在 x86 程序集中将整数打印到控制台

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？

目前我有这个代码：

;;---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---;;

我认为 dl 值应该是 ASCII 代码，但我不确定如何将加法后的 ax 值转换为 ASCII。

您基本上想要除以 10，打印余数（一位数），然后重复商。

    ; 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

附带说明一下，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，然后你把ax in dl。你基本上就是在垃圾ax在打印之前。

您还存在尺寸不匹配的情况，因为dl是8位宽并且ax是16位宽。

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.