Review Chap 2

• Bits: We represent the presence of voltages as “1” and the absence of a voltage as “0”. We refer to each 0 and each 1 as a “bit”, which is shortened from binary digit.
• Precisely, the electronic circuits in the computer differentiate voltages close to 0 from voltages far from 0.
• With k bits, we can distinguish 2^k distinct items at most.
• Data Types: 1. 2’s complement integers for representing positive and negative integers that we wish to perform arithmetic on, 2. ASCII codes for representing characters that we wish to input to a computer via the keyboard or output from the computer via a monitor(These are two data types which will be used in following chapters)

• Integer data types

  1. Unsigned Integers.（我寻思着这玩意我还是会的）with k bits,we can represent 2^k integers,ranging from 0 to 2^k -1.
  2. Signed Integers. （用中文）正数以0为先导，负数以1为先导，with k bits,we can represent 2^k -1 integers,ranging from - 2^（k -1）-1 to 2^（k -1）-1. 但是出现了-0（如在五位当中，10000表示-0而00000表示0，这就尴尬了）
  3. 1’s complement integers（反码，似乎不重要）
  4. 2’s complement integers （补码，很重要，一会展开讲）

  

• 关于补码的引入，书中从底层实现的角度给出了解释，进行归纳的话我认为主要是这两个因素：① 2's complement 中“溢出”与“取模”的关系；② ALU 模块只进行加法运算，而不能进行减法运算；（需要注意的是，所谓的符号位是给人看的，或者说只是一个特征。在实际过程中并不存在“符号位”，即该位也参与加法计算）

我觉得二进制加法我还不至于不会，这里省了。

• Conversion Binary to Decimal
  1. examine the leading bit b7 to determine whether it is positive or negative.If it is negative, do step 2 after flipping all the bits and adding 1.
  2. then simply: b6 2^6+b5 2^5+…+b0 *2^0
  3. get the result, then if it is negative, don’t forget the minus sign.

• Conversion Decimal to Binary

  1. obtain the binary representation of the magnitude of N by forming the equation N=b6⋅2^6+b5⋅2^5+…+b0⋅2^0. and repeating and following,until the left side of the equation is 0: a.If N is odd,the rightmost bit is 1.If N is even,the rightmost bit is 0. b.Subtract 1 or 0(according to whether N is odd or even)from N,remove the least significant term from the right side,and divide both side sof the equation by 2.
  2. if it is positive, add a leading 0 then finished.
  3. if it is negative, add a leading 0 and then form the negative of this 2’s complement representation, and then you are done.

• Specially for floating point data type

  • 1 bit for the sign (positive or negative)
  • 8 bits for the range (the exponent field)
  • 23 bits for precision (the fraction field) 数据部分，但是组织形式有一些微妙，“第一位”是0还是1有特殊的处理方式

  

  • They are mostly in normalized form: (不需要考虑补码）

    

• exponent=00000000——subnormal numbers

• exponent=11111111——infinity
• Sign-extension (SEXT)
• The value of a positive number does not change if we extend the sign bit 0 as many bit positions to the left as desired. Similarly, the value of a negative number does not change by extending the sign bit 1 as many bit positions to the left as desired.
• 也就是说按照符号位向高位扩展，且这种扩展是不会影响原来的值的。
• Overflow

• 最高位（比如符号位）进位后不会创建一个更高位来存储这个进位信息，而是直接丢弃这个进位，也就是“溢出”，在数学层面来看，就类似于取模，比如，对应1110 + 1001 = ~~1~~0111。