打印整数控制台x86汇编

Question

当我在16位汇编两个值相加，什么是打印结果到控制台的最佳方式？

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---;;

我觉得DL值应为ASCII码，但我不知道如何斧头附加值成ASCII后转换。

Answer 1

您基本上要除以10，打印余数（一个数位），然后与商数重复。

    ; 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

---

作为一个方面说明，你在你的代码中的错误就在这里：

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.

您把2在ah，然后你把ax在dl。你可以在打印前基本废弃，另外添购ax。

您还具有大小不匹配，因为dl为8个位宽并ax为16个位宽。

你应该做的是翻转的最后两行并修正尺寸不匹配：

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.

Answer 2

刚定影@Nathan费尔曼的代码

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

Answer 3

的基本算法是：

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

请注意，这将产生反顺序的数字，所以你可能会想更换的东西，卖场每个数字的“EMIT”的步骤，这样就可以在以后迭代反向覆盖已存储的数字。

而且，注意，为二进制数转换0和9（十进制）之间为ascii，只需添加ascii码为“0”（这是48）的数量。

Answer 4

mov dl, ax

这将不会作为dl和ax具有不同位大小工作。你想要做的是创造一个你除以10的16位值环，记得在堆栈上休息，然后继续整数除法结果的循环。当到达0的结果，清理由数字堆栈位，加入48到数字将它们转换成ASCII数字，然后打印出来。