1. CSE 246: Computer Arithmetic Algorithms and Hardware Design Instructor: Prof. Chung-Kuan Cheng Winter 2004 Lecture 9

2. CSE 2462 Topics:  Floating Point Numbers (IEEE P754)  Standard  Operations  Exceptional Situations  Rounding Modes

3. CSE 2463 Standard 2 32  Typically  Goal: Dynamic Range: largest #/ smallest #  If too large, holes between # ’ s

4. CSE 2464 Standard  ulp (unit in the last place)  Difference between two consecutive values of the significand. 3 Parts  x =  s  b e Sign Bit 8-bit exponent Significand

5. CSE 2465 Standard   a 1 a 2 a 3 a 4 a 5 a 6 a 7 a 8 b 1 b 2 b 3  b 22 b 23 1.* normalized number 0.* denormalized number 0 0.b 1 b 2 b 3  b 22 b 23  2 -126 1--------------------------------- 1. b 1 b 2 b 3  b 22 b 23  2 -126 2. 253 254 ------------------------------- 1. b 1 b 2 b 3  b 22 b 23  2 127 255   if b i = 0 for all i = 1,2, …,23, NaN otherwise NaN  Not a Number

6. CSE 2466 Standard 0.01x2 -3 = 0.00x2 -2  Same number, so normalize to remove redundancy  Smallest Number 0.00 … 01x2 -126 = 1.0x2 -23 x2 -126 = 1x2 -149  1.1101111001110011100101  Difference between 2 # ’ s small for normalized 0.0001 2 times compared to magnitudes 0.0010

7. CSE 2467 Standard - Example s. eeeeeeee nnnnnnnnnnnnnnnnnnnnnnn 0.00000000 00000000000000000000000 = 0.000 … 0x2 -126 1.00000000 00000000000000000000000 = 0 0.00000001 00000000000000000000000 = 1.000 … 0x2 -126 - minimal normalized # 0.00000001 00000000000000000000001 = 1.000 … 1x2 -126. 0.01111111 00000000000000000000001 = 1.000 … 1x2 0 0.10000000 00000000000000000000001 = 1.000 … 0x2 1

8. CSE 2468 Standard – Example Cont. 0.11111110 00000000000000000000001 = 1.000 … 1x2 127 0.11111110 11111111111111111111111 = 1.111 … 1x2 127 - Normalized Maximum 0.11111111 00000000000000000000000 =  N min = 1.0 x 2 -126 N max = (2 – 2 -23 )2 127

9. CSE 2469 Double Floating Point  a 1 a 2 … a 11 b 1 b 2 … b 52 000 … 00 0. b 1 b 2 … b 52 x 2 -1023 000 … 01 1. b 1 b 2 … b 52 x 2 -1022. 011 … 11 1. b 1 b 2 … b 52 x 2 0 100 … 00 1. b 1 b 2 … b 52 x 2 1. 111 … 10 1. b 1 b 2 … b 52 x 2 1023 111 … 11 =  if b i = 0 for all i = 1,2, …,52

10. CSE 24610 Overflow/Underflow N max N min SparserDenser Overflow Underflow

11. CSE 24611 Addition/Multiplication  s 1 xb e1 + (  s 2 xb e2 ) =  sxb e =  s 1 xb e1 +  s 2 /b e1-e2 x b e1 = (  s 1  s 2 /b e1-e2 ) x b e1  (  s 1 xb e1 ) x (  s 2 xb e2 ) =  (s 1 xs 2 )b e1+e2

12. CSE 24612 Exceptions a/0 =  if a > 0 a/  = 0if a != 0 a · 0 = 0 a ·  =  if a > 0 0 ·  = invalid operation (NaN) 0/0 = invalid operation (NaN) NaP op a = NaN a +  =   -  = NaN

13. CSE 24613 Rounding Mode  Adder Output = Cout z 1 z 0.z -1 z -2 … z - l GRS Guard Bit Round Bit Sticky Bit, OR of all bits below bit R 1.101 x 2 3 +1.110 x 2 3 11.011 x 2 3 1.1011x2 4 Normalize – need to round or

14. CSE 24614 Rouding 1.110 x 2 3 - 1.101 x 2 3 0.001 x 2 3 1.000 x 2 0 normalize 1.101 x 2 3 - 1.111 x 2 2 1.101 x 2 3 - 0.1111 x 2 3 0.1101 x 2 3 1.101 x 2 2 Guard bit

15. CSE 24615 Rounding  Round to the nearest even toward 0 1.1011 Toward +  1.1100 Toward -  1.1011

16. CSE 24616 Conventional Rounding Error Rounding Error 1.10100  1.101= 0 1.10101  1.101= -0.25 1.10110  1.110 = +0.5 1.10111  1.110 = +0.25 Average Error = 0.5/4 = 0.125