无需段寄存器即可写入地址 [英] Write to address without segment register

问题描述

我知道这段代码实际上会将数据写入ds:[100h]

I know this code will actually write data to ds:[100h]

mov [100h], ax

但是如何在不使用任何段寄存器作为段基的情况下直接写入线性地址100H?

But how can I write to linear address 100H directly without using any segment register for a segment base?

推荐答案

没有办法解决段寄存器.每个内存访问都与某个段寄存器相关.如果要写入绝对地址，请先将适当的段加载到段寄存器中:

There is no way to get around segment register; every memory access is relative to some segment register. If you want to write to an absolute address, first load a segment register with an appropriate segment:

        xor cx, cx
        mov es, cx        ; es = 0000
        mov [es:100h], ax ; [0000:0100] = ax

要在8086或80286系统上加载大于16位的线性地址，请尝试执行以下操作:

To load a linear address larger than 16 bit on an 8086 or 80286 system, try something like this:

addr    dd 0x12345        ; the address we want to load from
        ...
        mov bl, [addr]    ; load low part
        xor bh,bh
        mov cx, [addr+1]  ; load high part
        shl cx, 4         ; adjust high part for segment selector
        mov es, cx        ; load segment register
        mov [es:bx], ax   ; store ax

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