When I add two values in 16 bit assembly, what is the best way to print the result to console?

当我在16位汇编中添加两个值时，将结果打印到控制台的最佳方法是什么？

At the moment I have this code:

目前我有这样的代码：

;;---CODE START---;;
mov ax, 1   ;put 1 into ax
add ax, 2   ; add 2 to ax current value
mov ah,2  ; 2 is the function number of output char in the DOS Services.
mov dl, ax ; DL takes the value.
int 21h    ; calls DOS Services

mov ah,4Ch   ; 4Ch is the function number for exit program in DOS Services.
int 21h      ; function 4Ch doesn't care about anything in the registers.
;;---CODE END---;;

I think that dl value should be in ASCII code, but I'm not sure how to convert ax value after addition into ASCII.

我认为dl值应该是ASCII代码，但我不确定如何将相加后的AX值转换为ASCII代码。

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You basically want to divide by 10, print the remainder (one digit), and then repeat with the quotient.

您基本上想要除以10，打印余数(一位数)，然后重复商数。

    ; assume number is in eax
    mov ecx, 10

loophere:
    mov edx, 0
    div ecx

    ; now eax <-- eax/10
    ;     edx <-- eax % 10

    ; print edx
    ; this is one digit, which we have to convert to ASCII
    ; the print routine uses edx and eax, so let's push eax
    ; onto the stack. we clear edx at the beginning of the
    ; loop anyway, so we don't care if we much around with it

    push eax

    ; convert dl to ascii
    add dl, '0'

    mov ah,2  ; 2 is the function number of output char in the DOS Services.
    int 21h    ; calls DOS Services

    ; now restore eax
    pop eax

    ; if eax is zero, we can quit

    cmp eax, 0
    jnz loophere

---

As a side note, you have a bug in your code right here:

顺便说一句，您的代码中有一个错误：

mov ax, 1   ;put 1 into ax
add ax, 2   ; add 2 to ax current value
mov ah,2  ; 2 is the function number of output char in the DOS Services.
mov dl, ax ; DL takes the value.

You put 2 in ah, and then you put ax in dl. You're basically junking ax before printing it.

你把2放入ah，然后把ax放入d1。你基本上是在打印之前把斧头扔进垃圾桶。

You also have a size mismatch since dl is 8 bits wide and ax is 16 bits wide.

还有一个大小不匹配，因为dl是8位宽，而ax是16位宽。

What you should do is flip the last two lines and fix the size mismatch:

您应该做的是翻转最后两行并修复大小不匹配的问题：

mov ax, 1   ;put 1 into ax
add ax, 2   ; add 2 to ax current value

mov dl, al ; DL takes the value.
mov ah,2  ; 2 is the function number of output char in the DOS Services.

Just fixing the order of @Nathan Fellman 's code

只是修改@内森·费尔曼的S代码的顺序

PrintNumber proc
    mov cx, 0
    mov bx, 10
@@loophere:
    mov dx, 0
    div bx                          ;divide by ten

    ; now ax <-- ax/10
    ;     dx <-- ax % 10

    ; print dx
    ; this is one digit, which we have to convert to ASCII
    ; the print routine uses dx and ax, so let's push ax
    ; onto the stack. we clear dx at the beginning of the
    ; loop anyway, so we don't care if we much around with it

    push ax
    add dl, '0'                     ;convert dl to ascii

    pop ax                          ;restore ax
    push dx                         ;digits are in reversed order, must use stack
    inc cx                          ;remember how many digits we pushed to stack
    cmp ax, 0                       ;if ax is zero, we can quit
jnz @@loophere

    ;cx is already set
    mov ah, 2                       ;2 is the function number of output char in the DOS Services.
@@loophere2:
    pop dx                          ;restore digits from last to first
    int 21h                         ;calls DOS Services
    loop @@loophere2

    ret
PrintNumber endp

The basic algorithm is:

基本算法是：

divide number x by 10, giving quotient q and remainder r
emit r
if q is not zero, set x = q and repeat 

Note that this will yield the digits in the inverse order, so you are probably going to want to replace the "emit" step with something that stores each digit, so that you can later iterate in reverse over the stored digits.

请注意，这将以相反的顺序生成数字，因此您可能希望用存储每个数字的内容替换“emit”步骤，以便稍后可以对存储的数字进行反向迭代。

Also, note that to convert a binary number between 0 and 9 (decimal) to ascii, just add the ascii code for '0' (which is 48) to the number.

此外，请注意，要将0到9(十进制)之间的二进制数转换为ASCII，只需在数字上加上‘0’的ASCII代码(即48)。

mov dl, ax

This won't work as dl and ax have different bit sizes. What you want to do is create a loop in which you divide the 16 bit value by 10, remember the rest on the stack, and then continue the loop with the integer division result. When you reach a result of 0, clean up the stack digit by digit, adding 48 to the digits to turn them into ASCII digits, then print them.

这将不起作用，因为d1和ax具有不同的位大小。您要做的是创建一个循环，在该循环中，将16位值除以10，记住堆栈上的其余部分，然后使用整数除法结果继续循环。当结果为0时，逐位清理堆栈，在数字上加48以将其转换为ASCII数字，然后打印它们。

The accepted answer is wrong in that it displays the result reversed! That's why I write this answer today. My answer demonstrates the better technique of outputting the result in one gulp (via DOS.function 09h).

被接受的答案是错误的，因为它显示的结果是相反的！这就是我今天写下这个答案的原因。我的回答展示了一次输出结果的更好技术(通过DOS.Function 09H)。

Your code

mov ah,2  ; 2 is the function number of output char in the DOS Services.
mov dl, ax ; DL takes the value.
int 21h    ; calls DOS Services

Always load the DOS function number in the instruction right before the int 21h instruction.

Always check your sizes: you cannot move from a 16-bit register to an 8-bit register.

始终在INT 21h指令之前的指令中加载DOS功能编号。始终检查您的大小：您不能从16位寄存器转移到8位寄存器。

I think that dl value should be in ASCII code, but I'm not sure how to convert ax value after addition into ASCII.

Your numbers allow the use of the DOS.PrintCharacter function. To convert the value 3 into the character "3", you just need to add 48 so as to have an ASCII code in DL.

您的数字允许使用DOS.PrintCharacter函数。要将值3转换为字符“3”，您只需添加48即可在DL中使用ASCII代码。

When I add two values in 16-bit assembly, what is the best way to print the result to console?

Your example for a single-digit result

; Sum
  mov  al, 1
  add  al, 2

; Print
  add  al, '0'
  mov  dl, al
  mov  ah, 02h        ; DOS.PrintCharacter
  int  21h

; Exit
  mov  ax, 4C00h      ; DOS.TerminateWithExitcode
  int  21h

Solution for a multi-digit result

; Sum
  mov  ax, 7346
  add  ax, 913

; Print
  mov  bx, Buffer + 5 ; Position of the mandatory '$' terminator
  mov  cx, 10
More:
  xor  dx, dx
  div  cx
  dec  bx
  add  dl, '0'        ; Convert to ASCII
  mov  [bx], dl
  test ax, ax
  jnz  More
  mov  dx, bx
  mov  ah, 09h        ; DOS.PrintString
  int  21h

; Exit
  mov  ax, 4C00h      ; DOS.TerminateWithExitcode
  int  21h

; Storage
Buffer: db '.....$'  

NASM use mov bx, Buffer + 5

MASM use mov bx, OFFSET Buffer + 5

NASM使用MOV BX，缓冲区+5 MASM使用MOV BX，偏移缓冲区+5