dx
ret
; Name: divdw
; Desc: The division will not overflow, the dividend for the dword type,
; the divisor for the dword type, the result is dword type
; Args: (ax)=dword low 16-bit data type
; (dx)=dword high 16-bit data type
; (cx)=divisor
; Return: (dx)=high 16 result
; (ax)=low 16 result
; (cx)=remainder
divdw: push si
push bx
push ax
mov ax,dx
mov dx,0
div cx
mov si,ax
pop ax
div cx
mov cx,dx
mov dx,si
pop bx
pop si
ret
; Name: show_str
; Desc: At the special location with the special color, display the string
; Args: (dh)=row number(Range:0~24), (dl)=column number(Range:0~79)
; (cl)=color, ds:si points to the first address of the string
; Return: void
show_str: push ax
push bx
push cx
push si
push di
mov ax,0b800h
mov es,ax
mov al,0a0h
mul dh ;Offset of the row
mov bx,ax
mov al,2
mul dl ;Offset of the column
add bx,ax ;Ofset of display
mov ah,cl ;Set the color
mov ch,0 ;High 16-bit is 0
s: mov cl,[si] ;Set the character
jcxz eof
mov al,cl
mov es:[bx],ax
inc si
add bx,2
jmp short s
eof: pop di
pop si
pop cx
pop bx
pop ax
ret
code ends
end start | | |
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