Question

我想在可启动的程序集文件中分别打印两个十六进制数字。然而，打印第二个数字似乎将它添加到第一个数字。正如一些人所指出的那样，问题可能出在print_hex部分。

这是我的代码：

[org 0x7c00]

  mov dx, [number1]
  call print_hex  

  mov bx, space
  call print_string

  mov dx, [number2]
  call print_hex

  jmp $                 ; Hang once we're done

print_hex:
  pusha             ; save the register values to the stack for later

  mov cx,4          ; Start the counter: we want to print 4 characters
                    ; 4 bits per char, so we're printing a total of 16 bits

char_loop:
  dec cx            ; Decrement the counter

  mov ax,dx         ; copy bx into ax so we can mask it for the last chars
  shr dx,4          ; shift bx 4 bits to the right
  and ax,0xf        ; mask ah to get the last 4 bits

  mov bx, HEX_OUT   ; set bx to the memory address of our string
  add bx, 2         ; skip the '0x'
  add bx, cx        ; add the current counter to the address

  cmp ax,0xa        ; Check to see if it's a letter or number
  jl set_letter     ; If it's a number, go straight to setting the value
  add byte [bx],7   ; If it's a letter, add 7
                    ; Why this magic number? ASCII letters start 17
                    ; characters after decimal numbers. We need to cover that
                    ; distance. If our value is a 'letter' it's already
                    ; over 10, so we need to add 7 more.
  jl set_letter

set_letter:
  add byte [bx],al  ; Add the value of the byte to the char at bx

  cmp cx,0          ; check the counter, compare with 0
  je print_hex_done ; if the counter is 0, finish
  jmp char_loop     ; otherwise, loop again

print_hex_done:
  mov bx, HEX_OUT   ; print the string pointed to by bx
  call print_string

  popa              ; pop the initial register values back from the stack
  ret               ; return the function

print_string:     ; Push registers onto the stack
  pusha

string_loop:
  mov al, [bx]    ; Set al to the value at bx
  cmp al, 0       ; Compare the value in al to 0 (check for null terminator)
  jne print_char  ; If it's not null, print the character at al
                  ; Otherwise the string is done, and the function is ending
  popa            ; Pop all the registers back onto the stack
  ret             ; return execution to where we were

print_char:
  mov ah, 0x0e    ; Linefeed printing
  int 0x10        ; Print character
  add bx, 1       ; Shift bx to the next character
  jmp string_loop ; go back to the beginning of our loop

; global variables
  HEX_OUT: db '0x0000',0
  number1: dw 1
  number2: dw 2
  space: db " ",0

; Padding and stuff
  times 510-($-$$) db 0
  dw 0xaa55

它给出了输出：

0x0001 0x0003

我期待输出：

0x0001 0x0002

编辑：更新了代码和问题陈述，希望使其更加完整和可验证。

Answer 1

因为 print_hex 例程在 HEX_OUT 数据上使用add byte [bx],al，所以它必然取决于 HEX_OUT 的初始内容。 / p>

每次使用时只需重置内容：

print_hex:
 pusha 
 mov ax, "00"
 mov [HEX_OUT+2], ax
 mov [HEX_OUT+4], ax
 ...

HEX_OUT: db '0x0000',0

汇编：打印时无意中添加十六进制数

1 个答案: