打印出十进制数

时间:2015-04-06 20:23:43

标签: assembly x86

我的代码应该打印出一个十进制数字,但它不会,我不知道为什么。

我只是一个初学者,我已经阅读了很多建议,但我无法解决这个问题。谢谢你的反应! (要打印的数字是COUNT

        printout proc
xor dl, dl
mov dl, OFFSET COUNT
xor bx,bx
mov bh, 0ah
mov ax,dx
div bh 
mov bh,ah 
mov bl,al
xor dx,dx
mov dl,bl 
add dx,’0′ 
mov ah,02h 
int 21h
xor dx,dx
mov dl,bh
add dx,’0′ 
mov ah,02h 
int 21h
ret
endp

2 个答案:

答案 0 :(得分:2)

要阅读COUNT的值,请使用mov dl,byte [COUNT]mov dl,COUNT 这取决于您使用的汇编程序。

您还应该使用xor dh,dh代替xor dl,dl

您可以正确处理的最大数字是99.

您可以简化代码。

xor ah, ah
mov al, COUNT
mov bh, 0ah
div bh 
mov bx,ax 
mov dl,bl 
add dl,’0′ 
mov ah,02h 
int 21h
mov dl,bh
add dl,’0′ 
mov ah,02h 
int 21h
ret

答案 1 :(得分:0)

首先,您必须将数字转换为字符串,然后您可以"打印出来"。下一个代码是用EMU8086制作的,并展示了如何操作(评论将帮助您理解):

.stack 100h
.data

num  dw 10382      ;ANY NUMBER.
str  db 7 dup('$') ;STRING TO STORE NUMBER. 

.code          
;INITIALIZE DATA SEGMENT.
  mov  ax, @data
  mov  ds, ax

;CONVERT "NUM" TO STRING. RESULT COMES BACK IN "STR".
  mov  ax, num
  call number2string

;DISPLAY NUMBER.
  mov  ah, 9
  mov  dx, offset str ;NUMBER CONVERTED TO STRING.
  int  21h

;STOP UNTIL USER PRESS ANY KEY.
  mov  ah,7
  int  21h

;FINISH THE PROGRAM PROPERLY.
  mov  ax, 4c00h
  int  21h           

;------------------------------------------

;NUMBER TO CONVERT MUST ENTER IN AX.
;ALGORITHM : EXTRACT DIGITS ONE BY ONE, STORE
;THEM IN STACK, THEN EXTRACT THEM IN REVERSE
;ORDER TO CONSTRUCT STRING.

proc number2string
  mov  bx, 10 ;DIGITS ARE EXTRACTED DIVIDING BY 10.
  mov  cx, 0 ;COUNTER FOR EXTRACTED DIGITS.
cycle1:       
  mov  dx, 0 ;NECESSARY TO DIVIDE BY BX.
  div  bx ;DX:AX / 10 = AX:QUOTIENT DX:REMAINDER.
  push dx ;PRESERVE DIGIT EXTRACTED FOR LATER.
  inc  cx ;INCREASE COUNTER FOR EVERY DIGIT EXTRACTED.
  cmp  ax, 0  ;IF NUMBER IS
  jne  cycle1 ;NOT ZERO, LOOP. 
;NOW RETRIEVE PUSHED DIGITS.
  mov  si, offset str
cycle2:  
  pop  dx        
  add  dl, 48 ;CONVERT DIGIT TO CHARACTER.
  mov  [ si ], dl
  inc  si
  loop cycle2  

  ret
endp