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Overview of IEEE 754-1985 Floating Point Number System

Learn about IEEE 754-1985 floating point number system used in computers. Understand single and double precision formats, range of numbers represented, exponent biasing, special features, arithmetic operations like addition, subtraction, multiplication, division, rounding mechanisms, and accurate arithmetic.

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Overview of IEEE 754-1985 Floating Point Number System

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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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