1. Chapter 1 Data Storage(3) Yonsei University 1 st Semester, 2015 Sanghyun Park

2. Outline  Bits and their storage(prev. file)  Main memory(prev. file)  Mass storage(prev. file)  Representing information as bit patterns(prev. file)  Binary system(prev. file)  Storing integers  Storing fractions

3. Representing Integers  Unsigned integers can be represented in base ___  ______ integers = numbers that can be positive or negative  Sign and magnitude notation  Two’s complement notation  Excess notation

4. Sign and Magnitude Notation  Left hand bit is used to determine the ___ of the number  Left hand bit 0 = positive number0010 = +2  Left hand bit 1 = negative number1010 = -2  Using 4 bits with sign and magnitude, largest positive number represented is __ (-7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7)  ___ values overall can be represented  +2 added to -2 using regular binary addition does ___ _____ to 0

5. Two’s Complement Notation (1/2)  For a positive number, the two’s complement representation is _____  For a negative number, ___________ positive value and then ____ 1  3 in two’s complement is 011  -3 in two’s complement is 100 + 1 = 101

6. Two’s Complement Notation (2/2)  What is the decimal value of 1010 in two’s complement?  It is a negative number since left hand bit is 1  Complement it and add 1: 0101 + 1 = 0110 (+6)  Therefore the decimal value is -6

7. Two’s Complement Addition

8. Overflow Problem  Try adding 5 + 4 in 4-bit two’s complement notation  Result is negative value (-7)  Such an error is called ________  When using two’s complement notation, this might occur when adding two ________ values or when adding two ________ values  Normally integers are represented in 32-bit patterns, allowing for positive values as large as 2,147,483,647

9. Excess Notation (1/3)  Excess four notation  All positive numbers begin with 1  All negative numbers begin with 0  0 is represented as ___  Smallest negative number is ___  Largest positive number is ___  Why 4? (2 #of bits –1 ) = 2 3-1 = 4

10. Excess Notation (2/3)

11. Excess Notation (3/3)  What is the decimal value of 101 in excess four notation?  101 if interpreted unsigned is 5  101 in excess four notation is (5-4) = 1  What is the excess four notation for decimal value 3?  Add four: 3+4 = 7  Represent it in 3-bit binary pattern = 111

12. Excess Eight Notation

13. Storing Fractions (1/6)  Need to represent ______ and ______ of radix point  Use ___________ notation 1 bit:Sign bit (0 positive, 1 negative) 3 bits:Exponent (encodes position of radix point in excess four ) 4 bits:Mantissa (encodes number)

14. Storing Fractions (2/6)  What is the decimal value of 01101011 in floating-point notation?  First bit is 0, then positive  Exponent is 110 and mantissa is 1011  Extract mantissa and place a radix point on its left side 0.1011  Extract exponent and interpret as excess four notation  110 in excess four is +2 (make sure?)  +2 exponent means move radix point to the right by two bits (a negative exponent means move radix to left)  0.1011 becomes 10.11

15. Storing Fractions (3/6)  What is the decimal value of 10111100 in floating-point notation?  First bit is 1, then negative  Exponent is 011and mantissa is 1100  Extract mantissa and place a radix point on its left side 0.1100  Extract exponent and interpret as excess four notation  011 in excess four is -1 (make sure?)  -1 exponent means move radix point to the left by 1 bit  0.1100 becomes 0.01100

16. Storing Fractions (4/6)  What is the floating point notation of the number ?  Express the number in binary to obtain 1.001 (make sure?)  Copy bit pattern into mantissa field from left to right starting with the leftmost 1 in binary representation  Mantissa is 1001  Compute exponent to get 1.001 from.1001 (imagine mantissa with radix point at its left)  Need to move radix point to right one bit  Exponent is +1  Express exponent in excess four notation: 1+4 = 5 (101)  Therefore, 01011001

17. Storing Fractions (5/6)  What is the floating point notation of the number ?  Express number in binary to obtain 0.01  Copy bit pattern into mantissa field from left to right starting with the leftmost 1 in binary representation  Mantissa is 1000 (you append zeros to fill the 4-bit mantissa)  Compute exponent to get 0.01 from 0.1000  Need to move radix point to left one bit  Exponent is -1  Express exponent in excess four notation: -1+4 = 3 (011)  Therefore, 10111000

18. Storing Fractions (6/6)  What is the floating point notation of the number ?  We end up with the bit pattern 01101010, which represents 2 1 /2 instead of 2 5 /8  What has occurred is called a truncation error or round-off error