1 / 96

CSC 317 Computer Organization and Architecture

CSC 317 Computer Organization and Architecture. Chapter 2: Data Representation Spring 2007. Chapter 2 Objectives. Understand the fundamentals of numerical data representation and manipulation in digital computers. Master the skill of converting between various radix systems.

fenella
Download Presentation

CSC 317 Computer Organization and Architecture

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. CSC 317Computer Organization and Architecture Chapter 2: Data Representation Spring 2007

  2. Chapter 2 Objectives • Understand the fundamentals of numerical data representation and manipulation in digital computers. • Master the skill of converting between various radix systems. • Understand how errors can occur in computations because of overflow and truncation. • Understand the concepts of error detecting and correcting codes.

  3. 2.1 Introduction • A bit is the most basic unit of information in a computer. • It is a state of “on” or “off” in a digital circuit. • Or “high” or “low” voltage instead of “on” or “off.” • A byte is a group of eight bits. • A byte is the smallest possible addressable unit of computer storage. • A word is a contiguous group of bytes • Word sizes of 16, 32, or 64 bits are most common. • Usually a word represents a number or instruction. 3

  4. 2.2 Positional Numbering Systems • Computers use base 2 to represent numbers • The binary system is also called the base-2 system. • Our decimal system is the base-10 system. It uses powers of 10 for each position in a number. • Any integer quantity can be represented exactly using any base (or radix). • The decimal number 947 in powers of 10 is: 9  10 2 + 4  10 1 + 7  10 0 • The decimal number 5836.47 in powers of 10 is: 5  10 3 + 8  10 2 + 3  10 1 + 6  10 0 + 4  10 -1 + 7  10 -2 4

  5. 2.2 Positional Numbering Systems • An n-bit sequence of binary digits is an unsigned integer A whose value is: A = Σ 2i ai (0 ≤ i ≤ n-1) • The binary number 11001 in powers of 2 is: • 8 bits can represent unsigned numbers from 0 to 255 (28-1) • 1  2 4 + 1  2 3 + 0  2 2 + 0  2 1 + 1  2 0 • = 16 + 8 + 0 + 0 + 1 = 25 5

  6. 2.3 Decimal to Binary Conversions • In an earlier slide, we said that every integer value can be represented exactly using any radix system. • You can use either of two methods for radix conversion: the subtraction method and the division remainder method. • The subtraction method is more intuitive, but cumbersome. It does, however reinforce the ideas behind radix mathematics. • The division method employs the idea that successive division by a base is equivalent to successive subtraction by powers of the base. 6

  7. 2.3 Decimal to Binary Conversions • Suppose we want to convert the decimal number 190 to base 3. • We know that 3 5 = 243 so our result will be less than six digits wide. The largest power of 3 that we need is therefore 3 4 = 81, and 81  2 = 162. • Write down the 2 and subtract 162 from 190, giving 28. 7

  8. 2.3 Decimal to Binary Conversions • Converting 190 to base 3... • The next power of 3 is 3 3 = 27. We’ll need one of these, so we subtract 27 and write down the numeral 1 in our result. • The next power of 3, 3 2 = 9, is too large, but we have to assign a placeholder of zero and carry down the 1. 8

  9. 2.3 Decimal to Binary Conversions • Converting 190 to base 3... • 3 1 = 3 is again too large, so we assign a zero placeholder. • The last power of 3, 3 0 = 1, is our last choice, and it gives us a difference of zero. • Our result, reading from top to bottom is: 19010 = 210013 9

  10. 2.3 Decimal to Binary Conversions • Converting 190 to base 3... • First we take the number that we wish to convert and divide it by the radix in which we want to express our result. • In this case, 3 divides 190 63 times, with a remainder of 1. • Record the quotient and the remainder. 10

  11. 2.3 Decimal to Binary Conversions • Converting 190 to base 3... • 63 is evenly divisible by 3. • Our remainder is zero, and the quotient is 21. 11

  12. 2.3 Decimal to Binary Conversions • Converting 190 to base 3... • Continue in this way until the quotient is zero. • In the final calculation, we note that 3 divides 2 zero times with a remainder of 2. • Our result, reading from bottom to top is: • 19010 = 210013 12

  13. 2.3 Decimal to Binary Conversions • Fractional values can be approximated in all base systems. • Unlike integer values, fractions do not necessarily have exact representations under all radices. • The quantity ½ is exactly representable in the binary and decimal systems, but is not in the ternary (base 3) numbering system. 13

  14. 2.3 Decimal to Binary Conversions • Fractional decimal values have nonzero digits to the right of the decimal point. • Fractional values of other radix systems have nonzero digits to the right of the radix point. • Numerals to the right of a radix point represent negative powers of the radix: • 0.4710 = 4  10 -1 + 7  10 -2 • 0.112 = 1  2 -1 + 1  2 -2 • = ½ + ¼ • = 0.5+ 0.25 = 0.75 14

  15. 2.3 Decimal to Binary Conversions • As with whole-number conversions, you can use either of two methods: a subtraction method and an easy multiplication method. • The subtraction method for fractions is identical to the subtraction method for whole numbers. Instead of subtracting positive powers of the target radix, we subtract negative powers of the radix. • We always start with the largest value first, n -1, where n is our radix, and work our way along using larger negative exponents. 15

  16. 2.3 Decimal to Binary Conversions • The calculation to the right is an example of using the subtraction method to convert the decimal 0.8125 to binary. • Our result, reading from top to bottom is: • 0.812510 = 0.11012 • Of course, this method works with any base, not just binary. 16

  17. 2.3 Decimal to Binary Conversions • Using the multiplication method to convert the decimal 0.8125 to binary, we multiply by the radix 2. • The first product carries into the units place. 17

  18. 2.3 Decimal to Binary Conversions • Converting 0.8125 to binary . . . • Ignoring the value in the units place at each step, continue multiplying each fractional part by the radix. 18

  19. 2.3 Decimal to Binary Conversions • Converting 0.8125 to binary . . . • You are finished when the product is zero, or until you have reached the desired number of binary places. • Our result, reading from top to bottom is: • 0.812510 = 0.11012 • This method also works with any base. Just use the target radix as the multiplier. 19

  20. 2.3 Decimal to Binary Conversions • The binary numbering system is the most important radix system for digital computers. • However, it is difficult to read long strings of binary numbers-- and even a modestly-sized decimal number becomes a very long binary number. • For example: 110101000110112 = 1359510 • For compactness and ease of reading, binary values are usually expressed using the hexadecimal, or base-16, numbering system. 20

  21. 2.3 Decimal to Binary Conversions • The hexadecimal numbering system uses the numerals 0 through 9 and the letters A through F. • The decimal number 12 is C16. • The decimal number 26 is 1A16. • It is easy to convert between base 16 and base 2, because 16 = 24. • Thus, to convert from binary to hexadecimal, all we need to do is group the binary digits into groups of four. A group of four binary digits is called a hextet 21

  22. 2.3 Decimal to Binary Conversions • Using groups of hextets, the binary number 110101000110112 (= 1359510) in hexadecimal is: • Octal (base 8) values are derived from binary by using groups of three bits (8 = 23): Octal was very useful when computers used six-bit words. 22

  23. 2.4 Signed Integer Representation • Several representations exist for negative values: Sign Magnitude One's Complement Two's Complement 000 = +0 000 = +0 000 = +0 001 = +1 001 = +1 001 = +1 010 = +2 010 = +2 010 = +2 011 = +3 011 = +3 011 = +3 100 = -0 100 = -3 100 = -4 101 = -1 101 = -2 101 = -3 110 = -2 110 = -1 110 = -2 111 = -3 111 = -0 111 = -1 23

  24. 2.4 Signed Integer Representation • Sign Magnitude • Leftmost bit is sign bit: 0 for positive, 1 for negative • Remaining bits are magnitude • Drawbacks: • Sign bits give problems to addition and subtraction • Two representations for 0 • Rarely used 24

  25. 2.4 Signed Integer Representation • Two's Complement • Easier implementation of addition and subtraction • Leftmost bit still indicates sign • Positive numbers same as sign magnitude • Only one zero (all 0 bits) • Negating A: invert (complement) all bits of A and add 1 • Example: -55 start with +55: 0000...00110111 complement each bit: 1111...11001000 (1's complement) add 1: 1111...11001001 (2's complement) 25

  26. 2.4 Signed Integer Representation • Addition and Subtraction (numbers in 2's complement representation) • Add -55 and +58 (use 8 bits for simplicity) -55: 11001001 +58 00111010 ==> 100000011 underlined leftmost bit is an overflow (ignored here) • Overflow rule: • If two numbers are added and they are both positive or both negative, then overflow occurs if and only if the result has opposite sign 26

  27. 2.4 Signed Integer Representation • Subtraction uses addition • To subtract B from A, take 2's complement of B and add it to A • Digital circuit only need addition and complement circuits 27

  28. 2.4 Signed Integer Representation 28

  29. carry_in A 1-bit Full Adder S B carry_out 2.4 Signed Integer Representation • Given a full adder (FA), we can use it to add binary digits (up to 3) 29

  30. B31 B2 B1 B0 A31 A2 A1 A0 1-bit FA 1-bit FA 1-bit FA 1-bit FA S31 S2 S1 S0 2.4 Signed Integer Representation • Several FA's can be used to add binary numbers by feeding out the carry_out one FA to the carry_in of the FA of the left. add/sub C32 C31 C3 C2 C1 C0 32-bit Ripple Carry Adder/Subtractor (Better: Carry Lookahead Adder) Note: add/sub is ON (1) if we want A-B, otherwise is OFF 30

  31. 2.4 Signed Integer Representation • Multiplication • More complicate than addition • Done by a sequence of shifts and additions • Just like the paper-pencil approach. Use an example of a 4-bit multiplication. 1011 Multiplicand (11) 1101 Multiplier (13) 1011 0000 partial products 1011  1011  10001111 double-precision product (143) 31

  32. 2.4 Signed Integer Representation • Paper-pencil approach is inefficient • Solution: Do not wait until the end to add partial products. Algorithm (unsigned numbers): Do n times (e.g, 32 for MIPS) For each 1 on Multiplier, perform an add and a shift right For each 0 in the multiplier, perform only a shift right • Solutions for the multiplication of signed numbers: • Convert the numbers to positive numbers, multiple them as above. If signs were different, negate answer. • Apply Booth Algorithm

  33. 2.4 Signed Integer Representation • Booth’s algorithm • Multiplier and multiplicand are placed in registers Q & M • Q-1, 1-bit register placed to the right of Q0 Initialize A (third register) and Q-1 to 0 Do n times (n is the number of bits in Q): If Q0Q-1 = 01 then A <-- A + M If Q0Q-1 = 10 then A <-- A – M Arithmetic shift right A, Q, Q-1 33

  34. 2.4 Signed Integer Representation

  35. 2.4 Signed Integer Representation • Multiply 3 x 7 = 21 (00010101) Phase Comments M A Q Q-1 Initialize registers 0111 0000 0011 0 1 Q0Q-1 = 10, AA-M 1001 0011 0 ASR A, Q, Q-1 1100 1001 1 2 Q0Q-1 = 11, ASR A, Q, Q-1 1110 0100 1 3 Q0Q-1 = 01, AA+M 0101 0100 1 ASR A, Q, Q-1 0010 1010 0 4 Q0Q-1 = 00, ASR A, Q, Q-1 0001 0101 0 35

  36. 2.5 Floating-Point Representation • Integers can be considered as fixed point numbers • Decimal point at the far right • Floating-point (fp) numbers allow an arbitrary number of decimal places to the right of the decimal point. • For example: 0.5  0.25 = 0.125 • A 32-bit fp number: 000000000000000000011101.01001111 is equivalent to: 2 4 + 2 3 + 2 2 + 2 0 + 2 -2 + 2 -5 + 2 -6 + 2 -7 + 2 -8 = 29.30879375 36

  37. 2.5 Floating-Point Representation • Very large or very small fp numbers are expressed in scientific notation. • For example: 0.125 = 1.25  10-1 5,000,000 = 5.0  106 • Numbers written in scientific notation have three components: 37

  38. 2.5 Floating-Point Representation • Computer representation of a fp number: • The one-bit sign field is the sign of the stored value. • The size of the exponent field, determines the range of values that can be represented. • The size of the significand determines the precision of the representation. 38

  39. 2.5 Floating-Point Representation • The IEEE-754 single precision floating point standard uses an 8-bit exponent and a 23-bit significand. • The IEEE-754 double precision standard uses an 11-bit exponent and a 52-bit significand. For illustrative purposes, we will use a 14-bit model with a 5-bit exponent and an 8-bit significand. 39

  40. 2.5 Floating-Point Representation • The significand of a floating-point number is always preceded by an implied binary point. • Thus, the significand always contains a fractional binary value. • The exponent indicates the power of 2 to which the significand is raised. • (-1)sign X significand X 2exponent 40

  41. 2.5 Floating-Point Representation • Example: • Express 3210 in the simplified 14-bit floating-point model. • We know that 32 is 25. So in (binary) scientific notation 32 = 1.0 x 25 = 0.1 x 26. • Using this information, we put 110 (= 610) in the exponent field and 1 in the significand as shown. 41

  42. 2.5 Floating-Point Representation • The illustrations shown at the right are all equivalent representations for 32 using our simplified model. • Not only do these synonymous representations waste space, but they can also cause confusion. 42

  43. 2.5 Floating-Point Representation • Another problem with our system is that we have made no allowances for negative exponents. We have no way to express 0.5 (=2 -1)! (Notice that there is no sign in the exponent field!) All of these problems can be fixed with no changes to our basic model. 43

  44. 2.5 Floating-Point Representation • To resolve the problem of synonymous forms, the first digit of the significand must be 1 (normalized). This results in a unique pattern for each FP number. • In the IEEE-754 standard, this 1 is implied meaning that a 1 is assumed to the left of the binary point. • Hidden bit extends the significand by 1 bit • Number normalized to the form: 1.bbbbbb… • Biased exponents represent negative exponents • They are approximately midway point in range of values. • Bias of 127 for single precision,1023 for double precision. • Biased exponent: exponent + bias 44

  45. 2.5 Floating-Point Representation • Exponent ranges (in theory): • Unbiased: -127 to 128 ( -1023 to 1024) • Biased: 0 to 255 ( 0 to 2047) • Example: • Express -12.7510 in the revised single-precision IEEE-754 FP standard • Binary: -1100.11 = -1.10011 x 2 3 • biased exponent: 3 + 127 = 130 = 1000 0010 • significand: 1001 1000 0000 0000 0000 000 • (1 to the left of the binary point is hidden) • Bit 31 Final FP representation: Bit 0 • 1 10000010 10011000000000000000000 45

  46. 2.5 Floating-Point Representation

  47. 2.5 Floating-Point Representation • In IEEE-754 floating point standard: • An exponent of 0 and 255 (2047 for double precision) are used for special values (see next table). • Denormalized numbers extend the range of small numbers • Smallest normalized number: 1.0x2-126 • Smallest denormalized number: 0.000…1 x 2-126 = 2-149 • New actual range of exponents: • unbiased exponent range: -126 to 127 (-1022 to 1023) • biased exponent range: -1022 to 1023 (1 to 2046) 47

  48. 2.5 Floating-Point Representation F: fraction or significand

  49. 2.5 Floating-Point Representation Floating-point arithmetic • A FP operation may produce one of the conditions: • Exponent overflow: A positive exponent may exceed the maximum value. • Exponent underflow: A negative exponent is less than the minimum value. • Significand overflow: May happen during the addition of two significands of same sign. • Significand underflow: May happen during significand alignment. 49

  50. 2.5 Floating-Point Representation Addition and subtraction • Steps to add 2 FP numbers X and Y: Step 0: Restore hidden bits: Make explicit the implicit significand bit and change the sign of the subtrahend if it is a subtract operation • Step 1: Zero Check: Check if either operand is zero • Step 2: Significand alignment: Align significands so that exponents are equal Step 3: Addition: Add the two significands taking into account their signs. Significand or exponent overflow may exist. Step 4: Normalization: Normalize the result by shifting left significand digits and decrementing the exponent, which it may cause exponent overflow. 50

More Related