1 binary numbers. 2 main memory main memory holds information such as computer programs, numeric...

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Binary Numbers

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Main Memory • Main memory holds information such as computer

programs, numeric data, or documents created by a word processor.

• Main memory is made up of capacitors. • If a capacitor is charged, then its state is said to be

1, or ON.• We could also say the bit is set.• If a capacitor does not have a charge, then its state

is said to be 0, or OFF.• We could also say that the bit is reset or cleared.

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Main Memory (con’t)

• Memory is divided into cells, where each cell contains 8 bits (a 1 or a 0). Eight bits is called a byte.

• Each of these cells is uniquely numbered.• The number associated with a cell is known

as its address.• Main memory is volatile storage. That is,

if power is lost, the information in main memory is lost.

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Main Memory (con’t)

• Other computer components can

o get the information held at a particular address in memory, known as a READ,

o or store information at a particular address in memory, known as a WRITE.

• Writing to a memory location alters its contents.

• Reading from a memory location does not alter its contents.

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Main Memory (con’t)• All addresses in memory can be accessed in

the same amount of time.• We do not have to start at address 0 and

read everything until we get to the address we really want (sequential access).

• We can go directly to the address we want and access the data (direct or random access).

• That is why we call main memory RAM (Random Access Memory).

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Secondary Storage Media Disks -- floppy, hard, removable (random access) Tapes (sequential access) CDs (random access) DVDs (random access) Secondary storage media store files that contain

computer programs data other types of information

This type of storage is called persistent (permanent) storage because it is non-volatile.

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I/O (Input/Output) Devices Information input and output is handled by I/O

(input/output) devices. More generally, these devices are known as

peripheral devices. Examples:

monitor keyboard mouse disk drive (floppy, hard, removable) CD or DVD drive printer scanner

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Bits, Bytes, and Words A bit is a single binary digit (a 1 or 0). A byte is 8 bits A word is 32 bits or 4 bytes Long word = 8 bytes = 64 bits Quad word = 16 bytes = 128 bits Programming languages use these standard

number of bits when organizing data storage and access.

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Number Systems The on and off states of the capacitors

in RAM can be thought of as the values 1 and 0, respectively.

Therefore, thinking about how information is stored in RAM requires knowledge of the binary (base 2) number system.

Let’s review the decimal (base 10)

number system first.

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The Decimal Number System

• The decimal number system is a positional number system.

• Example: 5 6 2 1 1 X 100 = 1 103 102 101 100 2 X 101 = 20 6 X 102 = 600 5 X 103 = 5000

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The Decimal Number System (con’t)

• The decimal number system is also known as base 10. The values of the positions are calculated by taking 10 to some power.

• Why is the base 10 for decimal numbers?o Because we use 10 digits, the digits 0 through

9.

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The Binary Number System

• The binary number system is also known as base 2. The values of the positions are calculated by taking 2 to some power.

• Why is the base 2 for binary numbers?o Because we use 2 digits, the digits 0 and 1.

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The Binary Number System (con’t)

• The binary number system is also a positional numbering system.

• Instead of using ten digits, 0 - 9, the binary system uses only two digits, 0 and 1.

• Example of a binary number and the values of the positions:

1 0 0 1 1 0 1 26 25 24 23 22 21 20

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Converting from Binary to Decimal

1 0 0 1 1 0 1 1 X 20 = 1 26 25 24 23 22 21 20 0 X 21 = 0 1 X 22 = 4 20 = 1 1 X 23 = 8 21 = 2 0 X 24 = 0 22 = 4 0 X 25 = 0 23 = 8 1 X 26 = 64

24 = 16 7710

25 = 32 26 = 64

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Converting from Binary to Decimal (con’t)

Practice conversions:

Binary Decimal

11101 1010101 100111

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Converting From Decimal to Binary (con’t)

• Make a list of the binary place values up to the number being converted.• Perform successive divisions by 2, placing the remainder of 0 or 1 in each of

the positions from right to left.• Continue until the quotient is zero.

• Example: 4210

25 24 23 22 21 20

32 16 8 4 2 1 1 0 1 0 1 0

42/2 = 21 R = 021/2 = 10 R = 110/2 = 5 R = 05/2 = 2 R = 12/2 = 1 R = 01/2 = 0 R = 1

4210 = 1010102

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Converting From Decimal to Binary (con’t)

Practice conversions:

Decimal Binary

59 82 175

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Working with Large Numbers

0 1 0 1 0 0 0 0 1 0 1 0 0 1 1 1 = ?

• Humans can’t work well with binary numbers; there are too many digits to deal with.

• Memory addresses and other data can be quite large. Therefore, we sometimes use the hexadecimal number system.

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The Hexadecimal Number System The hexadecimal number system is also known as base

16. The values of the positions are calculated by taking 16 to some power.

Why is the base 16 for hexadecimal numbers ? Because we use 16 symbols, the digits 0 and 1 and

the letters A through F.

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The Hexadecimal Number System (con’t)

Binary Decimal Hexadecimal Binary Decimal Hexadecimal

0 0 0 1010 10 A

1 1 1 1011 11 B 10 2 2 1100 12 C 11 3 3 1101 13 D 100 4 4 1110 14 E 101 5 5 1111 15 F 110 6 6 111 7 7 1000 8 8 1001 9 9

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The Hexadecimal Number System (con’t)

Example of a hexadecimal number and the values of the positions:

3 C 8 B 0 5 1 166 165 164 163 162 161 160

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Example of Equivalent Numbers

Binary: 1 0 1 0 0 0 0 1 0 1 0 0 1 1 12

Decimal: 2064710

Hexadecimal: 50A716

Notice how the number of digits gets smaller as the base increases.

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Goals of Today’s Lecture Binary numbers

Why binary? Converting base 10 to base 2 Octal and hexadecimal

Integers Unsigned integers Integer addition Signed integers

C bit operators And, or, not, and xor Shift-left and shift-right Function for counting the number of 1 bits Function for XOR encryption of a message

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Why Bits (Binary Digits)? Computers are built using digital circuits

Inputs and outputs can have only two values True (high voltage) or false (low voltage) Represented as 1 and 0

Can represent many kinds of information Boolean (true or false) Numbers (23, 79, …) Characters (‘a’, ‘z’, …) Pixels Sound

Can manipulate in many ways Read and write Logical operations Arithmetic …

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Base 10 and Base 2 Base 10

Each digit represents a power of 10 4173 = 4 x 103 + 1 x 102 + 7 x 101 + 3 x 100

Base 2 Each bit represents a power of 2 10110 = 1 x 24 + 0 x 23 + 1 x 22 + 1 x 21 + 0 x 20 = 22

Divide repeatedly by 2 and keep remainders

12/2 = 6 R = 0 6/2 = 3 R = 0 3/2 = 1 R = 1 1/2 = 0 R = 1 Result = 1100

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Writing Bits is Tedious for People Octal (base 8)

Digits 0, 1, …, 7 In C: 00, 01, …, 07

Hexadecimal (base 16) Digits 0, 1, …, 9, A, B, C, D, E, F In C: 0x0, 0x1, …, 0xf

0000 = 0 1000 = 80001 = 1 1001 = 90010 = 2 1010 = A0011 = 3 1011 = B0100 = 4 1100 = C0101 = 5 1101 = D0110 = 6 1110 = E0111 = 7 1111 = F

Thus the 16-bit binary number

1011 0010 1010 1001

converted to hex is

B2A9

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Representing Colors: RGB Three primary colors

Red Green Blue

Strength 8-bit number for each color (e.g., two hex digits) So, 24 bits to specify a color

In HTML, on the course Web page Red: <font color="#FF0000"><i>Symbol Table Assignment

Due</i> Blue: <font color="#0000FF"><i>Fall Recess</i></font>

Same thing in digital cameras Each pixel is a mixture of red, green, and blue

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Storing Integers on the Computer Fixed number of bits in memory

Short: usually 16 bits Int: 16 or 32 bits Long: 32 bits

Unsigned integer No sign bit Always positive or 0 All arithmetic is modulo 2n

Example of unsigned int 00000001 1 00001111 15 00010000 16 00100001 33 11111111 255

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Adding Two Integers: Base 10 From right to left, we add each pair of digits We write the sum, and add the carry to the

next column

1 98

+ 2 64

Sum

Carry

0 11

+ 0 01

Sum

Carry

2

1

6

1

4

0

0

1

0

1

1

0

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Binary Sums and Carries

a b Sum a bCarry0 0 0 0 0 00 1 1 0 1 01 0 1 1 0 01 1 0 1 1 1

XOR AND

0100 0101

+ 0110 0111

1010 1100

69

103

172

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Fractional Numbers

Examples:456.7810 = 4 x 102 + 5 x 101 + 6 x 100 + 7 x 10-1+8 x 10-2

1011.112 = 1 x 23 + 0 x 22 + 1 x 21 + 1 x 20 + 1 x 2-1 + 1 x 2-2

= 8 + 0 + 2 + 1 + 1/2 + ¼

= 11 + 0.5 + 0.25 = 11.7510

Conversion from binary number system to decimal system

Examples: 111.112 = 1 x 22 + 1 x 21 + 1 x 20 + 1 x 2-1 + 1 x 2-2

= 4 + 2 + 1 + 1/2 + ¼ = 7.7510

Examples: 11.0112

22 21 20 2-1 2-2 2-3

4 2 1 ½ ¼ 1/8

2 1 0 -1 -2 -3

x x x x

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Fractional numbersExamples: 7.7510 = (?)2

1. Conversion of the integer part: same as before – repeated division by 27 / 2 = 3 (Q), 1 (R) 3 / 2 = 1 (Q), 1 (R) 1 / 2 = 0 (Q), 1 (R) 710 = 1112

2. Conversion of the fractional part: perform a repeated multiplication by 2 and extract the integer part of the result0.75 x 2 =1.50 extract 10.5 x 2 = 1.0 extract 1 0.7510 = 0.112

0.0 stop

Combine the results from integer and fractional part, 7.7510 = 111.112

How about choose some of

Examples: try 5.625

write in the same order

4 2 1 1/2 1/4 1/8=0.5 =0.25 =0.125

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Fractional Numbers (cont.)

Exercise 2: Convert (0.6)10 to its binary formSolution:

Exercise 1: Convert (0.625)10 to its binary form

Solution: 0.625 x 2 = 1.25 extract 1

0.25 x 2 = 0.5 extract 0

0.5 x 2 = 1.0 extract 1

0.0 stop

(0.625)10 = (0.101)2

0.6 x 2 = 1.2 extract 1

0.2 x 2 = 0.4 extract 0

0.4 x 2 = 0.8 extract 0

0.8 x 2 = 1.6 extract 1

0.6 x 2 =

(0.6)10 = (0.1001 1001 1001 …)2

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Fractional Numbers (cont.)

Exercise 3: Convert (0.8125)10 to its binary form

Solution: 0.8125 x 2 = 1.625 extract 1

0.625 x 2 = 1.25 extract 1

0.25 x 2 = 0.5 extract 0

0.5 x 2 = 1.0 extract 1

0.0 stop

(0.8125)10 = (0.1101)2

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Fractional Numbers (cont.)

Errors One source of error in the computations is due to

back and forth conversions between decimal and binary formatsExample: (0.6)10 + (0.6)10 = 1.210

Since (0.6)10 = (0.1001 1001 1001 …)2

Lets assume a 8-bit representation: (0.6)10 = (0 .1001 1001)2 , therefore0.6 0.10011001

+ 0.6 + 0.100110011.00110010

Lets reconvert to decimal system: (1.00110010)b= 1 x 20 + 0 x 2-1 + 0 x 2-2 + 1 x 2-3 + 1 x 2-4 + 0 x 2-5 + 0 x 2-6 + 1 x 2-7 + 0

x 2-8

= 1 + 1/8 + 1/16 + 1/128 = 1.1953125

Error = 1.2 – 1.1953125 = 0.0046875

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One’s and Two’s Complement One’s complement: flip every bit

E.g., b 01000101 (i.e., 69 in base 10) One’s complement is 10111010 That’s simply 255-69

Subtracting from 11111111 is easy (no carry needed!)

Two’s complement Add 1 to the one’s complement E.g., (255 – 69) + 1 1011 1011

- 0100 0101 1111 1111

1011 1010

b

one’s complement

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Putting it All Together Computing “a – b” for unsigned integers

Same as “a + 256 – b” Same as “a + (255 – b) + 1” Same as “a + onecomplement(b) + 1” Same as “a + twocomplement(b)”

Example: 172 – 69 The original number 69: 0100 0101 One’s complement of 69: 1011 1010 Two’s complement of 69: 1011 1011 Add to the number 172: 1010 1100 The sum comes to: 0110 0111 Equals: 103 in base 10

1010 1100

+ 1011 1011

1 0110 0111

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Signed Integers Sign-magnitude representation

Use one bit to store the sign Zero for positive number One for negative number

Examples E.g., 0010 1100 44 E.g., 1010 1100 -44

Hard to do arithmetic this way, so it is rarely used Complement representation

One’s complement Flip every bit E.g., 1101 0011 -44

Two’s complement Flip every bit, then add 1 E.g., 1101 0100 -44

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Overflow: Running Out of Room Adding two large integers together

Sum might be too large to store in the number of bits allowed

What happens?

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Bitwise Operators: AND and OR

Bitwise AND (&)

Mod on the cheap! E.g., h = 53 & 15;

&

0

1

0 1

0 0

0 1

Bitwise OR (|)|

0

1

0 1

0 1

1 1

0 0 1 1 0 1 0 153 0 0

0 0 0 0 1 1 1 115 0 0

0 0 0 0 0 1 0 15 0 0

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Bitwise Operators: Shift Left/Right Shift left (<<): Multiply by powers of 2

Shift some # of bits to the left, filling the blanks with 0

Shift right (>>): Divide by powers of 2 Shift some # of bits to the right

For unsigned integer, fill in blanks with 0 What about signed integers? Varies across machines…

Can vary from one machine to another!

0 0 1 1 0 1 0 053

1 1 0 1 0 0 0 053<<2

0 0 1 1 0 1 0 053

0 0 0 0 1 1 0 153>>2

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XOR Encryption Program to encrypt text with a key

Input: original text in stdin Output: encrypted text in stdout

Use the same program to decrypt text with a key Input: encrypted text in stdin Output: original text in stdout

Basic idea Start with a key, some 8-bit number (e.g., 0110 0111) Do an operation that can be inverted

E.g., XOR each character with the 8-bit number 0100 0101

^ 0110 0111

0010 0010

0010 0010 ^ 0110 0111

0100 0101

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Conclusions Computer represents everything in binary

Integers, floating-point numbers, characters, addresses, …

Pixels, sounds, colors, etc. Binary arithmetic through logic operations

Sum (XOR) and Carry (AND) Two’s complement for subtraction

Binary operations in C AND, OR, NOT, and XOR Shift left and shift right Useful for efficient and concise code, though

sometimes cryptic