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Floating Point

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 754-1985 (IEC 559). 4. IEEE 754-1985. Number representations: Single precision (32 bits)

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Floating Point

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  1. Floating Point Number system corresponding to the decimal notation 1,837 * 10 significandexponent A great number of corresponding binary standards exists. There is one common standard: IEEE 754-1985 (IEC 559) 4

  2. IEEE 754-1985 • 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

  3. Single Precision Format 1 8 23 Sign S S E M Exponent E: excess 127 binary integer Mantissa M: normalized binary significand w/ hidden integer bit: 1.M Excess 127; actual exponent is e = E - 127 N = (-1)S * (1.M [bit-string])*2e

  4. Example 1 S E M 1 01111110 10000000000000000000000 e = E - 127 e = 126 - 127 = -1 N = (-1)1 * (1.1 [bit-string]) *2-1 N = -1 * 0.11 [bit-string] N = -1 * (2-1 *1 + 2 -2 *1) N = -1 * (0.5*1 + 0.25*1) = -0.75

  5. Single Precision Range Magnitude of numbers that can be represented is in the range: 2-126 *(1.0) to 2-127 *(2-223) which is approximately: 1.8*10-38 to 3.4*1038

  6. IEEE 754-1985 • 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 10 -39 to 10 39 / 10 -308 to 10 308

  7. IEEE 754-1985 • Special features: • Correct rounding of “halfway” result (to even number). • Includes special values: • NaN Not 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.

  8. Add / Sub (s1 * 2e1) +/- (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.

  9. Multiplication (s1 * 2e1) * (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.

  10. P A x0 x1 x2 x3 x4 x5 g r s s s s P A x0 x1 x2 x3 x4 x5 r (STICKY or r)STICKY 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. The s bits are OR:ed together (“sticky bit”) STICKY guard round bit bit P A x1 x2 x3 x4 x5 g rSTICKY 0….0

  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

  12. Division (s1 * 2 ) / (s2 * 2 ) = (s1 / s2) * 2 e1 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

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