1. Recitation 4&5 and review 1 & 2 & 3
9-26/

2. Recitation 4

3. IEEE Floating Point Representation
V = (-1)s * M * 2E S: sign, s = 0 positive s = 1 negative M: Significand, 1 ≤ M ≤ 2 - € for Normalized 0 ≤ M ≤ 1- € for Denormalized € = smallest possible number greater than 0. E: Exponent and possibly negative.

4. IEEE Floating Point Representation
32 bit (Single precision)
s =1, k = 8, n =23
64 bit (Double precision)
s =1, k = 11, n =52
1 bit
k bits
n bits
111 s
exponent
fraction

5. IEEE Floating Point Representation
Normalized
Denormalized
Infinity
Nan
32 bits
111 s
exponent ≠ 0 & ≠ 255
fraction
111 s
exponent = 0
fraction
111 s
exponent = 255
fraction = 0
111 s
exponent = 255
fraction ≠ 0

6. IEEE Floating Point Representation
V = (-1)s * M * 2E
S: Sign bit
E = exponent – Bias (Normalized)
= 1 – Bias (Denormalized)
Bias = 2k-1 - 1
M = 1 + Fraction (Normalized)
= Fraction (Denormalized)
Fraction = .fn-1fn-2…f1f0 * 2-n
1 bit
k bits
n bits
111 s
exponent
fraction

7. Normalized 32 bit (Single precision) 64 bit (Double precision)
s =1, k = 8, n =23
Bias = = 127
Exponent ranges : 0 to 255 but not 0 and 255
E = Exponent – bias = -126 to +127
64 bit (Double precision)
s =1, k = 11, n =52
Bias = = 1023
Exponent ranges : 0 to 2047 but not 0 and 2047
E = Exponent – bias = to +1023

8. Normalized M = 1 + Fraction Fraction = .fn-1fn-2…f1f0 * 2-n
Here .fn-1fn-2…f1f0 = .11……1 < 1
M = 1 +Fraction < 2
= 1 ≤ M ≤ 2 – ε

9. Denormalized Exponent = 0 (All k bit is 0) 32 bit (Single precision)
s =1, k = 8, n =23
Bias = = 127
E = 1 – bias = -126
64 bit (Double precision)
s =1, k = 11, n =52
Bias = = 1023
E = 1 – bias = -1022

10. Denormalized M = Fraction Fraction = .fn-1fn-2…f1f0 * 2-n
Here .fn-1fn-2…f1f0 = .11……1 < 1
M = Fraction < 1
= 0 ≤ M ≤ 1 – ε

11. Example FP representation of (40.15625)10 in 32 bit Sign bit, s = 0
( )10 = ( )2
Normalize: * 25
Convert the exp to biased: = 132
(132)10 = ( )2
Result : s k n

12. Example
s = 0, positive
exp = = 137 (Normalize)
E = exp – Bias = 137 – = 10
M = 1 + Fraction
= 1 + .fn-1fn-2…f1f0 * 2-n = s k n

13. Example(Contd.) V = (-1)s * M * 2E = (-1)0 * M * 210 = M * 210
= * 210 = =

14. Examples Description Exponent Fraction Smallest denorm. 000…000
000…001
Largest denorm.
111…111
Smallest norm.
Largest norm.
111…110
Examples of Positive Floating Point Numbers

15. Rounding Modes 1.40 1.60 1.50 2.50 -1.50 Round-to-even 1 2 -2
Round-toward-zero -1
Round-down
Round-up 3
Rounding Modes

16. Recitation 5

17. Rounding Binary Numbers
Binary Fractional Numbers
– “Even” when least significant bit is 0
– “Half way” when bits to right of rounding position = 100…2
Examples
– Round to nearest 1/4 (2 bits right of binary point)
Value
Binary
Rounded
Action
Rounded Value
2 3/32
10.002
(<1/2- down) 2
2 3/16
10.012
(>1/2 - up)
2 1/4
2 7/8
11.002
( 1/2 - up) 3
2 5/8
10.102
( 1/2- down)
2 1/2

18. Round to Even
When rounding to even, consider the two possible choices and choose the one with a 0 in the final position.
Example: round to even at the 1/4 position: /2 /8 /4 /4 /8

19. Rounding practice

20. Floating Point Representation

21. An IEEE floating point representation uses 4 exp bits and 5 frac bits.

23. Review 1 & 2 & 3 Endian Bitwise & shift operation
Conversion between binary, decimal, hexadecimal

25. Recitation1: 10 problems 0x3A6B=(0011 1010 0110 1011)b
Problem2: Convert to binary: 0x3A6B
0x3A6B=( )b

26. Recitation1: 10 problems 935=a*162+b*161+c =a*256+b*16+c
Problem3: Convert from decimal to binary: 935
Problem4: Convert from decimal to hexadecimal: 935
935=a*162+b*161+c
=a*256+b*16+c
=3*256+10*16+7
=0x3A7
( )b

27. Recitation1: 10 problems (1011011101)b =(10 1101 1101)b =(2 D D )h
Problem5: Convert from binary to hexadeximal: = = 0x2DD
Problem6: Convert from binary to decimal: = 2* * = 733
( )b
=( )b
=( D D )h
=0x2DD
=2*162+13*16+13
=733

28. Little Endian vs Big Endian
Little endian: store the least significant byte in the smallest address.
store the most significant byte in the largest address.
Big Endian: store the most significant byte in the smallest address.
store the least significant byte in the largest address.
So, we can know how 0x is stored in memory.
address
1000
1001
1002
1003
Little Endian 67 45 23 01
address
1000
1001
1002
1003
Big Endian 01 23 45 67

29. Two’s Complement Addition
n = 5 bits (-3)10 + (-10)10 N = 3 =(00011)2 N = (10)10 = (01010)2 N* = -N = (11101)2 N* = -N = (10110) = N* = -N = (-13)10 N = (01100)2 = (13)10
Ignored

30. Bitwise Operators Or And Not Exclusive-Or (Xor)
A|B = 1 when either A=1 or B=1
And
A&B = 1 when both A=1 and B=1
Not
~A = 1 when A=0
Exclusive-Or (Xor)
A^B = 1 when either A=1 or B=1, but not both

31. Bitwise Operators Operate on Bit Vectors Operations applied bitwise
& | ^ ~

32. Shift Operations Left Shift: x << y Right Shift: x >> y
Argument x
Left Shift: x << y
Shift bit-vector x left y positions
Throw away extra bits on left
Fill with 0’s on right
Right Shift: x >> y
Shift bit-vector x right y positions
Throw away extra bits on right
Logical shift
Fill with 0’s on left
Arithmetic shift
Replicate most significant bit on left
Undefined Behavior
Shift amount < 0 or ≥ word size
<< 3
Log. >> 2
Arith. >> 2
Argument x
<< 3
Log. >> 2
Arith. >> 2