我想在Al寄存器中显示值，这是我的情况0Ah，这里是我的代码但是没有发生，我不确定，但我认为我的问题是我在寄存器中有一个hexa编号，但我只能打印Char或字符串，所以有一种方法将六进制数转换为字符串，例如
这是我使用的代码：
; You may customize this and other start-up templates;
; The location of this template is c:\emu8086\inc\0_com_template.txt
org 100h
.data segment
Array DB 0h,0h,0h,0h,0h
x db 0h
result db ?
.code segment
mov si,0
loop1:
;these 5 lines allows me to take
;input and let it be shown on the screen
mov ah,08h
int 21h
mov ah,02h
mov dl,al
int 21h
;those 3 lines allows me
;to start my code when
;enter button is pressed
mov bl,0Dh
cmp bl,dl
JZ start
;those 4 lines allow me
;to enter 4 chars and
;fill an array with them
;to use them later
mov array[si],dl
inc si
cmp si,5
jne loop1
start:
;putting each element in the
;array in a register to be
;able to deal with it
mov si,0
mov al,array[si]
mov bl,array[si+1]
mov cl,array[si+2]
mov dl,array[si+3]
;subtracting 30h from each
;number to turn it to its
;decimal form to deal with
;it as normal number
sub al,30h
sub bl,30h
sub cl,30h
sub dl,30h
;adding all numbers in
;variable called result
add al,cl
add al,bl
add al,dl
;printing
mov ah,02h
mov dl,al
int 21h
ret
首先，您忘记在计算中包含数组的最后一个元素。
要在计算中包含数组的最后一个元素的代码：
添加此行，将元素放在寄存器中：
mov bh, array[si+4] ; store the last element in BH register
添加此行，将寄存器内容转换为十进制：
sub bh, 30h ; convert it to decimal
添加此行添加所有寄存器的内容：
add al, bh ; add contents of AL with contents of BL
用十进制打印总和的代码：
mov ah, 0 ; clear AH because DIV instruction takes AX for division
mov cl, 10 ; store the divisor in CL
div cl ; AX/CL remainder will be in AH & quotient in AL
mov bx, ax ; move AX value to BX because later instruction are going to overwrite value of AH
mov ah,02h ; print quoteint
mov dl,bl
add dl,30h
int 21h
mov ah,02h ; print remainder
mov dl,bh
add dl,30h
int 21h
因为在你的情况下，添加所有5个元素后的最高数字将是2位数字（9 + 9 + 9 + 9 + 9 = 45），我们必须只划分一次，首先打印商，然后打印余数。如果总和包括超过2位数（如123），那么你必须连续重复相同的过程并将余数存储在堆栈中，直到你得零作为商。然后，您可以从堆栈中弹出数字并在执行必要的ASCII转换后显示它们。