Computer Engineering FloatingPoint page 1 Floating Point Number system corresponding to the decimal notation 1,837 * 10 significand exponent A great number of corresponding binary standards exists. There is one common standard: IEEE (IEC 559) 4

Computer Engineering FloatingPoint page 2 IEEE Number representations: –Single precision (32 bits) sign:1 bit exponent:8 bits fraction:23 bits –Double precision (64 bits) sign:1 bit exponent:11 bits fraction:52 bits

Computer Engineering FloatingPoint page 3 Single Precision Format 1823 Sign S Exponent E: excess 127 binary integer Mantissa M: normalized binary significand w/ hidden integer bit: 1.M Excess 127; actual exponent is e = E N = (-1) S * (1.M [bit-string])*2 e SEM

Computer Engineering FloatingPoint page 4 Example 1 SEM e = E e = = -1 N = (-1) 1 * (1.1 [bit-string]) *2 -1 N = -1 * 0.11 [bit-string] N = -1 * (2 -1 * *1) N = -1 * (0.5* *1) =

Computer Engineering FloatingPoint page 5 Single Precision Range Magnitude of numbers that can be represented is in the range: *(1.0) to *( ) which is approximately: 1.8* to 3.4 *10 38

Computer Engineering FloatingPoint page 6 IEEE Fraction part: 23 / 52 bits; 0  x < 1 Significand: 1 + fraction part. “1” is not stored; “hidden bit”. corresponds to 7 resp. 16 decimal digits. Exponent: 127 / 1023 added to the exponent; “biased exponent”. corresponds to to / to

Computer Engineering FloatingPoint page 7 IEEE Special features: –Correct rounding of “halfway” result (to even number). –Includes special values: NaNNot a number  Infinity -  - Infinity –Uses denormal number to represent numbers less than 2 -E min –Rounds to nearest by default; Three other rounding modes exist. –Sophisticated exception handling.

Computer Engineering FloatingPoint page 8 Add / Sub (s1 * 2 e1 ) +/- (s2 * 2 e2 ) = (s1 +/- s2) * 2 e3 = s3 * 2 e3 –s = 1.s, the hidden bit is used during the operation. 1: Shift summands so they have the same exponent: –e.g., if e2 < e1: shift s2 right and increment e2 until e1 = e2 2: Add/Sub significands using the sign bits for s1 and s2. –set sign bit accordingly for the result. 3: Normalize result (sign bit kept separate): –shift s3 left and decrement e3 until MSB = 1. 4: Round s3 correctly. –more than 23 / 52 bits is used internally for the addition.

Computer Engineering FloatingPoint page 9 Multiplication (s1 * 2 e1 ) * (s2 * 2 e2 ) = s1 * s2 * 2 e1+e2 so, multiply significands and add exponents. Problem: Significand coded in sign & magnitude; use unsigned multiplication and take care of sign. Round 2n bits significand to n bits significand. Normalize result, compute new exponent with respect to bias.

Computer Engineering FloatingPoint page 10 Accurate Arithmetic 1. Multiply the two significands to get the 2n-bits product: –But we have only 23/52 bit to store the result! Case 1: x0 = 0, shift needed: Case 2: x0 = 1, increment exponent, set g = r; r = STICKY or r. x0 x1 x2 x3 x4 x5g r s s s s PA x1 x2 x3 x4 x5 g r STICKY 0….0 PA x0 x1 x2 x3 x4 x5 r (STICKY or r) STICKY PA The s bits are OR:ed together (“sticky bit”) STICKY guard round bit

Computer Engineering FloatingPoint page 11 Rounding 2: For both cases: if r = 0, P is the correctly rounded product. if r = 1 and STICKY = 1, then P + 1 is the correctly rounded product if r = 1 and s = 0, (the “halfway case”), then P is the correctly rounded product if x5 (or g) is 0 P+1 is the correctly rounded product if x5 (or g) is 1

Computer Engineering FloatingPoint page 12 Division (s1 * 2 ) / (s2 * 2 ) = (s1 / s2) * 2 e1 e1-e2 so, divide significands and subtract exponents Problem: Significand coded in signed- magnitude - use unsigned division (different algoritms exists) and take care of sign Round n + 2 (guard and round) bits significand to n bits significand Compute new exponent with respect to bias