1 / 39

Chapter 2 รูปแบบของข้อมูลในคอมพิวเตอร์ Data representation in computer

Chapter 2 รูปแบบของข้อมูลในคอมพิวเตอร์ Data representation in computer. เนื้อหา. รูปแบบการจัดเก็บข้อมูลลงบนคอมพิวเตอร์ ตัวเลขฐานสองแบบมีเครื่องหมาย Overflow ของการกระทำทางคณิตศาสตร์ของตัวเลข เลขทศนิยมแบบ Fixed-Point เลขทศนิยมแบบ Floating-Point

radha
Download Presentation

Chapter 2 รูปแบบของข้อมูลในคอมพิวเตอร์ Data representation in computer

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Chapter 2รูปแบบของข้อมูลในคอมพิวเตอร์Data representation in computer

  2. เนื้อหา รูปแบบการจัดเก็บข้อมูลลงบนคอมพิวเตอร์ ตัวเลขฐานสองแบบมีเครื่องหมาย Overflow ของการกระทำทางคณิตศาสตร์ของตัวเลข เลขทศนิยมแบบ Fixed-Point เลขทศนิยมแบบ Floating-Point การกระทำทางคณิตศาสตร์กับเลขทศนิยมแบบ Floating-Point

  3. Number representation • ลองพิจารณาตัวเลข • เมื่อ • bi = 0 หรือ 1 • j • ค่าของ B หาได้จาก B = bn-1.....b1b0 Value of B = bn-1 x 2n-1 + .... + b1x 21 + b0 x 20

  4. Binary, signed-integer representations

  5. Addition/Subtraction of Signed numbers • 2’s complement is the most efficient method

  6. Addition/Subtraction of Signed numbers • Use only Adder • Subtraction:Perform 2’s complement with subtrahend

  7. Addition/Subtraction of Signed numbers

  8. Overflow in Integer arithmetic Overflow • 4-bit signed number ranges from -8...+7 • the result from addition • more than +7 or less than -8, overflow occurred

  9. Overflow • Overfolow detection rules: • 1. Overflow can occur only when adding 2 numbers that have the same sign • 2. When adding X and Y, overflow occurs when the sign of result is not the same as the sign of X and Y

  10. Number representation • We always represent a number in the 2’s complement system • 4 bit -8...+7 • 8 bit -128...+127 • 16 bit -32768....+32767 • 32 bit -2147483648.... +2147483647

  11. Sign extension • To represent 2’s complement signed number using larger number of bits, repeat the sign bits as many times as needed to the left • for example : convert 4 bits to 8 bits • 1001 (-7) 11111001(-7)

  12. Characters • ASCII : American Standard Code for Information Interchange ที่มาของรูป http://www.jimprice.com/ascii-0-127.gif

  13. Memory location and Addresses • Memory consists of many millions of storage cells,each of which can store a bit of information (0/1) • memory is organized into a group of n bits can be stored or retrieved in a single, basic operation • Each group of n bits is referred to as a word of information

  14. Memory location and Addresses • Bit, byte, word • A unit of 8 bit is called byte • Word length typically ranges from 16 to 64 bits • CPU access data in memory for 1 word at a time

  15. Word • 32-bit word can store • 32-bit 2’s complement number • four ASCII characters

  16. Memory accessing • To access the memory, addresses for each memory location is required • Addresses range from 0 through 2k-1 for 2kaddress space • 24-bit address generates address space of 224 or 16,777,216 locations.

  17. Byte addressable memory • most modern computer have successive addresses refer to successive byte location in the memory • Byte locations have address 0,1,2,3..... • Successive words are located at addresses 0,4,8,12,.... (for 32-bit machine)

  18. Big-Endian and Little-Endian assignments • 2 ways to assign byte address across words

  19. Little endian VS Big endian

  20. Little endian VS Big endian

  21. Word alignment • For Example: keeping value 201F539AH in memory

  22. Fixed-point number Sign bit -1 (1-2-(n-1)) B = b0.b-1b-2.....b-(n-1) F(B) = (- b0 x 20)+ (b-1 x 2-1)+(b-2 x 2-2)+... + (b-(n-1) x 2-(n-1))

  23. Fixed-point number F(B) = 1.1001011 F(B) = (- 1x 20)+ (1x 2-1)+(0x 2-2)+(0x 2-3)+(1x 2-4)+(0x 2-5)+(1x 2-6)+(1x 2-7) = -1+0.5+0+0+0.0625+0+0.015625+0.0078125 = -0.4140625

  24. Fixed-point number • 32-bit, signed, fixed point represent value range approximately 4.55*10-10 to 1 • This is not sufficient for scientific caclulation such as • Avogadro’s number 6.0247*1023 mole-1 • Planck’s constant 6.6254*10-27 erg s

  25. Floating-point number • General form for floating point number in decimal system +X1.X2X3X4X5X6X7*10+Y1Y2 • When the decimal point is placed to the right of the first(nonzero) significant digit, the number is said to be normalized exponent Significant digits

  26. IEEE standard floating-point format 23-bit mantissa fraction Sign bit 8-bit signed exponent in excess-127 representation Value represented = + 1.M x 2 E’-127

  27. IEEE standard floating-point format Value represented = + 1.M x 2 E E =signed exponent E’ = E + 127 1 < E’ < 254, 0 and 255 are used to represent special values -126 < E < 127

  28. Special Values E’ = 0 and M = 0 -----> 0 E’ = 255 and M = 0-----> E’ = 0 and M ≠ 0 -----> denormal number E’ = 255 and M ≠ 0 -----> NaN (not a number)

  29. floating-point format : Example 1.0010110….1 x 2-87

  30. Normalized vs unnormalized value 0.0010110…. x 29 Unnormalized value

  31. Normalized vs unnormalized value 1. 0110…. x 26 Normalized value

  32. Floating-point Add/Subtract rule • 1. Choose the number with the smaller exponent and shift its mantissa right a number of steps equal to the difference in exponents • 2. Set the exponent of the result equal to the larger exponent • 3. Perform addition/subtraction on the mantissas • 4. Normalize the resulting value, if necessary

  33. Floating-point Add/Subtract:Example 2.9400 x 102 + 4.3100 x 104 = 0.0294 x 104 + 4.3100 x 104 =4.3688 x 104

  34. Floating-point Multiply rule • 1. Add the exponents and subtract 127 • 2. Multiply the mantissas and determine the sign of the result • 3. Normalize the resulting value, if necessary

  35. Floating-point Multiply:Example 2.9400 x 102 x 4.3100 x 104 = (2.9400 x 4.3100) x 10 (2+4) =12.6714 x 106 =1.26714 x 107

  36. Floating-point Divide rule • 1. Subtract the exponents and add 127 • 2. Divide the mantissas and determine the sign of the result • 3. Normalize the resulting value, if necessary

  37. Floating-point Divide:Example 4.3100 x 104 ÷2.9400 x 102 = (4.3100 ÷2.9400) x 10 (4-2) =1.46598… x 102

  38. จบ บทที่ 2

More Related