我对A20线校验码的理解是否正确？

Question

我正在按照本教程了解如何检查和启用 A20 线路。我想我明白了，但是有人可以帮我澄清一下吗？

该教程中已有的评论开始; <comment>，
我的评论开始;<comment>

; The following code is public domain licensed

[bits 16]

; Function: check_a20
;
; Purpose: to check the status of the a20 line in a completely self-contained state-preserving way.
;          The function can be modified as necessary by removing push's at the beginning and their
;          respective pop's at the end if complete self-containment is not required.
;
; Returns: 0 in ax if the a20 line is disabled (memory wraps around)
;          1 in ax if the a20 line is enabled (memory does not wrap around)

check_a20:
    pushf                                  ;Backup the current flags onto the stack
                                           ;Backup the below registers onto the stack
    push ds                                ;|
    push es                                ;|
    push di                                ;|
    push si                                ;-----

    cli                                    ;Disable interupts

    xor ax, ax                             ; ax = 0
    mov es, ax                             ;es = ax

    not ax                                 ; ax = 0xFFFF
    mov ds, ax                             ; ds = ax

    mov di, 0x0500                         ;Boot signature part one (0x55)
    mov si, 0x0510                         ;Boot signature part two (0xAA)

    mov al, byte [es:di]                   ;al = value at AA:55
    push ax                                ;Backup ax register onto the stack

    mov al, byte [ds:si]                   ;al = value at 55:AA
    push ax                                ;Backup al onto the stack

    mov byte [es:di], 0x00                 ;Memory location AA:55 = 0
    mov byte [ds:si], 0xFF                 ;Memory location at 55:AA = 0xFF

    cmp byte [es:di], 0xFF                 ;Does value at AA:55 = 0xFF? If so, this means A20 is disabled

    pop ax                                 ;Restore saved ax register
    mov byte [ds:si], al                   ;Set 55:AA to al

    pop ax                                 ;Restore ax register
    mov byte [es:di], al                   ;set AA:55 to al

    mov ax, 0                              ;Return status of this function = 0 (Disabled)
    je check_a20__exit                     ;A20 is disabled. Go to check_a20__exit

    mov ax, 1                              ;Return status of this function = 1 (Enabled)

check_a20__exit:
                                           ;Backup registers
    pop si
    pop di
    pop es
    pop ds
    popf                                   ;Backup flags

    ret                                    ;Return

如果我不明白某些部分，您能解释一下原因吗？

Answer 1

该代码通过写入两个地址来检查FFFF:0510和是否0000:0500引用相同的地址，并查看写入一个地址是否会覆盖另一个地址。

事实证明，FFFF:0510可以表示线性地址0x100500而不是0x500，但前提是 A20 已启用。es:di因此，代码将全零写入(aka )处的字节0000:0500，并将全 1 写入ds:si(aka FFFF:0510) 处的字节。如果启用 A20，则两个段：偏移对引用不同的地址，第一个写入将保留，并且[es:di]将包含零。否则，这两对引用相同的地址，第二次写入将破坏第一个，并且[es:di]将包含0xff.

（顺便说一句，0x55和0xAA不是其中的一部分；我不确定你从哪里得到这些数字。启动签名通常位于 0x7dfe，IIRC。）