多少个字,可以在地址空间? [英] How many words can be in the address space?

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问题描述

下面是我的工作。

的问题::一种高速工作站有64位字和字节级的地址解析64位地址。有多少话可以在工作站的地址空间?

The Problem: A high speed workstation has 64 bit words and 64 bit addresses with address resolution at the byte level. How many words can in be in the address space of the workstation?

我在这个问题中定义的不同术语

I defined the different terms in the problem

         
  • 字长 - 数据处理器自然单位。字大小决定的信息,可以一次过处理的金额
  •       
  • 字节级寻址 - 支持一个字内访问各个字节硬件架构
  •        
  • <一个href="http://superuser.com/questions/168114/how-much-memory-can-a-64bit-machine-address-at-a-time">64位寻址 - 你已经有64位以指定保存运行时内存地址指令或数据
  •        
  • 地址空间 - 运行内存的程序的视图系统
  • Word Size - Processor natural unit of data. The word size determines the amount of information that can be processed in one go
  • Byte Level Addressing - Hardware architectures that support accessing individual bytes within a word
  • 64 Bit Addressing - You have have 64 bits to specify an address in Runtime memory that holds an instruction or data
  • Address Space - Running program's view of memory in the system

您将如何使用这些定义来解决这个问题?

How would you go about using all these definitions to solve this problem?

这是64位的,我知道,在技术上有在内存中,从64位字2 ^ 64的位置,一个处理器处理8个字节的时间。但我不知道如何使用这些信息来推断有多少话是在电脑的地址空间。

From 64 bits, I know that technically there are 2^64 locations in memory and from 64 bit words, that a processor processes 8 bytes a time. But I don't know how to use that information to conclude how many words are in the address space of the computer.

推荐答案

由于aruisdante的评论,我能想出解决办法。

Thanks to aruisdante's comment, I was able to figure this out.

基本上64位地址意味着有2 ^ 64共地址。因为字节寻址存储器用在这里,每个地址可以存储一个字节。

Basically 64 bit addresses means there are 2 ^ 64 total addresses. Because byte addressable memory is used here, each address will store one byte.

这意味着在总,在地址空间中,2 ^ 64字节可以存储。问题告诉你的是,该机拥有64位字或每个字是8个字节长。因此,必须在地址空间2 ^ 64/8或2 ^ 64/2 ^ 3 = 2 ^ 61字。

This means that in total, in the address space, 2 ^ 64 bytes can be stored. The problem tells you that the machine has 64 bit words or that each word is 8 bytes long. Therefore you have 2^64/8 or 2^64/2^3 = 2^61 words in the address space.

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