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Chapter 2 Representing and Manipulating Information

Prof. Qi TianCS 3843 Fall 2013

http://www.cs.utsa.edu/~qitian/CS3843/

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Summary of Lectures

• 09-30-2013 (Monday)– Section 2.4.4 Rounding Example– In-class Quiz 2 – Reminder: Midterm 1 on Monday Oct. 7, 2013– Practice Problems for Midterm One

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Summary of Lectures

• 09-27-2013 (Friday)– IEEE Rounding Methods– Practice Problem– Quiz 2 next Monday– Midterm 1 on Monday Oct. 7, 2013

• 09-25-2013 (Wednesday)– Examples for IEEE Floating Point Representation

• 09-23-2013 (Monday)– Section 2.4 IEEE Floating Point Representation (cont.)

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Summary of Lectures

• 09-20-2013 (Friday)– Section 2.4 IEEE Floating Point Representation (cont.)

• 09-18-2013 (Wednesday)– Section 2.4 IEEE Floating Point Representation

• 09-16-2013 (Monday)– Section 2.3.1 Unsigned Addition and Unsigned

Subtraction– Section 2.3.2 Two’s Complement Addition– Quiz 1

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Summary of Lectures

• 09-13-2013 (Friday)– Questions on P.9 of Assignment 1– Section 2.3.1 Unsigned Addition– Quiz 1

• 09-11-2013 (Wednesday)– Section 2.1.10 Shift Operations

• Practice Problem 4– Questions on Assignment 1

• 09-09-2013 (Monday)– Practice Problems 1-3

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Summary of Lectures

• 09-06-2013 (Friday)– Section 2.2.3 Representing Negative Numbers• Sign & Magnitude System• Two’s Complement System• One’s Complement System

• 09-04-2013 (Wednesday)– Section 2.1.2-2.1.10• Word size/data size/addressing and byte ordering• Boolean Algebra and Logical Operations in C

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Summary of Lectures

• 08-30-2013 (Friday)– Conversion between decimal and base R number• Integer• Fractional number

• 08-28-2013 (Wednesday)– Syllabus– Information Storage– Conversion between Binary and Hexadecimal Number

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Practice Problem 1 - Bit-level Operations in C

C expression Binary Representation Binary Result Hexadecimal result

~0x41 ~[0100 0001] [1011 1110] 0x BE

~0x00

0x69 & 0x55

0x69 | 0x55

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Practice Problem 1 - Bit-level Operations in C

C expression Binary Representation Binary Result Hexadecimal result

~0x41 ~[0100 0001] [1011 1110] 0x BE

~0x00 ~[0000 0000] [1111 1111] 0x FF

0x69 & 0x55 [0110 1001]&[0101 0101] [0100 0001] 0x 41

0x69 | 0x55 [0110 1001]|[0101 0101] [0111 1101] 0x 7D

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Practice Problem 2 - Boolean Operations (bit-level and logical operation in C)

Expression Value Expression Value

x & y x && y

x| y x || y

~x | ~y !x ||!y

x & !y x && ~y

Suppose x and y have byte values 0x66 and 0x39, respectively. Fill in the following table indicating the byte value of the different C expressions:

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Practice Problem 2 - Boolean Operations (bit-level and logical operation in C)

Expression Value Expression Value

x & y 0x20 x && y 0x 1

x| y 0x7F x || y 0x 1

~x | ~y 0x DF !x ||!y 0x 0

x & !y 0x 0 x && ~y 0x 1

Suppose x and y have byte values 0x66 and 0x39, respectively. Fill in the following table indicating the byte value of the different C expressions:

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Practice Problem 3 Representing Negative Numbers

Q1. Using a 8-bit word, find the binary representation of -27.

a) Using Sign and Magnitude System

b) Using 1’s Complement System

c) Using 2’s Complement System

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Practice Problem 3 Representing Negative Numbers

Q1. Using a 8-bit word, find the binary representation of -27.

a) Using Sign and Magnitude System N = 27 = 0001, 1011 -27 = 1001,1011b) Using 1’s Complement System N = 27 = 0001, 1011 = 1110, 0100c) Using 2’s Complement System

N* = 1110, 0101

N

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Practice Problem 3 Representing Negative Numbers

Q2. Using a 12-bit word, find the binary representation of -27.

a) Using Sign and Magnitude System

b) Using 1’s Complement System

c) Using 2’s Complement System

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Practice Problem 3 Representing Negative Numbers

Q2. Using a 12-bit word, find the binary representation of -27.

a) Using Sign and Magnitude System N = 27 = 0000, 0001, 1011 -27 = 1000, 0001,1011 (not sign extension from 8 bit)b) Using 1’s Complement System N = 27 = 0000, 0001, 1011 = 1111,1110, 0100 (sign extension from 8 bit)c) Using 2’s Complement System

N* = 1111, 1110, 0101 (sign extension from 8 bit)

N

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Section 2.2.1 Integer Representation

C data type Bytes Minimum Maximumchar 1 -128 127

unsigned char 1 0 255short (int) 2 -32,768 32,767unsigned short (int) 2 0 65,535int 4 -2,147,483,648 2,147,483,647unsigned (int) 4 0 4,294,967,295

long (int) 4 -2,147,483,648 2,147,483,647unsigned long (int) 4 0 4,294,967,295long long (int) 8 -9,223,372,036,854,775,808 9,223,372,036,854,775,807Unsigned long long (int) 8 0 18,446,744,073,709,551,615

Typical range for C integral data type on 32-bit machine

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Section 2.2 Integer Representation

Typical range for C integral data type on 64-bit machine

C data type Bytes Minimum Maximumchar 1 -128 127unsigned char 1 0 255short (int) 2 -32,768 32,767unsigned short (int) 2 0 65,535int 4 -2,147,483,648 2,147,483,647unsigned (int) 4 0 4,294,967,295long (int) 8 -9,223,372,036,854,775,808 9,223,372,036,854,775,807

unsigned long (int) 8 0 18,446,744,073,709,551,615long long (int) 8 -9,223,372,036,854,775,808 9,223,372,036,854,775,807

Unsigned long long (int) 8 0 18,446,744,073,709,551,615

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Section 2.1.5 ASCII Code

• A character is usually represented as a single byte by using the ASCII code.

• Strings are represented as arrays of characters terminated by the null character.

• ASCII – American Standard Code for Information Interchange– 7-bit code (128 ASCII Characters)– Some properties:

• Codes for digits are consecutive: ‘0’ = 48, ‘1’ =49, etc.• Codes for upper case letters are consecutive: ‘A’=65, ‘B’=66, etc.• Codes for lower case letters are consecutive: ‘a’=97, ‘b’=98, etc.• Maximum value is 127.

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ASCII Code

• A compact table in hex and decimal

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Section 2.1.10 Shift Operations

• X =[xn-1,xn-2,…, x1, x0]• Left shift: x << k (C expression)– Result: [xn-k-1, xn-k-2, …, x0, 0,…, 0]– Dropping off the k most significant bits, and filled the right

end with k zeros• Right shift: x >> k (C expression)– Logical shift: x >>L k

• Result: [0,…,0,xn-1, xn-2, …, xk]

– Arithmetic shift: x >>Ak• Result: [xn-1,…, xn-1, xn-1, …, xk]

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Section 2.1.10 Shift Operations

• x << k is equivalent to multiply by 2k, x*2k

• x >>A k is equivalent to divide by 2k , x/2k

• k < 32 for integer x• Many C compilers perform arithmetic right

shifts on negative values in which the vacated values are filled with the sign bit.

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Section 2.3.6 – Multiplying by constants

• A left shift by k bits is equivalent to multiplying by 2k.

• Using addition if a small number of 1 bits x * 49 = x * [110001]= x*[32 + 16 + 1] = x * [25+24+20] = (x*25)+(x*24)+(x*20)= (x<<5) + (x<<4) +x• Using subtraction if a large number of 1 bits in a row x * 78 = x*[1001110] = x*[26+24-2] = (x<<6)+(x<<4)-(x<<1)

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Practice Problem 4• For each of the following values of K, find ways to express x *K using only

the specified number of operations, where we consider both addition and subtractions to have comparable cost.

K Shifts Add/Subs Expression

6 2 1

31 1 1

-6 2 1

55 2 2

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Section 2.2.2 Unsigned Encodings

• There is only one standard way of encoding unsigned integers.

• Bit vector x = [xw-1, xw-2, …, x1, x0] with w bits• Binary to unsigned number:

B2Uw(x)=2w-1xw-1+2w-2 xw-2++21 x1+20 x0

• Each integer between 0 and 2w-1 has a unique representation with w bits.

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Section 2.2.2 Unsigned Encodings

• Examples:B2U4([0011])=0x23+0x22+1x21+1x20=3

B2U4([1011])=1x23+0x22+1x21+1x20=11

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Section 2.3.1 Unsigned Addition

• w bits, maximum value is 2w-1.• It might take w+1 bits to represent the value of

x+y.• Addition is done modulo 2w.• When x+y does not produce the correct result, we

say that overflow has occurred.• In C, overflow is not detected.• How can you test whether adding x and y will

produce an overflow?

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Section 2.3.1 Unsigned Addition

• What is

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Section 2.3.1 Unsigned Addition

• What isSol:

139+147>28 therefore =(139+147)-28=30

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Unsigned Addition and Substraction

Unsigned Addition

Unsigned Substraction

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Section 2.4 IEEE Floating Point

Number of bits

Precision s exp frac total

single 1 8 23 32

double 1 11 52 64

extended 1 15 64 80

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Example: A 6-bit format

• What is the bias: 22 -1 = 3. • How many different values can be represented with 6 bits: 26 = 64. • How many of these are NaN: 6 • How many of these are infinity: 2 • How many of these are positive, normalized: 6×4 = 24 • How many of these are negative, normalized: 6×4 = 24 • How many of values are zero (denormalized): 2 • How many of these are denormalized > 0: 3 • How many of these are denormalized < 0: 3

s exp (3 bits) frac (2 bits)

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A 6-bit format (continued)• What are the positive normalized values?

s = 0; exp = 1, 2, 3, 4, 5, or 6frac = 00, 01, 10, or 11, corresponding to 1.00, 1.01, 1.10, and 1.11 V = 1.frac * 2exp – 3

• Smallest positive normalized number: 0.25

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A 6-bit format (continued)• Denormalized values: M = frac * 2-2 = frac/4: 0,

.25, .5, .75value = M 2-2 = M/4.The values are 0, .0625, 0.125, and 0.1875Denormalized spacing: 0.0625

• Largest denormalized number: 0.1875

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A 6-bit format (continued)• Denormalized values: M = frac * 2-2 = frac/4: 0,

.25, .5, .75value = M × 2-2 = M/4.The values are 0, .0625, 0.125, and 0.1875Denormalized spacing: 0.0625

• Largest denormalized number: 0.1875

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A 6-bit format (continued)

•The denormalized values are equally spaced: spacing is 0.0625.•The normailzed values are not equally spaced.

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Example: Rounding

Mode 1.4 1.6 1.5 2.5 -1.5Round-to-evenRound-towards-zero

Round-downRound-up

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Example: Rounding

Mode 1.4 1.6 1.5 2.5 -1.5Round-to-even 1 2 2 2 -2Round-towards-zero 1 1 1 2 -1

Round-down 1 1 1 2 -2Round-up 2 2 2 3 -1

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Rounding Example

• American Patriot Missile Battery in the first Gulf-War, Feb. 25, 1991, failed to intercept an incoming Iraqi Scud missile.

• Disaster Result: 28 soldiers killed• Failure Analysis by the US General Accounting

Office (GAO):– The underlying cause was an impression in a

numeric calculation.