在x86汇编中打印十六进制值

Question

我需要制定一个例程，将内存地址转换为字节字符串。然后，该字符串将作为一个函数的输入，该函数将打印以空值结尾的字符串（我已经可以制作出）。例如，如果我的地址为0x1bf9，则需要在屏幕上打印文本“ 1bf9”。这本书还没有进入32位模式，但是有点暗示我们也需要它。这是我到目前为止的内容：

TABLE:
db "0123456789ABCDEF", 0

STRING:
db 0

hex_to_char:
    lea bx, TABLE
    mov ax, dx

    mov ah, al ;make al and ah equal so we can isolate each half of the byte
    shr ah, 4 ;ah now has the high nibble
    and al, 0x0F ;al now has the low nibble
    xlat ;lookup al's contents in our table
    xchg ah, al ;flip around the bytes so now we can get the higher nibble 
    xlat ;look up what we just flipped
    inc STRING
    mov [STRING], ah ;append the new character to a string of bytes
    inc STRING
    mov [STRING], al ;append the new character to the string of bytes

    ret

Answer 1

这会尝试增加文字标签，这是不正确的。同样，您的STRING内存位置仅分配一个字节（字符），而不分配更大的数字来容纳您想要的字符串的大小。

STRING:
    db 0

    inc STRING   ;THIS WON'T WORK
    mov [STRING], ah ;append the new character to a string of bytes
    inc STRING   ;THIS WON'T WORK
    mov [STRING], al ;append the new character to the string of bytes

中性注释：用于xlat的字符表不需要以零结尾。

另外，我建议您保存并恢复一些寄存器，作为asm编程的好习惯。这样，调用函数就不必担心寄存器在其“后面”被更改。最终，您可能想要这样的东西：

TABLE:
    db "0123456789ABCDEF", 0

hex_to_char:
    push bx

    mov   bx, TABLE
    mov   ax, dx

    mov   ah, al            ;make al and ah equal so we can isolate each half of the byte
    shr   ah, 4             ;ah now has the high nibble
    and   al, 0x0F          ;al now has the low nibble
    xlat                    ;lookup al's contents in our table
    xchg  ah, al            ;flip around the bytes so now we can get the higher nibble 
    xlat                    ;look up what we just flipped

    mov   bx, STRING
    xchg  ah, al
    mov   [bx], ax          ;append the new character to the string of bytes

    pop bx
    ret

    section .bss

STRING:
    resb  50                ; reserve 50 bytes for the string

编辑：根据彼得·科德斯的意见进行一些可取的调整。

Answer 2

[请在此页面上查看我的答案，以将EAX中的32位值转换为8个十六进制ASCII字节：Printing out a number in assembly language?