使用汇编器 8086 显示表的总和

我正在编写一些代码，允许我对表进行求和，然后使用汇编语言显示其结果。这是我到目前为止所想到的：

data segment 
  tab      db 9 dup(3 5 8 4 7 1 6 7 0)
  resultat db ?
data ends

code segment 
  xor di, di
  mov cx, 9
Prog: 
  mov al, tab[di]
  add ax, al
  inc di
loop prog
  mov resultat ,ax
  mov ah, 09h
  int 21h
end code

我对你的代码做了一些更改并窃取了 procnumber2string来自另一个答案（充满评论以帮助您理解）：

data segment 
tab db 3,5,8,4,7,1,6,7,0 ;ARRAY OF NUMBERS.
resultat db '     $'  ;STRING WITH 5 CHARS.
data ends
code segment 
mov ax,@data    ;INITIALIZE
mov ds,ax       ;DATA SEGMENT.

xor di,di 
mov cx,9
xor ax,ax       ;CLEAR THE SUM ACCUMULATOR.
Prog:
add al,tab[di]
inc di
loop prog     

call number2string ;CONVERT AX TO STRING IN RESULTANT.

lea dx,resultat
mov ah,09h
int 21h

mov ax, 4c00h  
int 21h        ;TERMINATE PROGRAM.

;------------------------------------------
;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.
;THE STRING IS STORED IN VARIABLE "RESULTAT".

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.
  lea  si, resultat
cycle2:  
  pop  dx        
  add  dl, 48 ;CONVERT DIGIT TO CHARACTER.
  mov  [si], dl
  inc  si
  loop cycle2  

  ret
endp  
;------------------------------------------

end code

该过程number2string将 ax 中的任意数字转换为字符串并将结果存储在变量中resultat（如数据段中定义的那样）。