在8086中如何生成物理地址? [英] How is a physical address generated in 8086?

问题描述

在 8086 架构中，内存空间的大小为1 MiB，并分成逻辑段，每个逻辑段最多64 KiB.

In the 8086 architecture, the memory space is 1 MiB in size and divided into logical segments of up to 64 KiB each.

即它具有 20 个地址行，因此使用以下方法:

i.e. it has 20 address lines thus the following method is used:

数据段寄存器左移4位，然后添加到偏移量寄存器

That the data segment register is shifted left 4 bits then added to the offset register

我的问题是:尽管所有寄存器只有16位，我们如何进行移位操作

My question is: How we do the shift operation although all the registers are only 16 bits

推荐答案

地址转换是由一个特殊的单元在内部完成的，而无需使用用户代码可用的寄存器来存储中间结果-它仅获取16位值并进行转换内部-不会在用户代码可以观察到的任何地方反映出来.

Address translation is done internally by a special unit without using the registers available to user code to store intermediate results - it just fetches 16-bit values and does the translation inside - it is not reflected anywhere where the user code could observe it.

这篇关于在8086中如何生成物理地址?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！