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

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## Chapter 2 รูปแบบของข้อมูลในคอมพิวเตอร์ Data representation in computer

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**Chapter**2รูปแบบของข้อมูลในคอมพิวเตอร์Data representation in computer**เนื้อหา**รูปแบบการจัดเก็บข้อมูลลงบนคอมพิวเตอร์ ตัวเลขฐานสองแบบมีเครื่องหมาย Overflow ของการกระทำทางคณิตศาสตร์ของตัวเลข เลขทศนิยมแบบ Fixed-Point เลขทศนิยมแบบ Floating-Point การกระทำทางคณิตศาสตร์กับเลขทศนิยมแบบ Floating-Point**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**Addition/Subtraction of Signed numbers**• 2’s complement is the most efficient method**Addition/Subtraction of Signed numbers**• Use only Adder • Subtraction:Perform 2’s complement with subtrahend**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**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**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**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)**Characters**• ASCII : American Standard Code for Information Interchange ที่มาของรูป http://www.jimprice.com/ascii-0-127.gif**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**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**Word**• 32-bit word can store • 32-bit 2’s complement number • four ASCII characters**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.**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)**Big-Endian and Little-Endian assignments**• 2 ways to assign byte address across words**Word alignment**• For Example: keeping value 201F539AH in memory**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))**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**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**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**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**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**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)**floating-point format : Example**1.0010110….1 x 2-87**Normalized vs unnormalized value**0.0010110…. x 29 Unnormalized value**Normalized vs unnormalized value**1. 0110…. x 26 Normalized value**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**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**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**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**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**Floating-point Divide:Example**4.3100 x 104 ÷2.9400 x 102 = (4.3100 ÷2.9400) x 10 (4-2) =1.46598… x 102

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