Chap 2 Representing and Manipulating Information (CS:APP)
--------------------------------------------------------------------------------------------------
Author: YuManZI
2014/06/23 1.1-3.5
2014/06/24 3.6-3.8
2014/06/27 4.1
2014/06/28 4.2-4.5
--------------------------------------------------------------------------------------------------
1. Information Storage
1.1 Virtual Memory: a machine-level program views memory as a very large array of bytes.
1.2 Data Sizes (bytes)
32 bit: char 1; short [int] 2; int 4; long [int] 4; long long [int] 8; char *(any pointer) 4; float 4; double 8;
64 bit: char 1; short [int] 2; int 4; long [int]8; long long [int] 8; char *(any pointer)8; float 4; double 8;
Contents within square brackets [] are optional. The main difference between 32 bit and 64 bit machines are following two points: a) different sizes of data type long; b) different sizes of pointers.
1.3 Byte Ordering: big endian & little endian. (Takinh 0x01234567 as example.)
Big endian, the most significant byte comes first(lower address), bytes from low address to high address are 01 23 45 67, respectively;
Little endian, the least significant byte comes first(lower address), bytes from low address to high address are 67 45 23 01, respectively.
1.4 Shift Operations in C
Left shift << k: dropping off the k most significant bits and filling the right end with k zero;
Logical right shift >>k: dropping off the k least significant bits and filling the lest end with k zero;
Arithmetic right shift >>k: dropping off the k least significant bits and filling the lest end with k themost significant bit.
Arithmetic right shift uses the most significant bit as filling unit, because of the two's complement representation of negative integers. In some languages(e.g. java), the number of shifting bits can never be more than bit sizes of data types.
2. Integral Data Types
2.1 Unsigned Encodings (w bits, x=[x_(w-1), x_(w-2),...,x_0])
B2U(x)=sigma{i=[0,w-1]}(x_i*2^i)
B2U means Bits to Unsigned
It can represent integers between [0..2^w-1]
2.2 Two's-Complement Encodings (same setting as 2.1)
B2T(x)=-x_(w-1)*2^(w-1) + sigma{i=[0,w-2]}(x_i*2^i)
B2T means Bits to Two's
It's a signed encoding, can represent integers between [-2^(w-1), 2^(w-1)-1], the difference between it and Unsigned encoddings are the weight of the significant bit, i.e. positive for unsigned and negative for two's-complement.
2.3 Conversions
Signed<-->Unsigned with identical size: the key is to keep bit representation stable;
Large size-> Small size with same type of signed or unsigned: truncate directly;
Chap 2 Representing and Manipulating Information (CS:APP)的更多相关文章
- chapter 2: Representing and manipulating information
C allows conversion between unsigned and signed. The rule is that the underlying bit representation ...
- 《CSAPP》读书杂记 - Chapter 2. Representing and Manipulating Information
1. 一段查看地址内容的代码 代码: #include <stdio.h> typedef unsigned char *byte_pointer; void show_bytes(byt ...
- 深入理解计算机系统 (CS:APP) Lab2 - Bomb Lab 解析
原文地址:https://billc.io/2019/04/csapp-bomblab/ 写在前面 CS:APP是这学期的一门硬核课程,应该是目前接触到最底层的课程了.学校的教学也是尝试着尽量和CMU ...
- CS:APP配套实验 Data Lab
刚刚完成注册博客,想写一篇随笔,方便以后自己回顾.如果恰好也能帮助到你,是我的荣幸. 这次随笔是记载我的计算机系统(CS:APP,Computer Systems:A Programer's Pers ...
- 图文并茂-超详解 CS:APP: Lab3-Attack(附带栈帧分析)
CS:APP:Lab3-ATTACK 0. 环境要求 关于环境已经在lab1里配置过了.lab1的连接如下 实验的下载地址如下 说明文档如下 http://csapp.cs.cmu.edu/3e/at ...
- ubuntu12.04 安装CS:APP Y86模拟器
下的第一UBUNTU12.04下Y86模拟器的安装:(參考http://archive.cnblogs.com/a/1865627/ 作适当改动) 1.安装bison和flex词法分析工具 sudo ...
- 《CS:APP》二进制炸弹实验(phase_1-3)
<深入理解计算机系统>第三章的bomb lab,拆弹实验:给出一个linux的可执行文件bomb,执行后文件要求分别进行6次输入,每一次输入错误都会导致炸弹爆炸,程序终止.需要通过反汇编来 ...
- 深入理解计算机系统 (CS:APP) 缓冲区漏洞实验 – Buffer Lab 解析
原文地址:https://billc.io/2019/05/csapp-cachelab/ 写在前面 这是 CSAPP 官网上的第 4 个实验 buflab,也是学校要求的第三个实验.这个实验比上一个 ...
- 信息的表示和处理 及 CS:APP 15213 datalab
信息的表示和处理 在通用计算机中中,字节作为最为最小 的可寻址的内存单元,而不是访问内存中单独的位. 寻址和字节顺序 big endian (大端法),数据最高字节部分地址在地址处,和人的感觉逻辑相似 ...
随机推荐
- [BZOJ 1177] Oil
Link:https://www.lydsy.com/JudgeOnline/problem.php?id=1177 Solution: 相当于将大矩形分为3块,取每块中最大的正方形 对于此类分成几块 ...
- POJ 2315:Football Game(博弈论)
[题目链接] http://poj.org/problem?id=2315 [题目大意] 两名球员轮流从N个球中挑出不多于M个射门,每个球半径都是R,离球门S. 每次只能踢出L以内的距离.进最后一个球 ...
- [LOJ6437]PKUSC
旋转多边形是没有前途的,我们考虑旋转敌人,那么答案就是所有人的可行区间长度之和除以$2\pi$ 首先对每个敌人找到那些旋转后会落到多边形上的角度,实际上就是圆和一些线段求交,解方程即可,注意判一下落在 ...
- [UOJ206]Gap
子任务$1$:直接找到最大值后从两边开始找,一步一步从左右到中间确定所有数,调用次数是$\left\lceil\dfrac n2\right\rceil$ 子任务$2$:先找到最大值$mx$和最小值$ ...
- 获取配置文件中的key的value
获取App.config文件的key的value System.Configuration.ConfigurationManager.AppSettings["keyName"] ...
- javascript函数,构造函数。js对象和json的区别。js中this指向问题
函数是由事件驱动的或者当它被调用时执行的可重复使用的代码块.好处:在出现大量程序相同的时候,可以封装为一个function,这样只用调用一次,就能执行很多语句.(1)语法:函数就是包裹在花括号中的代码 ...
- Java计算两个日期相差的天数
import java.text.ParseException; import java.text.SimpleDateFormat; import java.util.Calendar; impor ...
- 使用Python开发SQLite代理服务器(转载)
转载:https://mp.weixin.qq.com/s?timestamp=1498531736&src=3&ver=1&signature=Eq6DPvkuGJi*G5s ...
- [转载]Oracle Merge的使用
FROM: http://zhangqchang.blog.163.com/blog/static/464989732009219114653226/ 摘至网上的几个例子 一.************ ...
- Linux(CentOS)修改IP地址
登陆连接centos系统,输入 ifconfig 可以查看到当前本机的IP地址信息 一 临时修改IP地址: 1.假如查询IP为1.118,输入 ifconfig eth0 (默认是第一个网卡) 后面接 ...