linux - x86 汇编字符串缓冲区编号到 ASCII

Question

我正在编写一个 x86 汇编程序来输出一个十六进制数。该程序是使用 nasm 组装的，图像文件由 qemu 运行。该程序的行为让我很困惑。正如下面的工作程序所建议的那样，我不必将 0x30 添加到数字就可以打印该数字的字符。

    ; Boot sector code offset: 0x7c00
[org 0x7c00]

    mov dx, 0x1fb6 ; The hexadecimal to be printed
    call print_hex ; call the function
    jmp $ ; jump infinitely

%include "print_string.asm" ; Include the print_string function

print_hex:
    pusha ; push all registers to stack
    mov ax, 0x4 ; rotate through the number four times
print_hex_loop:
    cmp ax, 0x0 ; compare the counter with 0
    jle print_hex_end ; if it is zero then jump to the end  
    mov cx, dx ; move dx to cx
    and cx, 0x000F ; take the lower four binary digits of cx
    cmp cx, 0xa ;compare the digits with 0xa
    jge print_hex_letter ; if it is larger than a, jump to printing character
    add cx, 0x0 ; otherwise print the ascii of a number
    jmp print_hex_modify_string ; jump to routine for modifing the template
print_hex_letter:
    add cx, 0x7 ; print the ascii of a letter
print_hex_modify_string:
    mov bx, HEX_OUT ; bring the address of HEX_OUT into dx
    add bx, 0x1 ; skip the 0x
    add bx, ax ; add the bias
    add byte [bx], cl ; move the character into its position
    shr dx, 4 ; shift right 4 bits
    sub ax, 0x1 ; subtract 1 from the counter
    jmp print_hex_loop ; jump back to the start of the function
print_hex_end:
    mov bx, HEX_OUT ; move the address of HEX_OUT to bx
    call print_string ; call the function print_string
    popa ; pop all registers from stack
    ret ; return to calling function

HEX_OUT:
    db '0x0000',0 ; The template string for printing

    times 510-($-$$) db 0 ; fill zeros
    dw 0xaa55 ; MAGIC_FLAG for boot

boot_sect.asm

print_string:
    pusha
    mov ah, 0x0e
    mov al, [bx]
print_string_loop:
    cmp al, 0x0
    je print_string_end
    int 0x10
    add bx, 0x1
    mov al, [bx]
    jmp print_string_loop
print_string_end:
    popa
    ret

打印字符串.asm

该程序的输出符合我的预期，但是当我尝试在数字上添加 0x30 以获取数字的 ASCII 代码时，输​​出是乱码。是不是有什么技巧，还是我遗漏了一些关键点？

谢谢!

Answer 1

原始问题的答案:

因为你做add byte [bx], cl将数字写入缓冲区，而缓冲区已经包含'0'，所以第一次它会正常工作.第二次调用 print_hex 将再次产生乱码，因为 HEX_OUT 内容已经被修改(琐事:第一次打印哪个十六进制数也可以正确打印第二个值？ ).

---

现在只是为了好玩，我添加了我可能如何为自己做print_hex。也许它会为您的 x86 ASM 编程提供额外的想法，我试图对其进行大量评论以解释我为什么以我正在做的方式做事:

首先我会分离格式化函数，这样我最终可以在别处重用它，所以输入既是数字又是目标缓冲区指针。我使用 LUT(查找表)进行 ASCII 转换，因为代码更简单。如果您关心大小，可以在代码中以更少的字节进行分支并使用较慢的 pusha/popa 来保存寄存器。

format_hex:
    ; dx = number, di = 4B output buffer for "%04X" format of number.
    push    bx              ; used as temporary to calculate digits ASCII
    push    si              ; used as pointer to buffer for writing chars
    push    dx
    lea     si,[di+4]       ; buffer.end() pointer
format_hex_loop:    
    mov     bx,dx           ; bx = temporary to extract single digit
    dec     si              ; si = where to write next digit
    and     bx,0x000F       ; separate last digit (needs whole bx for LUT indexing)
    shr     dx,4            ; shift original number one hex-digit (4 bits) to right
    mov     bl,[format_hex_ascii_lut+bx]    ; convert digit 0-15 value to ASCII
    mov     [si],bl         ; write it into buffer
    cmp     di,si           ; compare buffer.begin() with pointer-to-write
    jb      format_hex_loop ; loop till first digit was written
    pop     dx              ; restore original values of all modified regs
    pop     si
    pop     bx
    ret
format_hex_ascii_lut:       ; LUT for 0-15 to ASCII conversion
    db      '0123456789ABCDEF'

然后为了方便起见，也可以添加一个 print_hex 函数，提供它自己的缓冲区以使用“0x”和 nul 终止符进行格式化:

print_hex:
    ; dx = number to print
    push    di
    push    bx
    ; format the number
    mov     di,HEX_OUT+2
    call    format_hex
    ; print the result to screen
    lea     bx,[di-2]       ; bx = HEX_OUT
    ; HEX_OUT was already set with "0x" and nul-terminator, otherwise I would do:
    ; mov       word [bx],'0x'
    ; mov       byte [bx+6],0
    call    print_string
    pop     bx
    pop     di
    ret
HEX_OUT:
    db      '0x1234',0      ; The template string for printing

最后是引导代码中的示例用法:

    mov     dx,0x1fb6       ; The hexadecimal to be printed
    call    print_hex
    mov     dx,ax           ; works also when called second time
    call    print_hex       ; (but would be nicer to print some space between them)

    jmp     $               ; loop infinitely

(我确实在一定程度上验证了这段代码(它会编译和运行)，尽管只是通过它的不同部分并且在 32b 环境中(修补几行使其成为 32b)，所以一些错误可能已经溜进去了。我没有 16b 环境来验证它是否是完整的引导代码。)