68000 汇编 – 通过堆栈传递字符串连接参数

我正在开发一个Motorola 68000汇编程序，该程序使用子例程连接两个字符串。挑战在于通过堆栈实现输入和输出的参数传递，因此我专注于正确设置和恢复堆栈。

我在Sep Roland和Erik Eidt的帮助下开发了程序逻辑。之后，我研究了如何使用堆栈传递参数，这就是为什么我的代码有大量注释的原因。

任务要求：

• 用68000 汇编语言实现一个使用堆栈传递参数的子程序。
• 该子程序接受两个输入字符串：
  • A ="Hello"
  • B ="World"
• 它将它们连接成输出字符串C，结果为：
  • C ="HelloWorld"
• 主程序应该：
  1. 通过推送参数来准备堆栈。
  2. 调用子程序。
  3. 函数返回后正确恢复堆栈。

我的实现：

          ORG $8000
   
;DATA
StringA DC.B 'Hello',0    ; First string with a null terminator
StringB DC.B 'World',0    ; Second string with a null terminator
StringC DS.B 256          ; Buffer for the concatenated string 

START: 

; The stack pointer (A7) starts at address $8000. 
; In the 68000 architecture, A7 always points to the memory address where 
; the next value will be saved (push operation).

      pea.l StringC ; Equivalent to [move.l #StringC, -(a7)]
                    ; The stack pointer (A7) is decremented by 4 (pushing a longword = 4 bytes)
                    ; Initial A7 = $8000, now A7 = $7FFC
      
      pea.l StringB  ; A7 = $7FF8
      pea.l StringA  ; A7 = $7FF4

; Therefore, the stack (from lowest to highest address) contains:
; A7 = $7FF4  |StringA address| 
; A7 = $7FF8  |StringB address| 
; A7 = $7FFC  |StringC address| 
; A7 = $8000 (original SP value before the push operations)

      bsr.s CopyStrings     ; Call the first subroutine, saving the PC (Program Counter)
                            ; onto the stack
                                
; When executing bsr.s, the processor:
; - Saves the return address (PC) on the stack (another 4 bytes subtracted from A7).
; - Then branches to CopyStrings.

; Upon returning from the subroutine (rts), the stack pointer A7 will remain 
; where the subroutine left it. However, we need to clean up the three parameters 
; (StringA, StringB, StringC) that we previously pushed.

      addq.l #8,a7  ; Restore 8 bytes of the stack
      addq.l #4,a7  ; Restore the remaining 4 bytes (total 12 bytes)

      SIMHALT 

CopyStrings:
      ; At the entry of the subroutine, the stack looks like this:
      ; A7    |Return Address | 
      ; A7+4  |StringA Address| 
      ; A7+8  |StringB Address| 
      ; A7+12 |StringC Address|
      
      move.l 4(a7),a0  ; Retrieve the address of StringA 
      move.l 8(a7),a1  ; Retrieve the address of StringB
      move.l 12(a7),a2 ; Retrieve the address of StringC 
      
CopyA: 
      move.b (a0)+,(a2)+  ; Load a character from StringA into StringC
                          
      bne.s CopyA         ; If the character is not null, continue copying
      subq.l #1,a2        ; Move back 1 byte to overwrite the null terminator

CopyB:
      move.b (a1)+,(a2)+  ; Load a character from StringB into StringC
      bne.s CopyB         ; If the character is not null, continue copying
      rts                 ; Return from subroutine
    
     END START

问题：

1. 我通过堆栈传递参数的方法正确吗？
2. 我应该考虑哪些优化或最佳实践？

任何反馈都将不胜感激！