Data Representation and Computer Arithmetic Computer Systems Data Representation and Computer Arithmetic

Data Representation Binary codes used to represent numbers Conversion between number systems Binary  decimal Hexadecimal  decimal Between arbitrary number systems Representation of positive/negative numbers Floating point representation

Bit, Byte and Word Bit is the smallest quantity that can be handled by computer: 1 or 0. Byte is a group of 8 bits. Word is the basic unit of data that can be operated by computer: e.g., 16 bits. 11001010 00001101 Bit Byte Word Some architectures have 8, 32, or 64-bit words

Content of Word Bit Pattern May represent many things Actual meaning of a particular bit pattern is given by the programmer Computer itself cannot determine the meaning of the word Classic question: Can you tell the meaning of a word picked randomly from the memory?

Instruction: (op-code) A single word defines an action that is to be performed by CPU Numeric Quantity Character: A to Z, a to z, 0 to 9, *, -, +, etc. ASCII Code: - Representation of a character by 7 bits. - The ASCII for W is 101 01112, or 5716

ASCII Code

Pixel (Picture Element) The smallest unit to construct a picture 1bit/pixel for black-and-white pictures, >1 bits/pixel for gray scale or color pictures, e.g. 24bits/pixel. A letter is represented by a group of pixels on computer screen. Pixel (1)

Positional Notation Weight is associated with the location within a number. The 9 in 95 has weight 10 A number N in base b is represented by

Example:

Number Bases Decimal (b=10): {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} Binary (b=2): {0, 1} Octal (b=8): {0, 1, 2, 3, 4, 5, 6, 7} Hexadecimal (b=16): {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F}

How many binary bits are needed to represent an n-digits decimal? The value of an m-bit binary is up to The value of an n-digit decimal is up to so OR

Questions How many bits are needed to represent an n-digits octal? How many bits are needed to represent an n-digits hexadecimal?

Conversion of Integer - Decimal to Binary Divide a decimal successively by 2, record the remainder, until the result of the division is 0. 67/2 = 33 Rem.=1 33/2 = 16 Rem.=1 16/2 = 8 Rem.=0 8/2 = 4 Rem.=0 4/2 = 2 Rem.=0 2/2 = 1 Rem.=0 1/2 = 0 Rem.=1

Conversion of Integer - Decimal to Hexadecimal Divide a decimal successively by 16, record the remainder, until the result is 0 432/16 = 27 Rem.=0 27/16 = 1 Rem.=B 1/16 = 0 Rem.=1

Conversion of integer - Binary to Decimal Conversion of integer - Hexadecimal to Decimal

Conversion of integer - Hexadecimal Binary 0 1 0 1 0 1 1 1 5 7 1011 2 B 7 0010 0111 Conversion of integer: - Octal Binary 0 1 0 1 0 1 1 1 2 7 1 2 5 7 010 111 101

Conversion of Fraction - Decimal to Binary The fraction is repeatedly multiplied by 2, the integer part is stripped and recorded. 0.687510 = 0.10112 Question: Convert 0.625? 0.110? 0.687510 * 2 =1.375 1 0.37510 * 2 = 0.75 0 0.7510 * 2 = 1.5 1 0.510 * 2 = 1.0 1 0.010 * 2 = 0.0 end

Conversion of Fraction - Binary Hexadecimal 0.1010110012 = 0.1010 1100 10002 = 0.AC816 0.7BC16 = 0.0111 1011 11002 = 0.01111011112 Convert 0.1100101011 Convert 0.6D5?

Binary Coded Decimal (BCD) A decimal digit is coded into 4 bits  1 byte can store 2 digits Example: 1942 is encoded as 0001 1001 0100 0010 (2 bytes) Disadvantages Complex arithmetic Inefficient use of storage Advantage: Can represent “real” number Decimal BCD 0 0000 1 0001 2 0010 3 0011 4 0100 5 0101 6 0110 7 0111 8 1000 9 1001

Binary Addition Binary Subtraction ? 0 0 1 1 0 1 1 1 + 0 1 0 1 0 1 1 0 0 0 1 1 0 1 1 1 + 0 1 0 1 0 1 1 0 1 1 1 1 1 Carry 1 0 0 0 1 1 0 1 Binary Subtraction ? use 2’s complementary arithmetic (later … )

Signed Integer Representations Sign-and-Magnitude 1’s Complement 2’s Complement

Sign-and-Magnitude Representation Use the most significant bit (MSB) to indicate the sign of the number. The MSB is 0 for positive, 1 for negative. 8 bits represent -12710 to 12710 Examples: 00101100 for +4410 10101100 for -4410 Problem: 00000000 = +0, 10000000 = -0

1’s Complement Just flip the bits! The MSB is 0 for positive, 1 for negative. 8 bits represent -12710 to 12710 Examples: 00101100 for +4410 11010011 for -4410 Problem: 00000000 = +0, 11111111 = -0

2’s Complement The 2’s complement for N is Example: Using 5 bits (n = 5), if N = 7, then 12 -7 5 0 1 1 0 0 + 1 1 0 0 1 1) 0 0 1 0 1 2’s complement = 1’s complement + 1 7 0 0 1 1 1 1’s complement 1 1 0 0 0 + 1 2’s complement 1 1 0 0 1 = -7

2’s Complement (con.) To form the two’s complement of a number, simply invert the bits (1 to0, 0 to 1), and add 1  1’s complement + 1 1310 = 0 0 0 0 1 1 0 1 -1310 = 1 1 1 1 0 0 1 0 + 1 = 1 1 1 1 0 0 1 1 2’s comp. -34?

Properties of 2’s Complement One unique 0 MSB = 0, positive number MSB = 1, negative number The range is For 5 bits, the range is -1610 (10000) to +1510 (01111) The 2’s complement of the complement of X is X itself (Prove: )

2’s Complementary Arithmetic Subtraction is performed using addition A - B = A + (-B) X = 910 = 010012, Y = 610 = 001102 -X = -910 = 101112, -Y = -610 = 110102 X -Y 310 0 1 0 0 1 + 1 1 0 1 0 1) 0 0 0 1 1 - X - Y -1510 1 0 1 1 1 + 1 1 0 1 0 1) 1 0 0 0 1

Binary Adder/Subtractor

Arithmetic Overflow The overflow happens when positive + positive  negative negative + negative  positive If are MSBs, then the overflow bit v is expressed as 12 +13 25 0 1 1 0 0 + 0 1 1 0 1 1 1 0 0 1 -12 -13 -25 1 0 1 0 0 + 1 0 0 1 1 1) 0 0 1 1 1

Arithmetic Overflow In practice, Case I: A and B are + an-1 = __ bn-1 = __ cn = __ cn-1 = __ V = _____ Case II: A and B are – Most significant stage of full adder

Fixed-point Arithmetic 3.62510  0011.10102  00111010 + 6.510  0110.10002  + 01101000 10.12510 10100010 The fraction point is added 10100010 -> 1010.00102 What if we want to represent ? We need at least m > 3.3*n = 3.3*20 = 66 bits !

Floating Point Numbers Scientific Notation a is mantissa, r is radix, e is exponent is for binary

IEEE Floating Point Format S: Sign bit, 1 bit E: Exponent, 8 bits B: Bias, 8 bits, 12710=0111 1111 F: mantissa

IEEE Floating Point - Example Normalization, e.g., Use sign and magnitude representation for signed mantissa. Use biased exponent, e.g. in excess 127 If we have 8 bits for exponent, to represent -5, we add 12710 (0111 1111) to -5, the result is 12210=0111 10102.

IEEE Floating Point - Example Represent -2345.125 in 32 bits ? -2345.125 = -100100101001.0012 1) Normalization: 2) Negative mantissa, so S = 1 3) Exponent: E = 11 + 127 = 13810 = 100010102 4) Mantissa F = 00100101001 0010…000 1 10001010 00100101001001000…0 1.510 in IEEE format?

IEEE Floating Point Format – Special Cases Zero: exponent and fraction all 0s Denormalized: exponent all 0s, fraction non-zero for single precision: Infinity: exponents all 1s, fraction all 0s sign bit determines +infinity or -infinity Not a Number (NaN): exponents all 1s, A good reference is available at http://research.microsoft.com/~hollasch/cgindex/coding/ieeefloat.html

Bit Patterns and Logical Operations OR NOT Exclusive OR They are bit-wise operations.

AND Operation OR Operation x AND y is true, if and only if both bits x and y are true. A = 1 1 0 0 1 0 1 1 B = 0 1 1 0 1 1 0 1 C=A AND B = 0 1 0 0 1 0 0 1 OR Operation x OR y is true, if either bits x or y is true. A = 1 1 0 0 1 0 1 1 B = 0 1 1 0 1 1 0 1 C=A OR B = 1 1 1 0 1 1 1 1

EOR (Exclusive OR) Operation NOT Operation A 1 becomes 0, and a 0 becomes 1. A = 1 1 0 0 1 0 1 1 C = NOT A = 0 0 1 1 0 1 0 0 EOR (Exclusive OR) Operation It is true, if and only if just one of inputs is true. A = 1 1 0 0 1 0 1 1 B = 0 1 1 0 1 1 0 1 C = A EOR B = 1 0 1 0 0 1 1 0