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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 representation

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Floating point

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  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 754-1985 (IEC 559) 4

  2. IEEE 754-1985 • Number representation 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 Single extended and double extended numbers exists inside the floating point hardware

  3. IEEE 754-1985 SinglePrecision: 1 8 23 S E sign M exponent: excess 127 binary integer mantissa: sign + magnitude, normalized binary significand w/ hidden integer bit: 1.M actual exponent is e = E - 127 0 < E < 255 E-127 N = (-1) 2 (1.M) 0 = 0 00000000 0 . . . 0 -1.5 = 1 01111111 10 . . . 0 Magnitude of numbers that can be represented is in the range: -126 127 23 ) 2 (1.0) (2 - 2 to 2 which is approximately: -38 38 to 3.40 x 10 1.8 x 10

  4. 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 - 10 resp. 10 - 10 -39 39 -308 308

  5. 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 Rounds to nearest by default; Three other rounding modes exists. Sophisticated exception handling Emin

  6. Multiplication (s1 * 2 ) * (s2 * 2 ) = s1*s2 *2 so, multiply significands and add exponents Problem: Significand coded in signed- magnitude - use unsigned multiplication and take care of sign Round 2n bits significand to n bits significand Compute new exponent with respect to bias e1 e2 e1+e2

  7. P A x0 x1 x2 x3 x4 x5 g r s s s s P A x0 x1 x2 x3 x4 x5 r s s s s s Rounding 1. Multiply the two significands to get the 2n-bits product: Case 1: x0 = 0, shift needed: Case 2: x0 = 1, increment exponent, set g=r; r=s or r These four bits OR:ed together (“sticky bit”) guard round bit bit P A x1 x2 x3 x4 x5 g r s s s s

  8. Rounding 2: For both cases: if r = 0, P is the correctly rounded product. if r = 1 and s = 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

  9. Add / Sub e1 e2 e3 (s1 * e ) + (s2 * e ) = (s3 * e ) 1: Shift summands so they have the same exponent. (eg. if e2 < e1: shift s2 right and increment e2 until e1 = e2) 2: Add significands 3: Normalize number (shift s3 left and decrement e3 until MSB = 1) 4: Round s3 correctly (under the common assumption that more than 23 / 52 bits is internally used for addition) Subtraction use the same method s3

  10. 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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