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Data Representation

Data Representation. Summer 2014 COMP 2130 Intro Computer Systems Computing Science Thompson Rivers University. Course Contents. Introduction to computer systems: B&O 1 Data representations: B&O 2.1 – 2.4 Introduction to C programming: K&R 1 – 4

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Data Representation

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  1. Data Representation Summer 2014 COMP 2130 Intro Computer Systems Computing Science Thompson Rivers University

  2. Course Contents • Introduction to computer systems: B&O 1 • Data representations: B&O 2.1 – 2.4 • Introduction to C programming: K&R 1 – 4 • C: advanced topics: K&R 5.1 – 5.10, 6 – 7 • Introduction to IA32 (Intel Architecture 32): B&O 3.1 – 3.8, 3.13 • Compiling, linking, loading, and executing: B&O 7 (except 7.12) • Dynamic memory management – Heap: B&O 9.9.1 – 9.9.2, 9.9.4 – 9.9.5, 9.11 • Code optimization: B&O 5.1 – 5.6, 5.13 • Memory hierarchy, locality, caching: B&O 5.12, 6.1 – 6.3, 6.4.1 – 6.4.2, 6.5, 6.6.2 – 6.6.3, 6.7 • Virtual memory (if time permits): B&O 9.4 – 9.5 Data Representation

  3. Unit Learning Objectives • Convert a decimal number to binary, Octal, Hexadecimal. • Convert a hexadecimal number to binary number, or Octal and vice versa • Compute binary addition. • Compute binary subtraction using 2’s complement. • Determine the 2’s complement representation of a signed integer. • Understand the overflow of unsigned integers and signed integers. • Understand the 2’s complement representation for negative integers. • Subtract a binary number by using the 2’s complement addition. • Multiply two binary numbers. • Distinguish little endian byte order and big endian byte order • Use of left shift and right shift. • Binary division Data Representation

  4. 1. Information Storage • memory, address • Memory space is partitioned into more manageable units to store the different program objects, i.e., instructions, program data, and control information. • The actual machine level program simply treats each program object as a block of bytes, and the program itself as a sequence of bytes. • Example • int number = 28; • 4 bytes, i.e., 32 bits, will be allocated to the variable number. • The decimal number 28 will be stored in the 32 bits? How? • Binary number, not the decimal, 00000000 00000000 00000000 00011100 will be stored in the 32 bits. • Do programmers have to convert 28 to it’s binary number? Data Representation

  5. 2 1 0 -1 -2 100 10 1 0.1 0.01 500 10 2 0.7 0.04 The Decimal System • Base (also called radix) = 10 • 10 digits { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 } • Digit Position • Integer & fraction • Digit Weight • Weight = (Base) Position • Magnitude • Sum of “Digit x Weight” • Formal Notation 5 1 2 7 4 d2*B2+d1*B1+d0*B0+d-1*B-1+d-2*B-2 (512.74)10 Number Systems

  6. The Binary System • In computer systems, the most basic memory unit is a bit that contains 0 and 1. • The data unit of 8 bits is referred as a byte that is the basic memory unit used in main memories and hard disks. • All data are represented by using binary numbers. Data types such as text, voice, image and video have no meaning in the data representation. • 8 bits are usually used to express English alphabets. • A collection of nbits has 2npossible states. Is it true? • E.g., • How many different numbers can you express using 2 bits? • How many different numbers can you express using 4 bits? • How many different numbers can you express using 8 bits? • How many different numbers can you express using 32 bits? Number Systems

  7. 4 2 1 1/2 1/4 2 1 0 -1 -2 Binary Number System • Base = 2 • 2 digits { 0, 1 }, called binary digits or “bits” • Weights • Weight = (Base) Position • Magnitude • Sum of “Bit x Weight” • Formal Notation • Groups of bits 4 bits = Nibble 8 bits = Byte Kilobyte (KB) = 1,024 Bytes Megabyte (MB) = 1,024 Kilobytes Gigabyte (GB) =1,024 Megabytes Terabyte (TB) = 1,024 Gigabytes Petabyte (PB) = 1,024 Terabytes Exabyte (EB) = 1,024 Petabytes 1 0 1 0 1 1 *22+0 *21+1 *20+0 *2-1+1 *2-2 =(5.25)10 (101.01)2 1 0 1 1 1 1 0 0 0 1 0 1 Number Systems

  8. 64 8 1 1/8 1/64 2 1 0 -1 -2 Octal Number System • Base = 8 • 8 digits { 0, 1, 2, 3, 4, 5, 6, 7 } • Weights • Weight = (Base) Position • Magnitude • Sum of “Digit x Weight” • Formal Notation 5 1 2 7 4 5 *82+1 *81+2 *80+7 *8-1+4 *8-2 =(330.9375)10 (512.74)8

  9. 256 16 1 1/16 1/256 2 1 0 -1 -2 Hexadecimal Number System • Base = 16 • 16 digits { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F } • Weights • Weight = (Base) Position • Magnitude • Sum of “Digit x Weight” • Formal Notation 1 E 5 7 A 1 *162+14 *161+5 *160+7 *16-1+10 *16-2 =(485.4765625)10 (1E5.7A)16

  10. The Power of 2 Kilo Mega Giga Tera

  11. Conversion Chart • Uses two digits 0 and 1. How to expand binary numbers? • 02 = 0×20 = 010 • 12 = 1×20 = 110 • 102 = 1×21 + 0×20 = 210 • 112 = 1×21 + 1×20 = 310 • 1002 = 1×22 + 0×21 + 0×20 = 410 • 1012 = 1×22 + 0×21 + 1×20 = 510 • 1102 = 1×22 + 1×21 + 0×20 = 610 • 1112 = 1×22 + 1×21 + 1×20 = 710 • 10002 = 1×23 + 0×22 + 0×21 + 0×20 = 810 • 10012 = 1×23 + 0×22 + 0×21 + 1×20 = 910 • ... Number Systems

  12. Powers of 2 • 12 = 1×20 = 110 • 102 = 1×21 + 0×20 = 210 • 1002 = 1×22 + 0×21 + 0×20 = 410 • 10002 = 1×23 + 0×22 + 0×21 + 0×20 = 810 • 1 00002 = 1610 • 10 00002 = 3210 • 100 00002 = 6410 Number Systems

  13. Number Base Conversions Evaluate Magnitude Octal (Base 8) Evaluate Magnitude Decimal (Base 10) Binary (Base 2) Hexadecimal (Base 16) Evaluate Magnitude

  14. Decimal to Binary Conversion • Divide the number by the ‘Base’ (=2) • Take the remainder (either 0 or 1) as a coefficient • Take the quotient and repeat the division Example: (13)10 Coefficient Quotient Remainder 13 / 2 =6 1a0 =1 6 / 2 =3 0a1 =0 3 / 2 =1 1a2 =1 1 / 2 =0 1a3 =1 Answer: (13)10= (a3 a2 a1 a0)2= (1101)2 MSB LSB

  15. Decimal to Binary Conversion • Multiply the number by the ‘Base’ (=2) • Take the integer (either 0 or 1) as a coefficient • Take the resultant fraction and repeat the division Example: (0.625)10 Coefficient Integer Fraction a-1 =1 0.625 * 2 =1 . 25 0.25 * 2 =0 . 5a-2 =0 0.5 * 2 =1 . 0a-3 =1 Answer: (0.625)10= (0.a-1 a-2 a-3)2= (0.101)2 MSB LSB

  16. Decimal to Octal Conversion Example: (175)10 Coefficient Quotient Remainder 175 / 8 =21 7a0 =7 21 / 8 =2 5a1 =5 2 / 8 =0 2a2 =2 Answer: (175)10= (a2 a1 a0)8= (257)8 Example: (0.3125)10 Coefficient Integer Fraction a-1 =2 0.3125 * 8 =2 . 5 0.5 * 8 =4 . 0a-2 =4 Answer: (0.3125)10= (0.a-1 a-2 a-3)8= (0.24)8

  17. Binary − Octal Conversion • 8 = 23 • Each group of 3 bits represents an octal digit Example: Assume Zeros ( 1 0 1 1 0 . 0 1 )2 ( 2 6 . 2 )8 Works both ways (Binary to Octal & Octal to Binary)

  18. Binary − Hexadecimal Conversion • 16 = 24 • Each group of 4 bits represents a hexadecimal digit Example: Assume Zeros ( 1 0 1 1 0 . 0 1 )2 ( 1 6 . 4 )16 Works both ways (Binary to Hex & Hex to Binary)

  19. Octal − Hexadecimal Conversion • Convert to Binary as an intermediate step Example: ( 2 6 . 2 )8 Assume Zeros Assume Zeros ( 01 0 1 1 0 . 0 1 0 )2 ( 1 6 . 4 )16 Works both ways (Octal to Hex & Hex to Octal)

  20. Decimal, Binary, Octal and Hexadecimal

  21. Decimal Addition • Decimal Addition 1 1 Carry 5 5 + 5 5 1 1 0 = Ten ≥Base  Subtract a Base

  22. Binary Addition • Column Addition 1 1 1 1 1 1 = 61 = 23 1 1 1 1 0 1 + 1 0 1 1 1 = 84 1 0 1 0 1 0 0 ≥ (2)10

  23. Binary Subtraction • Borrow a “Base” when needed 1 2 = (10)2 2 0 2 0 0 2 1 = 77 = 23 0 0 1 1 0 1 − 1 0 1 1 1 0 1 1 0 1 1 0 = 54

  24. Binary Multiplication • Bit by bit 1 0 1 1 1 x 1 0 1 0 0 0 0 0 0 1 0 1 1 1 0 0 0 0 0 1 0 1 1 1 1 1 1 0 0 1 1 0

  25. Hexadecimal Number System • 010 = 00002 = 016 = 0x0 • 110 = 00012 = 116 = 0x1 • 210 = 00102 = 216 = 0x2 • 310 = 00112 = 316 = 0x3 • 410 = 01002 = 416 = 0x4 • 510 = 01012 = 516 = 0x5 • 610 = 01102 = 616 = 0x6 • 710 = 01112 = 716 = 0x7 • 810 = 10002 = 816 = 0x8 • 910 = 10012 = 916 = 0x9 • 1010 = 10102 = A16 = 0xA • 1110 = 10112 = B16 = 0xB • 1210 = 11002 = C16 = 0xC • 1310 = 11012 = D16 = 0xD • 1410 = 11102 = E16 = 0xE • 1510 = 11112 = F16 = 0xF • 14816= ???1014816= ???2 • 23c9d6ef = ???1023c9d6ef = ???2 4 bitscan be used for a hexadecimal number, 0, ..., F. Please memorize it! Number Systems

  26. Binary to/from Hexadecimal conversion • 4DA916= ???2 • 1001101101010012 = ???16 = 100 1101 1010 1001 = 4DA9 • 10 11102 = 0x??? = ???10 • 0100 1110 1011 1001 01002 = 0x??? = ???10 Number Systems

  27. Words • A word size – the nominal size of integer and pointer data. A pointer variable contains an address in the virtual address space. We will discuss pointer variable in the next unit. • The word size determines the maximum size of the virtual address space. • 32bit operating systems? • 64bit operating systems? Data Representation

  28. Addressing and Byte Ordering • A variable x of type int (allocates 4 bytes) • If the address of x: 0x100 (means it starts storing from 0x100) • This means the 4 bytes of x would be stored in memory locations 0x100, 0x101, 0x102, and 0x103. • Let’s assume x has the value 0x1234567, which needs to be stored in four bytes. • There are two conventions to store the values in the 4 consecutive byte memory locations. 0x01, 0x23, 0x45, and 0x67, or 0x67, 0x45, 0x23, and 0x01, depending on CPU architecture. • Little endian byte order – Intel-compatible machines 0x103 0x102 0x101 0x100 address 0x01 0x23 0x45 0x67 value • Big endian byte order – machines from IBM and Sun Microsystems 0x103 0x102 0x101 0x100 0x67 0x45 0x23 0x01 Data Representation

  29. The byte orderings are totally invisible for most application programmers. • Why are the byte orderings important? • Think of data exchange between two machines through a network. • Assembly programming • When type casting is used Data Representation

  30. 2. Integer Representations C Java Size char, unsigned char byte 1B short, unsigned short short 2Bs char in Java uses 2Bs. int, unsigned intint 4Bs long, unsigned long long 8Bs // there is no unsigned in Java float float 4Bs double double 8Bs Data Representation

  31. Unsigned Encodings • unsigned char All 8 bits are used. No sign bit. • The smallest number is 0 • The maximum number is 0xff. • unsigned short 16 bits • The smallest number is ___ • The maximum number is ____ • unsigned int 32 bits • The smallest number is ___ • The maximum number is ____ • unsigned long 64 bits • The smallest number is ____ • The maximum number is _____ Data Representations

  32. Representation of Unsigned Integers • 8-bit representation of unsigned char 255 11111111 254 11111110 ... ... 128 10000000 127 01111111 126 01111110 ... ... 2 00000010 1 00000001 0 00000000 • The maximum number is ? • The minimum number is ? • What if we add the maximum number by 1 ??? • What if we subtract the minimum number by 1 ??? +1 +1 +1 overflow Data Representations

  33. Binary Addition • We will discuss binary addition and binary subtraction, before we discuss the representation of signed integers. • How to add two binary numbers? Let’s consider only unsigned integers(i.e., positive numbers only) for a while. • Just like the addition of two decimal numbers. • E.g., 10010 10010 1111 + 1001 + 1011 + 1 11011 11101 ??? 10111 + 111 ??? carry Data Representations

  34. Binary Subtraction • How to subtract a binary number? • Just like the subtraction of decimal numbers. • E.g., 0112 02 02 1000 10 10 10010 10010 10010 -1-1 -11 -11 -11 1 ?1 ?11 1111 Try: 101010 How to do? 1 -101-10 Data Representations

  35. In the previous slide, 10010 – 11 = 1111 • What if we add 00010010 + 11111100 1 00001110 + 1 00001111 • Is there any relationship between 112 and 111002? • The 1’s complement of 112 is ??? Switching 0  1 • This type of addition is called 1’s complement addition. • Find the 8-bit one’s complements of the followings. • 11011 -> 00011011 -> • 10 -> 00000010 -> • 101 -> 00000101 -> Data Representations

  36. In the previous slide, 10010 – 11 = 1111 • What if we add 00010010 + 11111101 1 00001111 • Is there any relationship between 11 and 11101? • The 2’s complement of 11 is ??? • 2’s complement ≡ 1’s complement + 1 -> 11100 + 1 = 11101 • This type of addition is called 2’s complement addition. • Find the 16-bit two’s complements of the followings. • 11011 -> 0000000000011011 -> • 10 • 101 Data Representations

  37. Another example 101010 - 101 ??? • What if we use 1’s complement addition or 2’s complement addition instead as follow? Let’s use 8-bit representation. 00101010 00101010 + 11111010+ 11111011 1 00100100 1 00100101 + 1 00100101 • What does this mean? • A – B = A + (–B), where A and B are positive • Is the 1’s complement or the 2’s complement of B sort of equal to –B? Data Representations

  38. Can we use 8-bit 1’s complement addition for 12 – 102 = –12? 1 00000001 - 10+ 11111101<- 8-bit 1’s complement of 10 11111110 <- Is this correct? (1’s complement of 1?) • Let’s use 8-bit 2’s complement addition for 12 – 102. 00000001 +11111110<- 2’s complement of 10 11111111 <- Correct? (2’s complement of 1?) • 12 – 102 = 12 + (–102) = –12 • How to represent negative binary numbers, i.e., signed integers? Data Representations

  39. Representation of Negative Binaries • Representation of signed integers • 8 or 16 or 32 bits are usually used for integers. • Let’s use 8 bits for examples. • The left most bit (called most significant bit) is used as sign. • When the MSB is 0, non-negative integers. • When the MSB is 1, negative integers. • The other 7 bits are used for integers. • How to represent positive integer 9? • 00001001 • How about -9? • 10001001 is really okay? • 00001001 (9) + 10001001 (-9) = 10010010 (-18) It is wrong! • We need a different representation for negative integers. Data Representations

  40. How about -9? • 10001001 is really okay? • 00001001 (9) + 10001001 (-9) = 10010010 (-18) It is wrong! • We need a different representation for negative integers. • What is the 8-bit 1’s complement of 9? • 11110110 <- 8-bit 1’s complement of 9 • 00001001 + 11110110 <- 9 + 8-bit 1’s complement of 9 = 11111111 <- Is it zero? (1’s complement of 0?) • What is the 2’s complement of 9? • 11110111 <- 8-bit 2’s complement of 9 • 00001001 + 11110111 <- 9 + 8-bit 2’s complement of 9 = 1 00000000 <- It looks like zero. • 2’s complement representation is used for negative integers. Data Representations

  41. 12 – 102 = 12 + (–102) ??? What is the result in decimal? 00000001 + 11111110<- 2’s complement of 10, i.e., -102 11111111 <- 2’s complement of 1, i.e., -1 (= 1 – 2) • 1010102 – 1012 = 1010102 + (–1012) ??? • 100102 – 112 ??? • 102 – 12 ??? • -102 – 12 ??? • Is the two’s complement of the two’s complement of an integer the same integer? • What is x when the 8-bit 2’s complement of x is • 11111111 11110011 10000001 Data Representations

  42. 8-bit representation of signed char with 2’s complement 127 01111111 126 01111110 ... ... 2 00000010 1 00000001 0 00000000 -1 11111111 -2 11111110 -3 11111101 ... ... -127 10000001 -128 10000000 • The maximum number is ? • The minimum number is ? • What if we add the maximum number by 1 ??? • What if we subtract the minimum number by 1 ??? overflow +1 +1 -1 overflow +1 -1 -1 Data Representations

  43. 16-bit representation signed short with 2’s complement ... 01111111 11111111 ... 01111111 11111110 ... ... 3 00000000 00000011 2 00000000 00000010 1 00000000 00000001 0 00000000 00000000 -1 11111111 11111111 -2 11111111 11111110 -3 11111111 11111101 ... ... ... 10000000 00000001 ... 10000000 00000000 • The maximum number is ? What if we add the maximum number by 1 ??? • The minimum number is ? What if we subtract the minimum number by 1 ??? overflow +1 +1 +1 -1 -1 -1 overflow Data Representations

  44. Note that computers use the 8-bit representation, the 16-bit representation, the 32-bit representation and the 64-bit representation with 2’complement for negative integers. • In programming languages • char, unsigned char 8-bits • short, unsigned short 16-bits • int, unsigned int 32-bits • long, unsigned long 64-bits • When we use the 32-bit representation with 2’s complement, • The maximum number is ? • What if we add the maximum number by 1 ??? • The minimum number is ? • What if we subtract the minimum number by 1 ??? Data Representations

  45. Now we know how to represent negative integers. • 2’ complement addtion A + (–B) is computed for subtraction A – B. • Let’s suppose B is negative. Then –B is really a positive integer? For example, let’s consider 1 byte signed integer. 127 01111111 126 01111110 ... ... 2 00000010 1 00000001 0 00000000 -1 11111111 -2 11111110 -3 11111101 2’s complement of -3 is 00000011, i.e., 3. ... ... -127 10000001 -128 10000000 2’s complement of -128 is 10000000 again. • For any -127 < x < 127, x – x = 0. But (-128) – (-128) = ??? Data Representation

  46. Binary Multiplication • Bit by bit 1 0 1 1 1 x 1 0 1 0 0 0 0 0 0 1 0 1 1 1 0 0 0 0 0 1 0 1 1 1 1 1 1 0 0 1 1 0 Data Representation

  47. Binary Division • Binary division? 1111 <- quotient 101 1001101 -101 1001 -101 1000 -101 111 -101 10 <- remainder • Try 1101011 / 110 • Dividing negative binary numbers: division without sign, and then put the sign. This is why division is an expensive operation. Number Systems

  48. Binary division by a power of 2? 1001101 / 10 = 100110 1001101 / 100 = 10011 1001101 / 1000 = 1001 • What if we shift 1001101 right by 1 bit? • What if we shift 1001101 right by 2 bits? • What if we shift 1001101 right by 3 bits? • 1001101 / 101 ??? Complicated implementation required Number Systems

  49. 2.4 Floating Point • Fractional binary numbers • IEEE floating-point representation Data Representation

  50. 2-1 = 0.5 2-2 = 0.25 2-3 = 0.125 00101000.101 (40.625) + 11111110.110(-1.25) (2’s complement) 00100111.011 (39.375) Fractions: Fixed-Point • How can we represent fractions? • Use “binary point” to separate positive from negative powers of two -- like “decimal point.” • 2’s comp addition and subtraction still work. (Assuming binary points are aligned) No new operations -- same as integer arithmetic. Number Systems

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