将两个 n 字节整数相加以产生 6502 中的 n 字节答案？

Question

我在 6502 中遇到了另一个加法问题....

我正在尝试将两个 n 字节整数相加以产生 n 字节结果。我不完全确定我是否对 6502 芯片有足够的了解，因此对我当前代码的任何反馈都会非常有帮助。

我知道我应该使用 INX（递增 x 寄存器）和 DEY（递减 y 寄存器），但我不确定操作码的位置。

说明： 使用绝对索引寻址将两个 n 字节整数相加

Adding two n-byte integers using absolute indexed addressing 
The addends start at memory locations $xxxx, $yyyy, answer is at $zzzz
Byte length of the integers is at $AAAA (¢—>256)

START = $0500
              CLC
              ____
loop          LDA      $0400, x 
              ADC      $0410, x
              STA      $0412, x
             ____
             BNE      loop
             BRK

LDA、ADC、STA在循环外（第一次在汇编中使用循环）

编辑：

    Variables

A1 = $0600
B1 = $0700
B2 = $0800
Z1 = $0900

    [START] = $0500

            CLC             18  
            LDX             AE  
            LDY     A1      AC
loop:       LDA     B1, x   BD      
            ADC     B2, x   7D  
            STA     Z1, x   9D      
            INX     E8  
            DEY             88  
            BNE    loop     D0

Answer 1

;Adding two n-byte integers using absolute indexed addressing 
;The addends start at memory locations $xxxx, $yyyy, answer is at $zzzz
;Byte length of the integers is at $AAAA (¢—>256)

        CLC
        LDX #0        ; start at the beginning
        LDY $AAAA     ; load length into Y
loop:   LDA $xxxx, X  ; load first operand
        ADC $yyyy, x  ; add second operand
        STA $zzzz, x  ; store result
        INX           ; go on to next byte
        DEY           ; count how many are left
        BNE loop      ; if more, do more

Answer 2

;------------------------------------------------------;
;      Subroutine for multibyte addition               ;
;------------------------------------------------------;

; All words in Little Endian format
; First word for addition in $A0
; Second word for addition in $B0
; Result word in $C0
; word length in $D0
; Note! push 'A' register to stack before calling subroutine
; Provided below is an example 

    .org $0000

    LDA #$41           ; First word: 779AAA819541
    STA $A0
    LDA #$95
    STA $A1
    LDA #$81
    STA $A2
    LDA #$AA
    STA $A3
    LDA #$9A
    STA $A4
    LDA #$77
    STA $A5

    LDA #$F4           ; Second word: 5ED594552DF4
    STA $B0
    LDA #$2D
    STA $B1
    LDA #$55
    STA $B2
    LDA #$94
    STA $B3
    LDA #$D5
    STA $B4
    LDA #$5E
    STA $B5
    
    LDA #$06           ; Result: D6703ED6C335
    STA $D0
    JSR AddBytes
    BRK

AddBytes:
    LDX #00
    CLC
    PHP
Addloop:
    PLP
    LDA $A0,X
    ADC $B0,X
    STA $C0,X
    PHP
    INX
    CPX $D0
    BNE Addloop
    RTS