组装 8086 |数组的总和，打印多位数字 [英] Assembly 8086 | Sum of an array, printing multi-digit numbers

问题描述

我在 asm x8086 中编写了一个非常简单的代码，但遇到了错误.如果有人能帮我做一个简短的解释，我将不胜感激.

I've written a pretty simple code in asm x8086 and I'm facing an error. If anyone could help me with a brief explanation I would greatly appreciate it.

IDEAL
MODEL small
STACK 100h
DATASEG
; --------------------------
    array db 10h, 04h, 04h, 04h, 04h, 04h, 04h, 04h, 04h, 04h
    sum db 0
    ; --------------------------
CODESEG
start:
    mov ax, @data
    mov ds, ax
; --------------------------
    xor cx, cx
    mov al, 0
    mov bx, offset array
StartLoop:
    cmp cx, 10
    jge EndLoop
    add al, [bx]
    add [sum],al
    inc cx
    inc bx
    jmp StartLoop
EndLoop:
    mov ah, 09h
    int 21h

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

exit:
    mov ax, 4c00h
    int 21h
END start

推荐答案

将 add 更正替换为 mov，如您的评论(注意这一行:add al, [bx] 实际上是 mov al, [bx]) 只是在标签 EndLoop 处的函数调用是错误的！

With the correction for the add to be replaced by mov as noted in your comment (Note that the line: add al, [bx] is actually mov al, [bx]) there's just the function call at the label EndLoop that's wrong!

您想显示总和，并且正在使用 DOS 打印功能.此函数 09h 需要您未提供的 DS:DX 中的指针！
即使您这样做了，您仍然需要在其文本表示中转换 sum 数字.

You want to display the sum, and are using the DOS print function. This function 09h expects a pointer in DS:DX that you are not providing!
Even if you did, you would still have to convert the sum number in its text representation.

这里的一个快速解决方案是让自己满意，并以单个 ASCII 字符的形式显示结果.硬编码的和是 52，所以它是一个可显示的字符:

A quick solution here would be to content yourself and just display the result in the form of a single ASCII character. The hardcoded sum is 52 and so it is a displayable character:

EndLoop:
    mov dl, [sum]
    mov ah, 02h    ;Single character output
    int 21h

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

exit:
    mov ax, 4c00h
    int 21h

再进一步，我们可以显示52":

One step further and we can display "52":

mov al,[sum]
mov ah,0
mov dl,10
div dl        ---> AL=5   AH=2
add ax,3030h  ---> AL="5" AH="2"
mov dh,ah     ;preserve AH
mov dl,al
mov ah,02h
int 21h
mov dl,dh     ;restore
int 21h

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