Bài giảng Computer architecture - Chapter 3: Arithmetic for computers

Introduction
Computer words are composed of bits; thus, words can be represented as
binary numbers. Chapter 2 shows that integers can be represented either in
decimal or binary form, but what about the other numbers that
commonly occur?
For example:
■ What about fractions and other real numbers?
■ What happens if an operation creates a number bigger than can be
represented?
■ And underlying these questions is a mystery: How does hardware
really multiply or divide numbers? 
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  1. CE Arithmetic for Computers 1. Introduction 2. Addition and Subtraction 3. Multiplication 4. Division 5. Floating Point 6. Parallelism and Computer Arithmetic: Associativity 11
  2. CE Multiplication Example Although the decimal example above happens to use only 0 and 1, multiplication of binary numbers must always use 0 and 1, and thus always offers only these two choices: 1. Just place a copy of the multiplicand (1 ×multiplicand) in the proper place if the multiplier digit is a 1. 2. Place 0 (0 ×multiplicand) in the proper place if the multiplier digit is 0. 12
  3. CE Multiplication Sequential Version of the Multiplication Algorithm and Hardware Fig.1: First version of the division hardware Note: Three steps are repeated 32 times to obtain the product. If each step took a clock cycle, this algorithm would require almost 100 clock cycles to multiply two 32-bit numbers. Fig.2: The first multiplication algorithm 13
  4. CE Multiplication Sequential Version of the Multiplication Algorithm and Hardware Example: Answer: Step by step follow the multiplication algorithm 14
  5. CE Multiplication Sequential Version of the Multiplication Algorithm and Hardware Fig.3 Refine version of the multiplication hardware  Compare with the first version in the previous slide, the Multiplicand register, ALU, and Multiplier register are all 32 bits wide, with only the Product register left at 64 bits.  It just takes one clock cycle to get the Product. 15
  6. CE Multiplication Signed Multiplication  The easiest way to understand how to deal with signed numbers is to first convert the multiplier and multiplicand to positive numbers and then remember the original signs.  The algorithms should then be run for 31 iterations, leaving the signs out of the calculation.  It turns out that the last algorithm will work for signed numbers. 16
  7. CE Multiplication Faster Multiplication Fig.4 Fast multiplication hardware. 17
  8. CE Multiplication Multiply in MIPS  MIPS provides a separate pair of 32-bit registers to contain the 64-bit product, called Hi and Lo.  To produce a properly signed or unsigned product, MIPS has two instructions: multiply (mult) and multiply unsigned (multu).  To fetch the integer 32-bit product, the programmer uses move from lo (mflo). The MIPS assembler generates a pseudo instruction for multiply that specifies three general purpose registers, generating mflo and mfhi instructions to place the product into registers. 18
  9. CE Arithmetic for Computers 1. Introduction 2. Addition and Subtraction 3. Multiplication 4. Division 5. Floating Point 6. Parallelism and Computer Arithmetic: Associativity 19
  10. CE Division  The reciprocal operation of multiply is divide, an operation that is even less frequent and even more quirky.  It even offers the opportunity to perform a mathematically invalid operation: dividing by 0. Example: 20
  11. CE Division A division algorithm and hardware Fig.5 First version of the multiplication hardware Note: both the dividend and the divisor are positive and hence the quotient and the remainder are nonnegative. The division operands and both results are 32-bit values, and we will ignore the sign for now. Fig.6 The first division algorithm 21
  12. CE Division The division algorithm and hardware Example: Answer: Step by step follow the multiplication algorithm 22
  13. CE Division A faster division hardware Fig.7 The improved version of the division hardware 23
  14. CE Division Signed division  The simplest solution is to remember the signs of the divisor and dividend and then negate the quotient if the signs disagree.  The one complication of signed division is that we must also set the sign of the remainder. Remember that the following equation must always hold: Dividend = Quotient x Divisor + Remainder Remainder = Dividend – (Quotient x Divisor) Example: 24
  15. CE Division Divide in MIPS  You may have already observed that the same sequential hardware can be used for both multiply and divide in Fig.3 and Fig.7 .  The only requirement is a 64-bit register that can shift left or right and a 32-bit ALU that adds or subtracts. Hence, MIPS uses the 32-bit Hi and 32-bit Lo registers for both multiply and divide.  As we might expect from the algorithm above, Hi contains the remainder, and Lo contains the quotient after the divide instruction completes.  To handle both signed integers and unsigned integers, MIPS has two instructions: divide (div) and divide unsigned (divu).  The MIPS assembler allows divide instructions to specify three registers, generating the mflo or mfhi instructions to place the desired result into a general-purpose register. 25
  16. CE Arithmetic for Computers 1. Introduction 2. Addition and Subtraction 3. Multiplication 4. Division 5. Floating Point 6. Parallelism and Computer Arithmetic: Associativity 26
  17. CE Floating Point Definitions Some representations of the real number:  The alternative notation for the above last two numbers is called scientific notation, which has a single digit to the left of the decimal point.  A number in scientific notation that has no leading 0s is called a normalized number -9 Example: 1.0ten x 10 : normalized scientific number -8 0.1ten x 10 : not normalized scientific number -10 10.0ten x 10 : not normalized scientific number  The binary number shown in scientific notation is called floating point Floating point: Computer arithmetic that represents numbers in which the binary point is not fixed. 27
  18. CE Floating Point Floating-Point representation  A designer of a floating-point representation must find a compromise between the size of the fraction and the size of the exponent. This tradeoff is between precision and range: - Increasing the size of the fraction enhances the precision of the fraction. - Increasing the size of the exponent increases the range of numbers that can be represented.  Floating-point numbers are usually a multiple of the size of a word. The representation of a MIPS floating-point number: Where s is the sign of the floating-point number (1 meaning negative) exponent is the value of the 8-bit exponent field (including the sign of the exponent) fraction is the 23-bit number This representation is called sign and magnitude, since the sign is a separate bit from the rest of the number. In general, floating-point numbers are of the form: 28
  19. CE Floating Point Floating-Point representation  Overflow (floating-point): A situation in which a positive exponent becomes too large to fit in the exponent field.  Underflow (floating-point): A situation in which a negative exponent becomes too large to fit in the exponent field. To reduce chances of underflow or overflow is to offer another format that has a larger exponent. In C this number is called double, and operations on doubles are called double precision floating-point arithmetic; single precision floating-point is the name of the format in previous slide. Double precision: A floating-point value represented in two 32-bit words. Single precision: A floating-point value represented in a single 32-bit word. The representation of a double precision floating-point number: Where: s is the sign of the floating-point number (1 meaning negative) exponent is the value of the 11-bit exponent field (including the sign of the exponent) fraction is the 52-bit number 29
  20. CE Floating Point Floating-Point representation  These above formats go beyond MIPS. They are part of the IEEE 754 floating-point standard (IEEE 754), found in virtually every computer invented since 1980.  To pack even more bits into the significand (also coefficient or mantissa is part of a number in scientific notation), IEEE 754 makes the leading 1-bit of normalized binary numbers implicit. Hence, the number is actually 24 bits long in single precision (implied 1 and a 23-bit fraction), and 53 bits long in double precision (1 +52). Note: To be precise, we use the term significand to represent the 24- or 53-bit number that is 1 plus the fraction, and fraction when we mean the 23- or 52-bit number.  The representation of the rest of the numbers uses the form from before with the hidden 1 added where the bits of the fraction represent a number between 0 and 1 and E specifies the value in the exponent field. If we number the bits of the fraction from left to rights1, s2, s3, . . . ,then the value is: 30
  21. CE Floating Point Floating-Point representation  Negative exponents pose a challenge to simplified sorting. If we use two’s complement or any other notation in which negative exponents have a 1 in the most significant bit of the exponent field. Example: -1  1.0two x 2 would be represented with a negative exponent will look like a big number. +1  1.0two x 2 would be represented with a negative exponent will look like the smaller number. 31
  22. CE Floating Point Floating-Point representation  The desirable notation must therefore represent the most negative exponent as 00 . . . 00two and the most positive as 11 . . . 11two. This convention is called biased notation, with the bias being the number subtracted from the normal, unsigned representation to determine the real value.  IEEE 754 uses a bias of 127 for single precision, so an exponent of -1 is represented by the bit pattern of the value (-1 + 127ten ), or 126ten = 0111 1110two , and +1 is represented by (1+127), or 128ten = 1000 0000two .  The exponent bias for double precision is 1023.  Importance: Biased exponent means that value represented by a floating-point number is really: The range of single precision number is from as small as to as large as Fig.8: IEEE 754 encoding of floating-point numbers 32
  23. CE Floating Point Floating-Point representation Example 1: 33
  24. CE Floating Point Floating-Point representation Answer1: 34
  25. CE Floating Point Floating-Point representation Answer1: 35
  26. CE Floating Point Floating-Point representation Example 2: Converting Binary to Decimal Floating-Point Answer 2: 36
  27. CE Floating Point Floating-Point addition Let’s add numbers in scientific notation by hand to illustrate the problem in floating- 1 -1 point addition: 9.999ten x 10 + 1.610ten x 10 . Assume that we can store only four decimal digits of the significand and two decimal digits of the exponent. 37
  28. CE Floating Point Floating-Point addition Note: check for overflow or underflow the exponent still fits in its field 38
  29. CE Floating Point Floating-Point addition The algorithm for binary floating-point addition. Fig.9 The algorithm for binary floating-point addition. 39
  30. CE Floating Point Floating-Point addition Example: Try adding the number 0.5ten and -0.4375ten in binary using the algorithm in Fig.9 Answer: 40
  31. CE Floating Point Floating-Point addition Answer: 41
  32. CE Floating Point Floating-Point addition Hardware Architecture: Fig.10 Block diagram of an arithmetic unit dedicated to floating-point addition. 42
  33. CE Floating Point Floating-Point multiplication Example 1: Let’s try floating-point multiplication. We start by multiplying decimal numbers 10 -5 in scientific notation by hand: 1.110ten x 10 * 9.200ten x 10 . Assume that we can store only four decimal digits of the significand and two decimal digits of the exponent. Answer 1: 43
  34. CE Floating Point Floating-Point multiplication Note: check for overflow or underflow the exponent still fits in its field 44
  35. CE Floating Point Floating-Point multiplication 45
  36. CE Floating Point Floating-Point multiplication The algorithm for binary floating-point multiplication have 5 steps like the answer section in the Example 1 in this sector. Fig.11 The algorithm for binary floating-point multiplication. 46
  37. CE Floating Point Floating-Point multiplication Example 2: Binary Floating-Point multiplication. Let’s try multiplying the number 0.5ten and -0.4375ten. Answer 2: 47
  38. CE Floating Point Floating-Point multiplication Answer 2: 48
  39. CE Floating Point Floating-Point instruction in MIPS 49
  40. CE Floating Point Floating-Point instruction in MIPS  Floating-point comparison sets a bit to true or false, depending on the comparison condition, and a floating-point branch then decides whether or not to branch depending on the condition.  The MIPS designers decided to add separate floating-point registers ̶ called $f0, $f1, $f2, ̶ used either for single precision or double precision they included separate loads and stores for floating-point registers: lwcl and swcl.  The base registers for floating-point data transfers remain integer registers. The MIPS code to load two single precision numbers from memory, add them, and then store the sum might look like below: 50
  41. CE Floating Point Floating-Point instruction in MIPS Fig.12 MIPS floating-point architecture 51
  42. CE Floating Point Floating-Point instruction in MIPS Fig.13 MIPS floating-point architecture (cont.) 52
  43. CE Floating Point Floating- Point instruction in MIPS Fig.14 MIPS floating-point instruction encoding 53
  44. CE Floating Point Accurate Arithmetic  Unlike integers, which can represent exactly every number between the smallest and largest number, floating-point numbers are normally approximations for a number they can’t really represent.  The reason is that an infinite variety of real numbers exists between, say, 0 and 1, but no more than 253 can be represented exactly in double precision floating point. The best we can do is getting the floating-point representation close to the actual number. Thus, IEEE 754 offers several modes of rounding to let the programmer pick the desired approximation.  Rounding sounds simple enough, but to round accurately requires the hardware to include extra bits in the calculation. IEEE 754, therefore, always keeps two extra bits on the right during intermediate additions, called guard and round, respectively. 54