1. Chapter 2 Introduction to Number System & Data Formats

2. Chapter 2: (-133)*33+44*14.5 How to represent these data? How to calculate them? How about other characters?

3. Chapter 2: Introduction to Number System & Data Formats 2.1 Number System ( 数制 )2.1 Number System 2.2 Computer Data Formats

4. 2.1 Number System ( 数制 )

5. Binary 101 (1) 1 0 1 power( 幂 ) 2 2 2 1 2 0 (2) weight( 权 )4 2 1 Numeric value 1* 2 2 + 0* 2 1 + 1* 2 0 =5 (3)

6. 2.1 Number System( 数制 ) 2.1.1 Digits ( 数 )2.1.1 Digits ( 数 ) 2.1.2 Positional Notation ( 位计数法）2.1.2 Positional Notation ( 位计数法） 2.1.3 Conversion to Decimal 2.1.4 Converting From Decimal 2.1.5 Binary-Coded Hexadecimal 2.1.6 Operation of Binary 2.1.7 Three Forms of Binary Data

7. 2.1.1 Digits ( 数 )

8. A. Decimal( 十进制 ) (0 — 9) B. Binary (0----1) C. Octal( 八进制 ) (0 — 7) D. Hexadecimal( 十六进制 ) (0 — 9,A,B,C,D,E,F)

9. 2.1.1 Digits ( 数 ) Example: A base 13( 以 13 为基， 13 进制 )number contains 13 digits: 0 — 9,A(10),B(11),C(12)

10. 2.1.2 Positional Notation ( 位计数法）

11. Example 1: Decimal 132 1( 百位 ) 3( 十位 ) 2( 个位 ) hundreds tens units position position position power( 幂 ) 10 2 10 1 10 0 weight( 权 )100 10 1 In a base N number system, the radix is N( 基数 ) and the exponent (i) ( 指数 ) means the position.

12. 2.1.2 Positional Notation ( 位计数法） Example 2: Binary 101 1 0 1 power( 幂 ) 2 2 2 1 2 0 weight( 权 )4 2 1 Numeric value 1* 2 2 + 0* 2 1 + 1* 2 0 =5 In a base 2 number system, the radix is 2( 基数 ) and the exponent (i) ( 指数 ) means the position.

13. 2.1.2 Positional Notation ( 位计数法） Special example Binary 1111 1 1 1 1 power( 幂 ) 2 3 2 2 2 1 2 0 weight( 权 ) 8 4 2 1 Numeric value 8+4+2+1 =15 0-15 It can represent 16 different numbers.

14. 2.1.2 Positional Notation ( 位计数法） Example 3: Decimal 1.025 1. 0 2 5 power( 幂 ) 10 0 10 -1 10 -2 10 -3 (N i ) weight( 权 )1 0.1 0.01 0.001 In a base N number system, the radix is N and the exponent (i) ( 指数 ) means the position. If the position is right to the point( 小数点 ),the power is negative (N -i ).

15. 2.1.2 Positional Notation ( 位计数法） Summarize –A base N number system X m-1 X m-2 X m-3 …… X 0. Y 1 Y 2 Y 3 …… Y n –Numeric value( 数值 ): X m-1 * N m-1 +X m-2 * N m-2 + …… X 0 * N 0 +Y 1 *N -1 +Y 2 *N -2 + …… Y n * N -n

16. 2.1.3 Conversion to Decimal

17. Summarize –A base N number system X m-1 X m-2 X m-3 …… X 0. Y 1 Y 2 Y 3 …… Y n –Decimal X m-1 * N m-1 +X m-2 * N m-2 + …… X 0 * N 0 +Y 1 *N -1 +Y 2 *N -2 + …… Y n * N -n

18. 2.1.3 Conversion to Decimal Example: Hexadecimal: E6A.C2 Decimal: E*16 2 +6*16 1 +6*16 0 +C*16 -1 +2*16 -2 =14*16 2 +6*16 1 +6*16 0 +12*16 -1 +2*16 -2 =3690+0.75+0.0078125 =3690.7578125

19. 2.1.4 Converting From Decimal

20. Summarize –A base N number system X m-1 X m-2 X m-3 …… X 0. Y 1 Y 2 Y 3 …… Y n –Decimal X m-1 * N m-1 +X m-2 * N m-2 + …… X 0 * N 0 +Y 1 *N -1 +Y 2 *N -2 + …… Y n * N -n

21. 2.1.4 Converting From Decimal From conversion to decimal we can assume that if we can figure out how many N i are there in a decimal number, we may convert it back to a base N number system.

22. 2.1.4 Converting From Decimal Principle A: –To convert a decimal whole number portion( 整数部分 ) to another number system, divide by the radix N ( 基数 ) and save the remainders( 余数 ).

23. 2.1.4 Converting From Decimal Principle A: Step 1: Divide the decimal number by the radix (number base). Step 2: Save the remainder (first remainder is the least significant digit －－最先得到的余数是最低有效位 ) Step 3: Repeat steps 1 and 2 until the quotient( 商 ) is zero.

24. 2.1.4 Converting From Decimal Principle A: Example convert decimal number 14 to a base 4 number system 4 14 remainder = 2 (least significant) 4 3 remainder = 3 0 result = 32

25. 2.1.4 Converting From Decimal Principle A: Example convert decimal number 14 to a base 2 number system 2 14 remainder = 0 (least significant) 2 7 remainder = 1 2 3 remainder = 1 2 1 remainder = 1 0 result = 1110

26. 2.1.4 Converting From Decimal Principle B: –To convert a decimal fractional portion( 小数部分 ) to another number system, multiply it by the radix ( 基数 ) and the whole number portion( 整数部分 ) of the result is saved as a significant digit of the result.

27. 2.1.4 Converting From Decimal Principle B: Step 1: Multiply the decimal fraction by the radix (number base) Step 2: Save the whole number portion of the result (even if zero) as a digit. The first result is written immediately to the right of the radix point ( 得到的第一个整数 写在最靠近小数点左边的位置 ). Step 3: Repeat 1 and 2 until the fraction part of step 2 is zero

28. 2.1.4 Converting From Decimal Principle B: Example convert decimal number 0.125 to a base 4 number system.125 * 4 0.5 digit is 0 0.5 * 4 2.0 digit is 2 result is 0.02

29. 2.1.4 Converting From Decimal Principle B: –Note: Some numbers are never-ending. That is, a zero is never a remainder. – For example converting 0.15 to a base 4 number system. –Why ?

30. 2.1.4 Converting From Decimal Example: convert 0.15 to a base 4 number system..15 0.4 * 4 * 4 0.6 digit is 0 1.6 digit is 1 0.6 * 4 2.4 digit is 2 result is 0.0212 ……

31. 2.1.4 Converting From Decimal 0.0212 …… –Numeric value: –0*4 -1 + 2*4 -2 + 1*4 -3 + 2*4 -4 –=0 + 0.125 + 0.015625 + 0.0078125 –= 0.1484375 –≈ 0.15

32. 2.1.4 Converting From Decimal When converting to decimal –A base N number system 0. Y 1 Y 2 Y 3 …… Y n –Decimal Y 1 *N -1 +Y 2 *N -2 + …… Y n * N -n So it means that numerical value won’t exist exactly in a base N number system. But we can find one close to it.

33. 2.1.5 Binary-Coded Hexadecimal

34. How many different numbers can 4 binary numbers form ?

35. 2.1.5 Binary-Coded Hexadecimal BCH code hexadecimal digit BCH code 0 1 0000 0001 2 3 0010 0011 4 5 0100 0101 6 7 0110 0111 8 9 1000 1001 A B 1010 1011 C D 1100 1101 E F 1110 1111 9E = 1001 1110 BCD

36. 2.1.5 Binary-Coded Hexadecimal (BCH) BCH is used to represent hexadecimal data in binary code. Each hexadecimal digit is represented by a 4-bit binary number. Each BCH digit is separated by a space.

37. 2.1.5 Binary-Coded Hexadecimal Principle: We just use 4-bit binary to replace each hexadecimal digit and write down them in the same sequence. Note that there must be a space between each 4-bit binary.

38. 2.1.5 Binary-Coded Hexadecimal Example: Hexadecimal:2 A C. 5 BCH :0010 1010 1100. 0101 BCH :0010 1110. 0101 1011 Hexadecimal : 2 E. 5 B

39. 2.1.6 Operation of Binary All the numbers are unsigned.

40. 2.1.6 Operation of Binary In the above sliders we have discussed about binary, hexadecimal and decimal data. At the same time we learned the conversion between them. Now we will put our focus on the arithmetic operation of binary data. We take it for granted that all the data is unsigned.

41. 2.1.6 Operation of Binary Add Operation Rules: 0+0 = 0 0+1 = 1 1+0 = 1 1+1 = 0 submit( 进位 ) 1 1+1+1 = 1 submit( 进位 ) 1

42. 2.1.6 Operation of Binary Add Operation 1101 + 1011 1 1 1 1 submit bit 1 1 0 1 add bit + 1 0 1 1 added bit 1 1 0 0 0 result

43. 2.1.6 Operation of Binary Subtract Operation Rules 0-0 = 0 1-1 = 0 1-0 = 1 0-1 = 1 borrow( 借位 ) 1

44. 2.1.6 Operation of Binary Subtract Operation 11000100 – 00100101 1 1 1 1 1 1 ( 借位 ) 1 0 1 1 1 0 1 ( 借位后的被减数 ) 1 1 0 0 0 1 0 0 ( 被减数 ) - 0 0 1 0 0 1 0 1 ( 减数 ) 1 0 0 1 1 1 1 1

45. 2.1.6 Operation of Binary Multiply Operation Rules 0*0 = 0 0*1 = 0 1*0 = 0 1*1 = 1

46. 2.1.6 Operation of Binary Multiply Operation 1111 * 1101 1 1 1 1 * 1 1 0 1 1 1 0 0 0 0 1 1 1 1 1 1 0 0 0 0 1 1

47. 2.1.6 Operation of Binary Divide Operation 0 0 0 1 1 1 1 0 1 1 0 1 1 0 0 0 1 1 1 0 1 1 0 0 1 0 1 1 0 1 1 0 1 0 1 1 1 1 1 0 0 0 1 1 0 1 1 0 1 1 0 1 1 1 1 1 0 1 1 0 1 0 1 0

48. 2.1.6 Operation of Binary All the numbers are unsigned, so they are positive. How can we represent negative number in binary in a computer?

49. 2.1.7 Three Forms of Binary Data

50. Signed data can be represented in three forms. Original binary data ( 原码 ) Radix-1 complements ( 反码 ) Radix complements ( 补码 ) We use byte-sized data in our examples.

53. 2.1.7 Three Forms of Binary Data Original binary data ( 原码 ) 0 0 0 0 105 ---- 0 1101001 -105 ---- 1 1101001 0 positive 1 negative Numerical value (-127~+127)

54. 2.1.7 Three Forms of Binary Data Radix-1 complements ( 反码 ) A. positive Radix-1 complements is the same as original binary data. B. negative Represent the data in original form, then invert each bit of the number from 0 to 1 or from 1 to 0. The sign position is always 1.

55. 2.1.7 Three Forms of Binary Data Original binary data ( 原码 ) 105 ---- 0 1101001 -105 ---- 1 1101001 Radix-1 complements ( 反码 ) 105 ---- 0 1101001 -105 ---- 1 0010110 0 ---- 0 0000000 -0 ---- 1 1111111

56. 2.1.7 Three Forms of Binary Data Radix complements ( 补码 ) A. Positive Radix complements is the same as original binary data. B. negative If we add 1 to the least significant bit of the radix-1 complements, then we get radix complements. The sign position is always 1.

57. 2.1.7 Three Forms of Binary Data Radix complements ( 补码 ) 反码 radix-1 complement 补码 radix complement 105 0 1101001 0 1101001 -105 1 0010110 1 0010111 0 0 0000000 0 0000000 -0 1 1111111 0 0000000

58. 2.1.7 Three Forms of Binary Data Example: -8 +8=00001000 1000 (write number to first 1) 1111 (invert the remaining bits) -8= 1111 1000

59. 2.1.7 Three Forms of Binary Data － 25 ＋ 25 ＝ 0001 1001 － 25 （补）＝ 1110 0111 － 39 ＋ 39 ＝ 0010 0111 － 39 （补）＝ 1101 1001

60. 2.1.7 Three Forms of Binary Data Negative signed numbers are stored in the two ’ s complement form. ( 负数以 2 进制的补 码形式存储 ) A simpler technique for two ’ s implement: –Write the number exactly as it appears from right to left until the first one. Write down the first one, and then invert all the remaining bits.

61. 2.1.7 Three Forms of Binary Data If we use radix complement to represent a negative data, we can change subtract operation to add operation. Z = X – Y equal to Z 补 =X 补 + (-Y) 补

62. 2.1.7 Three Forms of Binary Data Z = X – Y equal to Z 补 =X 补 + (-Y) 补 Example 1: 64 -10 64-10 = 64 补 + (-10) 补 64 补 = 01000000 (-10) 补 =11110110 01000000 + 11110110 (right) 100110110 result = 00110110 = 54 Nature lost

63. 2.1.7 Three Forms of Binary Data Example 2: 34-68 34-68 = 34 补 +( -68) 补 34 补 =00100010 ( -68) 补 =10111100 00100010 + 10111100 11011110 result = 11011110 = -34(right) 11011110( 补 )= 11011101( 反 ) = - 00100010( 原 ) =-34

64. 10001101( 补 ) = 10001100( 反 ) = - 01110011( 原 ) =-115

65. 2.1.7 Three Forms of Binary Data The example shows that our test “ Z = X – Y equal to Z 补 =X 补 + (-Y) 补 ” is right, but why? The number of hour in a clock is 12. So we say the hour is mode 12. Suppose it’s 10 o’clock, if we turn the hand 8 hour clockwise, we get 6 o’clock. If we turn the hand 4 hour anticlockwise, we get 6 o’clock too.

66. 2.1.7 Three Forms of Binary Data If Z = nK + Y (n integer) Then Z and Y has the same mode to K. (the above example K = 12, Z = -4, Y=8) So for a certain mode K, subtracting Y from X (Y<K) equals to adding (-Y+K). 对于某一固定的模, X 减 Y (Y<K), 总可以 用加上该数的负数与其模之和（即补数） 来代替。

67. 2.1.7 Three Forms of Binary Data You can turn to the reference book “ Microcomputer Principle and Application ” 《微机原理和应用》 page 9 for details.

68. 2.1.7 Three Forms of Binary Data Z = X – Y equal to Z 补 =X 补 + (-Y) 补 Example 1: 64 -10 64-10 = 64 补 + (-10) 补 64 补 = 01000000 (-10) 补 =11110110 01000000 + 11110110 (right) 100110110 result = 00110110 = 54 Nature lost

69. 2.1.7 Three Forms of Binary Data Example 3: 127 +8 127= 01111111 8=00001000 01111111 + 00001000 10000111 result = 135 Overflow Error!

70. 加法运算溢出 – 当次高位向最高位产生进位，而最高位不向 前产生进位时， O = 1( 溢出） – 当次高位向最高位无进位，而最高位向前有 进位时， O ＝ 1 （溢出） – 当最高位向前有进位时， C=1, 否则 C ＝ 0 – 0 1 1 0 0 1 0 0 – ＋ 0 1 1 0 0 1 0 0 – 1 1 0 0 1 0 0 0 O=1 C=0 加法运算溢出 – 当次高位向最高位产生进位，而最高位不向 前产生进位时， O = 1( 溢出） – 当次高位向最高位无进位，而最高位向前有 进位时， O ＝ 1 （溢出） – 当最高位向前有进位时， C=1, 否则 C ＝ 0 – 0 1 1 0 0 1 0 0 – ＋ 0 1 1 0 0 1 0 0 – 1 1 0 0 1 0 0 0 O=1 C=0 次高位最高位

71. 加法运算溢出 – 当次高位向最高位产生进位，而最高位不向 前产生进位时， O = 1( 溢出） – 当次高位向最高位无进位，而最高位向前有 进位时， O ＝ 1 （溢出） – 当最高位向前有进位时， C=1, 否则 C ＝ 0 – 1 0 1 0 1 0 1 1 – ＋ 1 1 1 1 1 1 1 1 – 1 1 0 1 0 1 0 1 0 O=0 C=1

72. 加法运算溢出 – 当次高位向最高位产生进位，而最高位不向 前产生进位时， O = 1( 溢出） – 当次高位向最高位无进位，而最高位向前有 进位时， O ＝ 1 （溢出） – 当最高位向前有进位时， C=1, 否则 C ＝ 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 +0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 O=0 C=0

73. 减法运算溢出 – 当次高位不向最高位借位，而最高位向前借位， O=1 – 当次高位向最高位借位，而最高位不向前借位， O=1 – 当最高位向前借位， C=1, 否则 C=0 – 1 0 0 0 1 0 0 0 – － 0 1 0 0 1 0 0 0 – 0 1 0 0 0 0 0 0 O=1 C=0

74. 2.2 Computer Data Formats

75. 2.2.1 ASCII Data 2.2.2 BCD Data 2.2.3 Byte-Sized Data 2.2.4 Word-Sized Data 2.2.5 Doubleword-Sized Data 2.2.6 Real Numbers

76. 2.2.1 ASCII Data

77. ASCII (American Standard Code for Information Interchange) represent alphanumeric characters in the memory of a computer system. ASCII use 8-bit(00H — FFH) represent alphanumeric characters in computer.

78. 2.2.1 ASCII Data Table 1-7 on page 35 shows what is 00H — 7FH representing. Table 1-8 on the same page shows extended ASCII code. Including what is 8FH — FFH representing. How to represent characters in other countries?

79. 2.2.1 ASCII Data Many Windows-based applications use the Unicode system to store alphanumeric data and thus resolve the problem. Unicode system use 16-bit data to represent character. 0000H — 00FFH is the same as ASCII code 0100H — FFFFH is used to store other special characters from other countries.

80. 2.2.2 BCD Data

81. BCD code decimal digit BCD code 0 1 0000 0001 2 3 0010 0011 4 5 0100 0101 6 7 0110 0111 8 9 1000 1001

82. 1MB 1M BYTE 8 Bit

83. 2.2.2 BCD Data BCD data has two forms. One is packed and the other is unpacked. In all case, store the least-significant data first.( 规则 : 先存储最低有效位 ) Packed BCD data are stored as two digits per byte 37-----0011 0111-----37H Unpacked BCD data are stored as one digit per byte 37-----0000 0011 0000 0111 -----0307H

84. 2.2.3 Byte-Sized Data

85. The weights of each binary-bit position Unsigned byte Signed byte 1286432168421 -1286432168421

86. 2.2.3 Byte-Sized Data One byte contains 8 bits. Byte-sized data are stored as unsigned and signed integers. Unsigned integers range in value from 00H to FFH (0-255) Signed integers range in value from -128 to +127.

87. 2.2.3 Byte-Sized Data Example of defining a byte-sized data: –unsigned –0000 FE DATA1 DB 254 –0001 87 DATA2 DB 87H –0002 47 DATA3 DB 71 –Signed –0003 9C DATA4 DB -100 –0004 64 DATA5 DB +100 Store address Store value All the data formats are based on a 8088 microproces sor, whose data width is 8 bits.

88. 2.2.4 Word-Sized Data

89. One word contains 16 bits. That is it is formed with two bytes of data. Word-sized data are stored as unsigned and signed integers. The difference between them is the weight of the highest position. For unsigned integers it is 32768 while signed is -32768.

90. In Intel microprocessor, the least significant byte is stored in the lowest-numbered memory location, and the most significant byte is stored in the highest-numbered memory location. DATA DW 1234H 3002H 3001H 3000H 2FFFH 2.2.4 Word-Sized Data 12H 34H Memory width : 8 bit

91. 2.2.4 Word-Sized Data Example: Unsigned 0000 09F0 DATA1 DW 2544 0002 87AC DATA2 DW 87ACH 0004 02C6 DATA3 DW 710 Signed 0006 CBA8 DATA4 DW -13400 0008 00C6 DATA5 DW +198

92. 2.2.5 Doubleword-Sized Data

93. One doubleword contains 32 bits. That is it is formed with two words or four bytes of data. Doubleword-sized data are stored as unsigned and signed integers. The difference between them is the weight of the highest position.

94. In Intel microprocessor, the least significant byte is stored in the lowest-numbered memory location, and the most significant byte is stored in the highest-numbered memory location. DATA DD 12345678H 3003H 3002H 3001H 3000H DATA DD 5678H ??? 2.2.5 Doubleword-Sized Data 12H 34H 56H 78H

95. 2.2.5 Doubleword-Sized Data Example : Unsigned 0000 0003E1C0 DATA1 DD 254400 0004 87AC1234 DATA2 DD 87AC1234H 0008 00000046 DATA3 DD 70 Signed 000C FFEB8058 DATA4 DD -1343400 0010 000000C6 DATA5 DD +198

96. 2.2.5 Doubleword-Sized Data Note ： Byte-sized, word-sized, doubleword-sized data are standard form. We can use DB to define those nonstandard data. Example : Define 123456H DB 56H,34H,12H DD 123456H What is the difference?

97. 2.2.6 Real Numbers

98. In high-level language we often encounter number like 24.5. We call it real number or floating-point number( 浮点数 ). As we have noticed, 24.5 (2.45*10 1 ) contained two parts: significand ( 尾数 ) and exponent ( 指数 ). 4-byte real number is called single- precision and 8-byte double-precision.

99. 2.2.6 Real Numbers Single-precision using a bias of 7FH Double-precision using a bias of 3FFH S 31 Exponent 30-23 Significant 22-0 S 63 Exponent 62-52 Significant 51-0

100. 2.2.6 Real Numbers The exponent is stored as bias exponent. For single-precision: bias exponent = exponent + 7FH For double-precision: bias exponent = exponent + 3FFH

101. 2.2.6 Real Numbers Table 1.10 single-precision number(P41) decimalbinarynormalizedsignBiased exponent significant +1211001.1*2 3 00111,1 111+11 100 0000 0000 0000 0000 0000H -121100-1.1*2 3 1100000 10 +10011001 00 1.1001*2 6 0011111 11+110 100 1000 0000 0000 0000 0000H

102. 2.2.6 Real Numbers Define single-precision data(32 bit) DD / REAL4 Define double-precision data(64 bit) DQ / REAL8 Define extended precision data REAL10

103. 2.2.6 Real Numbers Examples: Single-precision 0000 3F9DF3B6 NUMB1 DD 1.234 0004 C1BB3333 NUMB2 REAL4 -23.4 Double-precision 0008 405ED9999999999A NUMB3 DQ(REAL8) 123.4 Extended-precision 0010 4005F6CCCCCCCCCCCCCD NUMB4 REAL10 123.4

104. 2.2.6 Real Numbers Two exceptions to the rules for floating- point numbers. 0.0 is stored as all zeros. Positive infinity. Negative infinity. S0S0 Exponent 30-23 (ALL ONES) Significant 22-0 (ALL ZEROS) S1S1 Exponent 30-23 (ALL ONES) Significant 22-0 (ALL ZEROS) S0S0 Significant 22-0 (ALL 0) Exponent 30-23 (ALL 0)