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Binary Numbers

Binary Numbers. 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 .

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Binary Numbers

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  1. Binary Numbers

  2. 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 isset. • 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.

  3. 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.

  4. Main Memory (con’t) • Other computer components can • get the information held at a particular address in memory, known as a READ, • 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.

  5. 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).

  6. 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.

  7. 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

  8. 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.

  9. 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.

  10. 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

  11. 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? • Because we use 10 digits, the digits 0 through 9.

  12. 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? • Because we use 2 digits, the digits 0 and 1.

  13. 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: • 1001101 • 26 25 24 23 22 21 20

  14. Converting from Binary to Decimal • 10011011 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 = 20 X 24 = 0 • 22 = 40 X 25 = 0 • 23 = 81 X 26 = 64 • 24 = 16 7710 • 25 = 32 • 26 = 64

  15. Converting from Binary to Decimal (con’t) Practice conversions: BinaryDecimal 11101 1010101 100111

  16. 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 • 101010 • 42/2 = 21 R = 0 • 21/2 = 10 R = 1 • 10/2 = 5 R = 0 • 5/2 = 2 R = 1 • 2/2 = 1 R = 0 • 1/2 = 0 R = 1 • 4210 = 1010102

  17. Converting From Decimal to Binary (con’t) Practice conversions: DecimalBinary 59 82 175

  18. 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.

  19. 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.

  20. The Hexadecimal Number System (con’t) BinaryDecimalHexadecimalBinaryDecimalHexadecimal 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

  21. The Hexadecimal Number System (con’t) • Example of a hexadecimal number and the values of the positions: • 3C8B051 • 166 165 164 163 162 161 160

  22. 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.

  23. 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

  24. 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 • …

  25. 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

  26. 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 Thus the 16-bit binary number 1011 0010 1010 1001 converted to hex is B2A9

  27. 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

  28. 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

  29. 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 0 1 1 + 0 0 1 Sum Carry 1 9 8 + 2 6 4 Sum Carry 4 0 6 1 2 1 1 0 0 1 0 1

  30. Binary Sums and Carries a b Sum a b Carry 0 0 0 0 0 0 0 1 1 0 1 0 1 0 1 1 0 0 1 1 0 1 1 1 XOR AND 0100 0101 69 + 0110 0111 103 1010 1100 172

  31. 2 1 0 -1 -2 -3 22 21 20 2-1 2-2 2-3 4 2 1 ½ ¼ 1/8 x x x x 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

  32. write in the same order Fractional numbers Examples: 7.7510 = (?)2 • Conversion of the integer part: same as before – repeated division by 2 7 / 2 = 3 (Q), 1 (R)  3 / 2 = 1 (Q), 1 (R)  1 / 2 = 0 (Q), 1 (R) 710 = 1112 • Conversion of the fractional part: perform a repeated multiplication by 2 and extract the integer part of the result 0.75 x 2 =1.50  extract 1 0.5 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 4 2 1 1/2 1/4 1/8 =0.25 =0.125 =0.5

  33. 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 Fractional Numbers (cont.) Exercise 2: Convert (0.6)10to its binary form Solution: 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

  34. 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

  35. Fractional Numbers (cont.) • Errors • One source of error in the computations is due to back and forth conversions between decimal and binary formats Example: (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 .10011001)2 , therefore 0.6 0.10011001 + 0.6  + 0.10011001 1.00110010 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

  36. 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 1111 1111 - 0100 0101 b one’s complement 1011 1010

  37. 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

  38. 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

  39. 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?

  40. | & 0 1 0 1 0 0 0 0 0 1 1 1 0 1 1 1 53 15 5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 1 1 1 1 1 0 0 1 1 1 Bitwise Operators: AND and OR • Bitwise OR (|) • Bitwise AND (&) • Mod on the cheap! • E.g., h = 53 & 15;

  41. 53 0 0 1 1 0 1 0 0 53<<2 1 1 0 1 0 0 0 0 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! 53 0 0 1 1 0 1 0 0 53>>2 0 0 0 0 1 1 0 1

  42. 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 0010 0010 ^ 0110 0111 ^ 0110 0111 0010 0010 0100 0101

  43. 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

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