IEEE 754 Floating Point PDF

Computer Organization (Number representation & Arithmetics) KR Chowdhary Professor & Head Email: [email protected] webpage: krchowdhary.com Department of Computer Science and Engineering MBM Engineering College, Jodhpur November 14, 2013...

Computer Organization (Number representation & Arithmetics) KR Chowdhary Professor & Head Email: [email protected] webpage: krchowdhary.com Department of Computer Science and Engineering MBM Engineering College, Jodhpur November 14, 2013 KR Chowdhary Int & flt. pt. Arithmetic 1/ 16 Number Formats ◮ Factors for selection of number representation: 1. Types of numbers to be represented: integers, real numbers, complex numbers 2. Range of values to be encountered 3. Precision of numbers required 4. Cost of hardware required to store and process the numbers. ◮ Formats: Fixed point and floating point format. First have small range and simple hardware requirements. Second has complex hardware and large range. KR Chowdhary Int & flt. pt. Arithmetic 2/ 16 Binary Numbers −1101.01012 = −13.312510, ∵ 0.01012 = 21 × 0 + 14 × 1 + 81 × 0 + 16 1 × 1 = 0.312510 Non-negative integer representation: 8 bit binary represents 010... 25510. If we consider that the binary number is A = an−1 an−1... a0 , then, n −1 A= ∑ 2 i ai i =0 Signed magnitude representation: First bit = 0 ⇒ +ve, and 1 ⇒ −ve. ∴ + 18 = 00010010, in 8-bit representation, and −18 = 10010010 n −2 A= ∑ 2i ai , if an−1 = 0 i =0 and n −2 A = − ∑ 2i ai , if an−1 = 1. i =0 KR Chowdhary Int & flt. pt. Arithmetic 3/ 16 Negative Number’s Representation Drawback of above notation: 1) require consideration of sign and magnitude, 2) there is double notation for 0 +010 = 00000000, and −010 = 10000000, hence this notation is never used. Two’s Complement representation: ◮ -ve numbers are represented as two’s complement. ◮ Most significant bit is used as sign bit, while other bits are treated differently. For n-bit binary number, the range is −2n−1 to +2n−1 − 1. ◮ Number of representations of zero is one only. ◮ Addition and subtractions are straight forward. If the result is -ve, then actual result’s magnitude is found by obtaining its two’s complement. ◮ Boundary numbers are complement of each other, i.e., −2n−1 and +2n−1 − 1. KR Chowdhary Int & flt. pt. Arithmetic 4/ 16 Sign magnitude v/s two’s complement Decimal Sign Two’s Representation Magnitude Complement +7 0111 0111 +6 0110 0110 +5 0101 0101 +4 0100 0100 +3 0011 0011 +2 0010 0010 +1 0001 0001 +0 0000 0000 -0 1000 0000 -1 1001 1111 -2 1010 1110 -3 1011 1101 -4 1100 1100 -5 1101 1011 -6 1110 1010 -7 1111 1001 -8 - 1000 KR Chowdhary Int & flt. pt. Arithmetic 5/ 16 IEEE754 Floating-point Format A decimal number 9760000 can be represented in floating point as 0.97 × 107, and a fractions 0.0000976 can be represented as 0.976 × 10−4. ±S × B ±E here S is called significand (i.e., mantissa), B is base: 10 for decimal, 2 for binary. 1 23 bits Single precision 8 Sign Expo- Double precision nent Significand bit 1 11 20 bits 32 bits = 52 bits Sign Expo- bit nent Significand 1st word 2nd word Figure 1: IEEE754 Floating-point Formats Meaning: (−1)sign × (1 + mantissa) × 2expo.−bias KR Chowdhary Int & flt. pt. Arithmetic 6/ 16 IEEE754 Floating-point Format Biased Representation: ◮ Twos complement exponent is not effective for sorting(e.g. -1 = 1... 11, 1=0... 01). A bias is introduced so that 0... 0 is smallest representable exponent, and 1... 1 is largest. ◮ For k-bit exponent, a bias value of 2k −1 − 1 is subtracted from the exponent to obtain the true value of exponent. ◮ For single precision, bias = 127, for double precision it is 1023. ◮ Range: SP: −2 × 10−38 to 2 × 1038, DP: −2 × 10−308 to 2 × 10308 Fraction: ◮ in scientific notations, there is no leading 0, hence an implicit leading 1 is taken in mantissa. ◮ Significand is taken as 24-bits, the most significant bit is always taken as 1. Thus, 32-bit biased format is ±1.bbbb · · · × 2±E where b are binary, 0 and 1. KR Chowdhary Int & flt. pt. Arithmetic 7/ 16 IEEE754 Floating-point Format ◮ Let 32-bit number in biased exponent format is: 10110101110100010000000000000000. This number is −1.1010001 × 2x , where x = 01101011 = 10710, true value of exponent is x + 2′ s complement of 127 (=100000012) = -00010100=-20. ∴ true value of above FP representation is −1.6328125 × 2−20 ◮ 23 bits mantissa provides precision equivalent to 8 decimal digits, and 53 bit to 16 digits. ◮ Extended single precision and double precision: ◮ Exponent = 0, mantissa = 0: exact zero. With sign bit, it is ±0 ◮ Exponent of all 1’s and mantissa 0: ∞ is represented (over-flow due to division by 0). With sign bit it is±∞. ◮ Exponent = 0, nonzero mantissa is denormal numbers. M is non-zero 23 bit number. Value is ±0.M × 2−126. They are smaller than smallest normal no. ◮ E = 255, m 6= 0: Nan(Not √ a number format) indicator, e.g. due to operation of 0/0 or −1. ◮ Exception handling? KR Chowdhary Int & flt. pt. Arithmetic 8/ 16 Floating Point Addition/subtraction Algorithm-to-add 1. Align the two numbers by right shifting the smaller number until it matches the bigger one 2. set exponents of both equal to larger 3. Add/subtract significands (mantissas) 4. Normalize the sum 5. Check for overflow or underflow and raise an exception if necessary 6. If not normalized repeat from 3 KR Chowdhary Int & flt. pt. Arithmetic 9/ 16 Floating Point Multiplication Algorithm-to-multiply 1. Add the exponents and subtract bias 2. Multiply significands 3. Normalize if necessary 4. Check for over-/underflow and raise exception if needed 5. Round significand 6. If not normalized repeat from step 3 7. Set the sign to positive if input signs are equal, negative if they differ How to divide? KR Chowdhary Int & flt. pt. Arithmetic 10/ 16 Faster Division Cannot perform unrolling due to the strict dependency between stages, so 1. “Guess”more than one bit at a time using a precomputed Lookup Table with inputs from Remainder and Divisor 2. Do not add back the Divisor if the remainder is negative. Instead, add the Dividend to the shifted Remainder in the next step (nonrestoring division) 3. Do not store the result of the subtraction if it is negative (nonperforming division) KR Chowdhary Int & flt. pt. Arithmetic 11/ 16 Floating-point Arithmetic Arithmetic: ◮ Let X = Xs × B XE , and Y = Ys × B YE ◮ Then, X + Y = (XS × B XE −YE + YS ) × B YE , if XE ≤ YE X − Y = (XS − YS × B YE −XE ) × B XE , if XE ≥ YE X × Y = (XS × YS ) × B XE +YE XS X ÷Y = YS B XE −YE ◮ Let X = 0.3 × 102 = 30 and Y = 0.2 × 103 = 200. Then, X + Y = (0.3 × 102−3 + 0.2) × 103 = (0.03 + 0.2) × 103 = (0.23) × 103 = 230. KR Chowdhary Int & flt. pt. Arithmetic 12/ 16 ASCII and Extended Binary Coded Decimal Interchg Code Figure 2: ASCII-table ◮ ASCII encodes 128 specified characters-numbers 0-9, letters a-z, A-Z, some punctuation symbols, some control codes originated with Teletype machines, a blank space, into 7-bit binary integers. ◮ ASCII represent text in computers, communications equipment, and other devices that use text. ◮ EBCDIC is 8-bit character encoding used mainly on IBM mainframe and IBM midrange systems KR Chowdhary Int & flt. pt. Arithmetic 13/ 16 Practice Exercises 1. Express the decimal 0.5, -0.123 as signed 6-bit fractions. 2. What is the maximum representation error, e, if only 8 significant bits after decimal point are used? 3. Assuming a 6-bit exponent, 9-bit normalized fractional mantissa, and exponent is represented in biased format, add the number below:A = 0 100001 111111110, B=0 011111 0010110101. Assume an implicit 1 to the left of mantissa. 4. Assuming all numbers are in 2’s complement representation, which of the following numbers is divisible by 11111011? (A) 11100111 (B) 11100100 (C) 11010111 (D) 11011011 KR Chowdhary Int & flt. pt. Arithmetic 14/ 16 Practice Exercises 5. The hexadecimal representation of 6578 is: (a) 1AF (b) D78 (c) D71 (d) 32F 6. What is answer for (1 + 1 × 1020) − (1 × 1020)? Justify. 7. What is answer for 1 + (1 × 1020 − 1 × 1020)? Justify. 8. How the rounding/truncation is carried out by the CPU? 9. How overflow can be detected, if the carry bit is not used? I.e., decide it only based on the values of A, B, C = A + B and C = A − B. Assume that two’s complement is used for negative numbers. 10. The range of integers that can be represented by an n bit 2’s complement number system is: (a) −2n−1 to (2n−1 − 1) (b) −(2n−1 − 1) to (2n−1 − 1) (c) −2n−1 to 2n−1 (d) −(2n−1 + 1) to (2n−1 − 1) KR Chowdhary Int & flt. pt. Arithmetic 15/ 16 Practice Exercises 11. The following is a scheme for floating point number representation using 16 bits. Bit Position 15 14... 9 8... 0 S e m sign exponent mantissa Let s, e, and m be the numbers represented in binary in the sign, exponent, and mantissa fields, respectively. Then the floating point number represented is: (−1)s (1 + m × 2−9)2e −31 , if the number 6= 111111, and 0 otherwise. What is the maximum difference between two successive real numbers representable in this system? (A) 2−40 (B) 2−9 (C) 222 (D) 231 12. What are the values of expressions ∞/0, 0/∞, ∞/∞. Justify your answer. KR Chowdhary Int & flt. pt. Arithmetic 16/ 16