Module 2: Computer Arithmetic


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1 Module 2: Computer Arithmetic 1 B O O K : C O M P U T E R O R G A N I Z A T I O N A N D D E S I G N, 3 E D, D A V I D L. P A T T E R S O N A N D J O H N L. H A N N E S S Y, M O R G A N K A U F M A N N P U B L I S H E R S
2 Complement Addition Subtraction Multiplication Division Arithmetic Operations 2
3 Introduction Numbers are represented by binary digits (bits): How are negative numbers represented? What is the largest number that can be represented in a computer world? What happens if an operation creates a number bigger than can be represented? What about fractions and real numbers? 3 A mystery: How does hardware really multiply or divide numbers?
4 Binary Numbers 4
5 Binary Numbers Binary Number System System Digits: 0 and 1 Bit (short for binary digit): A single binary digit LSB (least significant bit): The rightmost bit MSB (most significant bit): The leftmost bit 5 Upper Byte (or nybble): The righthand byte (or nybble) of a pair Lower Byte (or nybble): The lefthand byte (or nybble) of a pair
6 Binary Equivalents Binary Numbers 1 Nybble (or nibble) = 4 bits 1 Byte = 2 nybbles = 8 bits 1 Kilobyte (KB) = 1024 bytes 1 Megabyte (MB) = 1024 kilobytes = 1,048,576 bytes 1 Gigabyte (GB) = 1024 megabytes = 1,073,741,824 bytes 6
7 Binary Numbers 7 Base 2
8 Binary Addition Rules of Binary Addition = = = = 0, and carry 1 to the next more significant bit Example: Carry
9 Binary Subtraction Rules of Binary Subtraction 00 = = 1, and borrow 1 from the next more significant bit 10 = = 0 Example: borrowed
10 Binary Multiplication Rules of Binary Multiplication 0 x 0 = 0 0 x 1 = 0 1 x 0 = 0 1 x 1 = 1, and no carry or borrow bits Example: 23 x 3 10 Another Method: Binary multiplication is the same as repeated binary addition
11 Binary division 11
12 2 s Complement Two's complement representation allows the use of binary arithmetic operations on signed integers, yielding the correct 2's complement results. Positive Numbers: Positive 2's complement numbers are represented as the simple binary. Negative Numbers: Negative 2's complement numbers are represented as the binary number that when added to a positive number of the same magnitude, will equals zero. 12
13 13 To represent positive and negative numbers look at the MSB (or the sign bit) MSB = 0 means positive MSB = 1 means negative
14 Step 1: Calculation of 2's Complement 14 invert the binary equivalent of the number by changing all of the ones to zeroes and all of the zeroes to ones (also called 1's complement) Step 2: Then add one. Example: = = Step1 : Step2 :
15 2's Complement Addition Two's complement addition follows the same rules as binary addition. Example: 5 + (3) 5 = = Ignore (2)
16 2's Complement Addition Two's complement addition follows the same rules as binary addition. Example: 3 + (5) 3 = = (2)
17 2's Complement Subtraction 17 Two's complement subtraction is the binary addition of the minuhend to the 2's complement of the subtrahend (adding a negative number is the same as subtracting a positive one). Example : (12) minuhend subtrahend
18 2's Complement Subtraction 18 Example : (12) 7 = = (5) CHECK! 5 = Step1 : Step2 : (5)
19 2's Complement Multiplication Two's complement multiplication follows the same rules as binary multiplication. Example : (4) 4 = (16) Ignore 4 = = x (16) 19 CHECK! 16 = Step1 : Step2 : (16)
20 2's Complement Division Two's complement division is repeated 2's complement subtraction. 20 The 2's complement of the divisor is calculated, then added to the dividend. For the next subtraction cycle, the quotient replaces the dividend. This repeats until the quotient is too small for subtraction or is zero, then it becomes the remainder. The final answer is the total of subtraction cycles plus the dividend remainder Example : 6/ 3 = 2 divisor quotient
21 2's Complement Division Example: 6/3 Example: 7/3 6 + (3) = (3) = 0 6 / 3 = 2 Cycle 1 Cycle 2 The number of cycle (3) = (3) = 1 Cycle 1 Cycle 2 7 / 3 = 2 remainder 1
22 2's Complement Division 22 Example: 42/ (6) = (6) = (6) = (6) = 18 Cycle 1 Cycle 2 Cycle 3 Cycle (6) = (6) = (6) = 0 42 / 6 = 7 Cycle 5 Cycle 6 Cycle 7
23 Sign Extension Extending a number representation to a larger number of bits. Example: 2 in 8 bit binary to 16 bit binary In signed numbers, it is important to extend the sign bit to preserve the number (+ve or ve) Example: 2 in 8 bit binary to 16 bit binary Sign bit Sign bit extended Sign bit
24 Detecting Overflow in Two Complement Numbers Overflow occurs when adding two positive numbers and the sum is negative, or vice versa A carry out occurred into the sign bit Overflow conditions for addition and subtraction 24
25 Overflow Rule for addition If 2 Two's Complement numbers are added, and they both have the same sign (both positive 7 or both = 1001 negative), 6 then = overflow 1010 occurs if and only if the result has the opposite sign. Adding two positive numbers must give a positive result Adding two negative numbers must give a negative result Overflow occurs Overflow never occurs when adding operands with different signs. Overflow occurs if 25 (+A) + (+B) = C ( A) + ( B) = +C (3) Example: Using 4bit Two's Complement numbers ( 8 x +7) (7) + (6) = (13) but Overflow (largest ve number is 8) The sign bit has changed to +ve
26 Overflow Rule for Subtraction If 2 Two's Complement numbers are subtracted, and their signs are different, then overflow occurs if and only if the result has the same sign as the subtrahend. subtrahend Overflow occurs if (+A) ( B) = C ( A) (+B) = +C result Example: Using 4bit Two's Complement numbers ( 8 x +7) Subtract 6 from +7 (i.e. 7 (6)) 26 result has the same sign as the subtrahend overflow happens
27 Overflow Rule for Subtraction If 2 Two's Complement numbers are subtracted, and their signs are different, then overflow occurs if and only if the result has the same sign as the subtrahend. Overflow occurs if (+A) ( B) = C ( A) (+B) = +C 7 = 0111 Example: Using 4bit Two's Complement numbers ( 8 x +7) Subtract 6 from +7 (i.e. 7 (6)) 27 Overflow occurs 6 = (3) Result has same sign as subtrahend
28 Addition A little summary Add the values, discarding any carryout bit Subtraction Negate the subtrahend and add, discarding any carryout bit Overflow (96) Not 160 because the sign bit is 1. (largest +ve number in 8 bits is 127) Occurs when adding two positive numbers produces a negative result, or when adding two negative numbers produces a positive result. Adding operands of unlike signs never produces an overflow Notice that discarding the carry out of the most significant bit during Two's Complement addition is a normal occurrence, and does not by itself indicate overflow As an example of overflow, consider adding ( = 160) 10, which produces a result of in 8bit two's complement.
29 Binary Multiplication 29
30 Two's Complement Multiplication There are a couple of ways of doing two's complement multiplication by hand. Remember that the result can require 2 times as many bits as the original operands. 30
31 Sign Extend Method" for Two's Complement Multiplication In 2's complement, sign extend both integers to twice as many bits. Then take the correct number of result bits from the least significant portion of the result A 4bit example: 2 x (3) 2= = = Sign extend to 8 bit x Correct answer underlined
32 " Partial Product Sign Extend Method " for Two's Complement Multiplication 32 Another way is to sign extend the partial products 1100 to the correct number of bits. x 0011 Sometimes we do have to make some adjustments Example 1: (4) x 34 = = 0011 Sign extend to the correct number of bits (12)
33 " Partial Product Sign Extend Method " for Two's Complement Multiplication Sometimes we do have to make some adjustments. If (+ve) x (+ve) then OK Normal stuff If (+ve) x (ve) then get additive inverse of both And then Sign extend partial products Example: 3 x (4) 33 If (ve) x (+ve) then Sign extend partial products 3=0011 ;4 = =1101; 4 =0100 If (ve) x 1101 (ve) then get x additive 0100 inverse of both (12) Like the slide before (4)x3
34 " Partial Product Sign Extend Method " for Two's Complement Multiplication Sometimes we do have to make some adjustments. 34 If (+ve) x (+ve) then OK 3= = If (+ve) x (ve) then x 0011 get additive inverse of both 0011 And then 0011 Sign extend partial products (9) Example: 3 x (4) If (ve) x (+ve) then Sign extend partial products If (ve) x (ve) then get additive inverse of both Example: (3)x(3)
35 Signed Multiplication Another way to deal with signed numbers. 35 First convert the multiplier and multiplicand to positive numbers and then remember the original signs Leave the sign out of the calculation To negate the product only if the original signs disagree
36 1st Version of Multiplication Hardware Actual implementations are far more complex, and use algorithms that generate more than one bit of product each clock cycle. 32bit multiplicand starts at right half of multiplicand register Algorithm Flows of 1st Version Multiplication Multiplier0 = 1 Start 1. Test Multiplier0 Multiplier0 = 0 Multiplicand 64 bits Shift left 1a. Add multiplicand to product and place the result in Product register 64bit ALU Multiplier Shift right 32 bits 2. Shift the Multiplicand register left 1 bit Product 64 bits Write Control test 3. Shift the Multiplier register right 1 bit Product register is initialized at 0 Multiplicand register, product register, ALU are 64bit wide; multiplier register is 32bit wide 32nd repetition? No: < 32 repetitions Yes: 32 repetitions Done
37 Example of Multiplication 4 bits Example : 2 x 3 =? Multiplicand (MC) Multiplier (MP) Product (P) 2 x x 0011 Steps: 1a test multiplier (0 or 1) If 1 then P = P + MC If 0 then no operation 2 shift MC left 3 shift MP right All bits done? If still <max bit, repeat If = max bit, stop 37 Multiplier0 = 1 1a. Add multiplicand to product and place the result in Product register Start Test Multiplier0 2. Shift the Multiplicand register left 1 bit 3. Shift the Multiplier register right 1 bit 32nd repetition? Multiplier0 = 0 P = P + MC = = MC = MP = No: < 32 repetitions Max bit? NO repeat Yes: 32 repetitions Done
38 Iteration Step Multiplier (MP) Multiplicand (MC) Product (P) 0 Initial value a:1 P = P + MC 2: Shift MC left 3: Shift MP right 1a:1 P = P + MC 2: Shift MC left 3: Shift MP right 1a:0 no operation 2: Shift MC left 3: Shift MP right 1a:0 no operation 2: Shift MC left 3: Shift MP right Try with 5 x
39 Iteration Step Multiplier (MP) Multiplicand (MC) Product (P) 0 Initial value Example : 5 x 4 1a:0 no operation 2: Shift MC left : Shift MP right a:0 no operation 2: Shift MC left : Shift MP right a:1 P = P + MC : Shift MC left : Shift MP right a:0 no operation 2: Shift MC left : Shift MP right 0000 Try with 2 x (3) 20 40
40 Iteration Step Multiplier (MP) Multiplicand (MC) Product (P) 0 Initial value a:1 P = P + MC Example : 2 x (3) get additive inverse of both 2: Shift MC left : Shift MP right a:1 P = P + MC : Shift MC left : Shift MP right a:0 no operation 2: Shift MC left : Shift MP right a:0 no operation 2: Shift MC left : Shift MP right
41 Binary Division 42
42 1st Version of Division Hardware Divisor starts at left half of divisor register 32bit divisor starts at left half of divisor register Divisor 64 bits Shift right Quotient register is initialized to be 0 43 Flows of 1st Version Division Start 1. Subtract the Divisor register from the Remainder register and place the result in the Remainder register Remainder > 0 Test Remainder Remainder < 0 64bit ALU Remainder Write Control test Quotient Shift left 32 bits 2a. Shift the Quotient register to the left, setting the new rightmost bit to 1 2b. Restore the original value by adding the Divisor register to the Remainder register and place the sum in the Remainder register. Also shift the Quotient register to the left, setting the new least significant bit to 0 64 bits Remainder register is initialized with the dividend at right 3. Shift the Divisor register right 1 bit Divisor register, remainder register, ALU are 64bit wide; quotient register is 32bit wide 33rd repetition? No: < 33 repetitions Yes: 33 repetitions Done Algorithm
43 Example of Division 4 bits 44 Start 1. R = R  D Example : 7 / 2 =? Dividend (DD) Divisor (D) 7 / / 0010 Steps: 1 Remainder (R) = R D 2 test new R (>=0 or <0) 2a  If >=0 then R = no operation; Q = Shift left (add 1 at LSB) 2b  If <0 then R = D + R Q = Shift left (add 0 at LSB) 3 shift D right All bits done? If still <(max bit + 1), repeat If = (max bit+1), stop Quotient (Q) If R = R D = +ve 2a. Shift the Quotient register to the left, setting the new rightmost bit to 1 1. Subtract the Divisor register from the Remainder register and place the result in the Remainder register Remainder > 0 2a. R >= 0; R = no operation Q = Shift left (add 1 at LSB) 4. If not yet 33 repeat to Step 1 (new iteration) Test Remainder 3. D = Shift right 1 3. Shift the Divisor register right 1 bit 33rd repetition? If R = R D = ve Remainder < 0 2b. R < 0; R = D + R Q = Shift left (add 0 at LSB) 2b. Restore the original value by adding the Divisor register to the Remainder register and place the sum in the Remainder register. Also shift the Quotient register to the left, setting the new least significant bit to 0 No: < 33 repetitions Yes: 33 repetitions Done
44 Iteration Step Quotient (Q) Divisor Divisor Remainder ( (D) starts at left half Example: 7/2 R) of divisor register 0 Initial value R = R  D R = R D = R + (D) Negate b. R < 0; R = D + R Q = Shift left (add 0 at LSB) D = Shift right R = R  D b. R < 0; R = D + R Q = Shift left (add 0 at LSB) D = Shift right R = R  D b. R < 0; R = D + R Q = Shift left (add 0 at LSB) D = Shift right
45 Iteration Step Quotient (Q) Divisor (D) Remainder(R) R = R  D b. R < 0; R = D + R Q = Shift left (add 0 at LSB) D = Shift right R = R  D b. R >=0; R = no operation Q = Shift left (add 1 at LSB) D = Shift right R = R  D b. R >=0; R = no operation Q = Shift left (add 1 at LSB) D = Shift right Example: 7/2 = 3 remainder 1 47
46 Example: 6/3 Iteration Step Quotient (Q) Divisor (D) Remainder(R) 0 Initial value R = R  D b. R < 0; R = D + R Q = Shift left (add 0 at LSB) D = Shift right R = R  D b. R < 0; R = D + R Q = Shift left (add 0 at LSB) D = Shift right R = R  D b. R < 0; R = D + R Q = Shift left (add 0 at LSB) D = Shift right
47 Iteration Step Quotient (Q) Divisor (D) Remainder (R) R = R  D b. R < 0; R = D + R Q = Shift left (add 0 at LSB) D = Shift right R = R  D b. R >=0; R = no operation Q = Shift left (add 1 at LSB) D = Shift right R = R  D b. R < 0; R = D + R Q = Shift left (add 0 at LSB) D = Shift right Example: 6/3 = 2 49
48 Signed Division Signed divide: make both divisor and dividend positive and perform division negate the quotient if divisor and dividend were of opposite signs make the sign of the remainder match that of the dividend this ensures always dividend = (quotient divisor) + remainder quotient (x/y) = quotient ( x/y) e.g. 7 = = 3 2 1) 50
49 Module 2 Part 2 51 FLOATING POINT
50 Floating Point Representation Floating point aids in the representation of very big or very small fixed point numbers x Fixed point Floating point 976,000,000,000, x x 1014
51 Floating Point Numbers 53 Significand/Fraction Exponent x Base/radix Significand and Exponent can be +ve or ve. Decimal numbers use the radix 10, binary use 2
52 Normalized and Unnormalized In generalized normalization (like in mathematics), a floating point number number is said to after be normalized the if the number after the radix radix point point is a nonzero is a value. Unnormalized floating number is when the number after the radix point is 0. Example: x x x x 10 5 unnormalized nonzero value. unnormalized normalized normalized
53 Normalization Process Normalization is the process of deleting the zeroes until a Move nonzero radix point value to is detected. the right (in this case 2 Example points) : 56 Move radix point to the left (in this x case points) x x x x x 10 6 A rule of thumb: moving the radix point to the right subtract exponent moving the radix point to the left add exponent
54 Floating Point Format for Binary Numbers In the beginning, the general form of floating point is : +/ 0.Mantissa x r +/ exponent In binary: +/ (sign) Exponent +/ (sign) Mantissa 1 word Mantissa = Significand The 2 sign bits are not good for design as it incurs extra costs
55 Biased Exponent A new value to represent the exponent without the sign bit is introduced 59 This will eliminate the sign for the exponent value that is the exponent will be positive. (indicative) +/ (sign) Biased Exponent Mantissa 1 word +/ (1 bit) E t (n bit) Mantissa, m
56 Biased Exponent 60 +/ (1 bit) E t (n bit) Mantissa, m Biased value, b = 2 n1 Normalized exponent, e = E t  b Biased exponent, E t = e + b Where, E t = biased exponent n = bits of exponent format (i.e. the word format) This is used unless the IEEE standard is mentioned then is a different calculation
57 Conversion to Floating Point Number Change to binary (if given decimal number) Normalized the number Change the number to biased exponent Form the word (3 fields) 61
58 Example 3 62 Transform to floating point word using the following format (radix 2) Sign Biased exponent Significand 1 bit 4 bit 12 bit Step 1 : Change to binary This is the given word format x 2 = x 2 = x 2 = = = = x
59 Example 3 Step 2 : Normalized the number x Normalized exponent, e Step 3: Change the number to biased exponent 1 bit 4 bit 12 bit Biased value, b = 2 n1 = = 8d = 1000b Biased exponent, E t = e + b = = x x
60 Example x Step 4 : Form the word (3 fields) 1 bit 4 bit 12 bit Padding Rule of thumb: the biased exponent is always padded to the left  the significand (or mantissa) is always padded to the right
61 FloatingPoint Representation Value in general form: (1) S x F x 2 e In an 8bit representation, we can represent: From 2 x to 2 x This is called singleprecision = 1 word If anything goes above or under, then overflow and underflow happens respectively. One way to reduce this is to offer another format with a larger exponent use double precision (2 words) From 2 x to 2 x
62 IEEE 754 FloatingPoint Standard 69 This 1 is made implicit to pack more bits into the significand
63 Normalized Scientific Notation In IEEE standard normalization (used in computers), a floating point number is said to be normalized if there is only a single nonzero before the radix point. Example: 70 there is only a single nonzero before the radix point normalized x B normalized x 2 011
64 Challenge of Negative Exponents Placing the exponent before the significand simplifies sorting sign of floatingpoint Exponent numbers using Significand integer comparison 1 bit instructions. 8 bit 23 bit However, using 2 s complement in the exponent field makes a negative exponent look like a big number. 72
65 Biased Notation 73 Bias In single precision is 127 In double precision 1023 Biased notation (1)sign x (1 + Fraction) x 2 (exponentbias)
66 74 To convert a decimal number to single (or double) precision floating point: Step 1: Normalize Step 2: Determine Sign Bit Step 3: Determine exponent IEEE 754 Conversion Step 4: Determine Significand
67 IEEE 754 Conversion : Example 1 Convert 10.4 d to single precision floating point. Step 1: Normalize x 2 = x 2 = x 2 = x 2 = x 2 = x 2 = For continuous results, take the 1 st pattern before it repeats itself 0.4 = = x x 2 3
68 IEEE 754 Conversion : Example 1 76 Step 2: Determine Sign Bit (S) Because (10.4) is positive, S = 0 3 is from 2 3 Step 3: Determine exponent Because its single precision bias = 127 Exponent = 3 + bias = = 130 d = b Step4: Determine Significand Drop the leading 1 of the significand x Then expand (padding) to 23 bits sign Exponent Significand
69 IEEE 754 Conversion : Example 2 Convert d to single precision floating point. Step 1: Normalize x 2 = x 2 = x 2 = = x x 21
70 IEEE 754 Conversion : Example 2 78 Step 2: Determine Sign Bit (S) Because (0.75) is negative, S = 1 Step 3: Determine exponent Because its single precision bias = 127 Exponent = 1 + bias = = 126 d = b Step4: Determine Significand Drop the leading 1 of the significand 1.1 x Then expand (padding) to 23 bits sign Exponent Significand
71 IEEE 754 Conversion : Example 3 Convert d to double precision floating point. Step 1: Normalize x 2 = x 2 = x 2 = = x x 21
72 IEEE 754 Conversion : Example 3 80 Step 2: Determine Sign Bit (S) Because (0.75) is negative, S = 1 Step 3: Determine exponent Because its double precision bias = 1023 Exponent = 1 + bias = = 1022 d = b Step4: Determine Significand Drop the leading 1 of the significand 1.1 x Then expand (padding) to 52 bits sign Exponent (11) Significand (52)
73 Converting Binary to Decimal FloatingPoint What decimal number is represented by this single precision float? Extract the values: Sign = 1 81 Sign (1 bit) Exponent(8 bit) Significand(23 bit) Exponent = b = 129 d Significand Remember: Biased notation (1)sign x (1 + Fraction) x 2 (exponentbias) The Fraction = (0 x 21 ) + (1 x 22 ) + (0 x 23 ) = ¼ = ( ) The number =  (1.25 x 2 (exponentbias) ) =  (1.25 x 2 2 ) =  (1.25 x 4) = 5.0
74 Module 2 Part 3 82 FLOATINGPOINT OPERATIONS
75 FloatingPoint Addition Flows 83
76 Decimal FloatingPoint Addition 84 Assume 4 decimal digits for significand and 2 decimal digits for exponent Step 1: Align the decimal point of the number that has the smaller exponent Step 2: add the significand Step 3: Normalize the sum check for overflow/underflow of the exponent after normalisation Step 4: Round the significand If the significand does not fit in the space reserved for it, it has to be rounded off Step 5: Normalize it (if need be)
77 Decimal FloatingPoint Addition Example: d x d x 101 Step 1: Align the decimal point of the number that has the smaller exponent Make d x 101 to x = 1 x = 2 move 2 to left d x 10 1 Step 2: add the significand x d x x
78 Decimal FloatingPoint Addition Example: d x d x 101 Step 3: Normalize the sum x x Step 4: Round the significand (to 4 decimal digits for significand) x x 10 2 Step 5: Normalize it (if need be) No need as its normalized
79 Decimal FloatingPoint Addition Example: 0.5 d + ( d ) Adjusts the numbers Step 1: Align the decimal point of the number that has the smaller exponent b x b x b x b x 22 Make 1.11 b x 22 to x = 1 x = 1 move 1 to left b x 21
80 Decimal FloatingPoint Addition Example: 0.5 d + ( d ) Step 2: add the significand Step 3: Normalize the sum Step 4: Round the significand (to 4 decimal digits for significand) Fits in the 4 decimal digits Step 5: Normalize it (if need be) No need as its normalized x x
81 FloatingPoint Multiplication Flows 90
82 FloatingPoint Multiplication 91 Assume 4 decimal digits for significand and 2 decimal digits for exponent Step 1: Add the exponent of the 2 numbers Step 2: Multiply the significands Step 3: Normalize the product check for overflow/underflow of the exponent after normalisation Step 4: Round the significand If the significand does not fit in the space reserved for it, it has to be rounded off Step 5: Normalize it (if need be) Step 6: Set the sign of the product
83 FloatingPoint Multiplication 92 Example: (1.110 d x ) x (9.200 d x 105 ) Assume 4 decimal digits for significand and 2 decimal digits for exponent Step 1: Add the exponent of the 2 numbers 10 + (5) = 5 If biased is considered 10 + (5) = 132 Step 2: Multiply the significands x x 10 5
84 FloatingPoint Multiplication Example: (1.110 d x ) x (9.200 d x 105 ) Step 3: Normalize the product x x 10 6 Step 4: Round the significand (4 decimal digits for significand) x 10 6 Still normalized Step 5: Normalize it (if need be) Step 6: Set the sign of the product x 10 6
85 FloatingPoint Multiplication Example: (1.000 b x 21 ) x ( b x 22 ) 95 Assume 4 decimal digits for significand and 2 decimal digits for exponent Step 1: Add the exponent of the 2 numbers 1 + (2) = 3 If biased is considered 1 + (2) = 124 Step 2: Multiply the significands x x 103
86 FloatingPoint Multiplication Example: (1.000 b x 21 ) x ( b x 22 Step 3: Normalize the product x 103 already normalized 96 Step 4: Round the significand (4 decimal digits for significand) x 103 Still normalized Step 5: Normalize it (if need be) Step 6: Set the sign of the product b x /32 d
87 FloatingPoint ALU Sign Exponent Significand Sign Exponent Significand 97 Small ALU Compare exponents Exponent difference Control Shift right Shift smaller number right Big ALU Add 0 1 Increment or decrement 0 1 Shift left or right Normalize Rounding hardware Round Sign Exponent Significand
88 Accurate Arithmetic If a calculation exceeds the limits of the floating point scheme then CPU will flag this error. 98 If number is too tiny to be represented
89 Accurate Arithmetic : Truncation & Rounding 99 Some number have infinite decimal points (the irrational numbers) 1/3 d = Truncation is done to fit the decimal points into manageable units. Truncation is where decimal values beyond the truncation point are simply discarded and it can cause error in floating point calculations. Rounding :If you have a number such as then if you have to round this to 3 significant digits, the number becomes 3.46 A small error called the rounding error has occurred ***Note : the CPU will not flag any error when truncation and rounding occurs, as it is acting within its limits. programmers must assess if this will lead to significant errors
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