给定16条地址线(总线)和8位字长的寻址能力是什么? [英] What is the addressability given number of 16 address wires (bus) and 8-bit word size?

问题描述

计算机具有16条地址线(地址总线?)和8位字长.什么是寻址能力?

A computer has 16 address wires (address bus?) and 8-bit word size. What is the addressability?

我发现地址空间是2 ^ 16 = 65536，但是我仍然不知道如何计算可寻址性.

I figured out that address space is 2^16 = 65536, but I still don't know how to calculate addressability.

我知道可寻址性是每个空间占用的字节数，但是如何计算呢?任何帮助将不胜感激，尤其是一些将字长/地址总线与可寻址性相关联的通用公式.

I know addressability is the bytes each space occupies but how do I figure this out? Any help would be appreciated, especially some general formula associating word size/address bus with addressability.

对不起，这个问题太简单了.

I'm sorry if this question is super simple.

推荐答案

由于可以在地址总线上放置2 ^ 16个唯一值，并且希望能够以字节粒度寻址内存，因此每个值都映射到一个字节.

Since there are 2^16 unique values you can put on the address bus, and you want to be able to address your memory with byte granularity, each value maps to a single byte.

如果您的机器始终加载64B高速缓存行，并且您的RAM已设置为从请求的地址传递64B突发，则只需10条地址线即可覆盖相同的64k内存. CPU会在内部整理负载实际需要的字节，而无需放入. (或具有16条地址线，可寻址性为2 ^ 16 * 64B).

If your machine always loaded say 64B cache lines, and your RAM was set up to deliver 64B bursts from a requested address, you'd only need 10 address lines to cover the same 64k of memory. The CPU would sort out which byte the load actually wanted internally, without needing to put the . (Or with 16 address lines, 2^16 * 64B addressability).

如果要使用位可寻址的存储器，字长= 1b，则2 ^ 16位仅为2 ^ 13字节.

If you wanted bit-addressable memory, word-size = 1b, 2^16 bits is only 2^13 bytes.

在现实生活中，内存控制器将地址分为两半传输到DRAM，这将地址线的数量减少了一半.请参阅乌尔里希·德雷珀(Ulrich Drepper)的《每个程序员应该了解的内存》 ，其中提到了这一点，但重点是缓存行为以及如何对其进行优化.

In real life, memory controllers transfer addresses to DRAM in two halves, which cuts the number of address lines in half. See Ulrich Drepper's What Every Programmer Should Know About Memory which mentions that, but focuses on cache behaviour and how to optimize for it.

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