1. Datorteknik FloatingPoint bild 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. Datorteknik FloatingPoint bild 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. Datorteknik FloatingPoint bild 3 IEEE 754-1985 1823 sign exponent: excess 127 binary integer SE M mantissa: sign + magnitude, normalized binary significand w/ hidden integer bit: 1.M Single Precision: actual exponent is e = E - 127 N = (-1) 2 (1.M) E-127 0 < E < 255 0 = 0 00000000 0... 0 -1.5 = 1 01111111 10... 0 Magnitude of numbers that can be represented is in the range: 2 -126 (1.0) to2 127 (2 - 2 23 ) which is approximately: 1.8 x 10 -38 to3.40 x 10 38

4. Datorteknik FloatingPoint bild 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 -3939 -308308

5. Datorteknik FloatingPoint bild 5 IEEE 754-1985 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 Rounds to nearest by default; Three other rounding modes exists. Sophisticated exception handling Emin

6. Datorteknik FloatingPoint bild 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 e1e2e1+e2

7. Datorteknik FloatingPoint bild 7 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 x0 x1 x2 x3 x4 x5g r s s s s PA x1 x2 x3 x4 x5 gr s s s s PA x0 x1 x2 x3 x4 x5r s s s s s PA These four bits OR:ed together (“sticky bit”) guard round bit

8. Datorteknik FloatingPoint bild 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. Datorteknik FloatingPoint bild 9 Add / Sub (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 e1e2e3 s3

10. Datorteknik FloatingPoint bild 10 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