Chapter 2 Bits, Data Types and Operations

Abstract

Covered in Lecture 1 2022.7.11 and Lecture 2 2022.7.12
Topics:
1. the bit binary digit codes
2. data type(unsigned integer, signed integer, logical variable, float number, ASCII)

n bits, can represent \(2^n\) numbers, ranging from 0 to \(2^{n-1}\)

Bits, Bytes

• bit: only 1/0
• byte: 1 byte = 8 bits

Data Type

Integer

Unsigned Integer

n bits, can represent \(2^n\) numbers. range: \([0,2^{n-1}]\)

Signed Integer

• signed-magnitude(原码): a leading 0 signifies a positive integer, a leading 1 signify a negative integer. In a 4-bit example,\(0110=6,\ 1110=-6\),can represent \([-7, 7]\).
  It has the problem of "positive zero" and "negative zero".

• 1's Complement(反码): For a non-negative number, its opposite number is obtained after bitwise inversion. In a 4-bit example, \(0110=6,\ 1001=-6\), can represent [-7, 7], also has the problem of "positive zero" and "negative zero".
  It has also the problem above.

• 2's Complement(补码): The highest bit of the 2's complement is the sign bit. The sign bit is 0 for non-negative numbers, and 1 for negative numbers. For an n-bit signed number, the weight of the sign bit is \(-2^{n-1}\). range: \([-2^{n-1},2^{n-1}-1]\).
  Obtain 2's Complement: 按位取反再加一

• Extension

• sign extension: Fill sign bit when extending.(不会改变数字大小)

• zero extension: Fill 0 when extending.

  Example

  两个数 0100 1100 和 1011.
  0100 1100(76) + 1011(-5) = 11111011
  但 11111011 != 71
  因为两个数字长度不同，要对1011符号扩充为 1111 1011(-5)，再相加即可。

• Overflow
  The only possible overflow situations:

• positive + positive == negative, that is, carry to the sign bit, and the sign bit becomes 1 after adding

• negative + negative == positive，the sign bit becomes 0 after adding.

• Conversion between binary and decimal
  用2乘十进制小数，将积的整数部分取出，再用 2 乘余下的小数部分，再将积的整数部分取出，如此直到积中的整数部分为 0/1，此时 0/1 为二进制的最后一位。或者达到所要求的精度为止。
  然后把取出的整数部分按顺序排列起来，先取的整数作为二进制小数的高位有效位，后取的整数作为低位有效位。

Logical Variables

• bit vector: number string of 0 or 1.

• functions: AND, OR, Exclusive-OR (XOR), Equivalence, NAND, NOR

  Info

  \(X\cdot Y\Leftrightarrow X\ AND\ Y\)
  \(X+Y\Leftrightarrow X\ OR\ Y\)
  \(X\oplus Y\Leftrightarrow X\ XOR\ Y\)
  \(\overline X \Leftrightarrow NOT\ X\)
  Be careful of the diffrence of +(AND) and +(plus).

Floating Point

	S	exp	frac
Float	1	8	23
Double	1	11	52

Normalized form: \(N=(-1)^S\times M\times 2^E\)

• S: sign. \(S=1\) indicates the number is negative.
• M: 尾数. Normally, \(M=1.frac\).
• E: 阶码. Normally, \(E=exp-Bias\) where \(Bias=127\) for floating point numbers. \(Bias = 1023\) for double.

Note

• 当 \(exp=0\) 时, 规定 \(M=0.frac\).
  其中 \(frac=0\) 时, 表示的数字为 0.(可能有 +0/-0)
• 当 \(exp=1111\ 1111\)时
  • 若 \(frac=0\), 则表示 \(+\inf /\) \(-\inf\).
  • 若 \(frac\neq 0\), 则表示 NaN(Not a number). e.g. \(1/0, \inf/\inf\).

How to represent decimal number in the floating point type?

• Convert decimal to binary.
• Convert binary to floating point.

ASCII Code

Ameican Standard Code for Informationo Interchange.

Hexadecimal Notation

convert 4(resp. 3) binary bits to 1 hexadecimal(resp. octal) bits. \(A=1010,\ B=1011,\ C=1100,\ D=1101,\ E=1110,\ F=1111\)

评论