Lecture 2: 8/29/2002CS149D Fall CS149D Elements of Computer Science Ayman Abdel-Hamid Department of Computer Science Old Dominion University Lecture 2: 8/29/2002

CS149D Fall Outline Number Systems Decimal/Binary Octal/Hexadecimal Binary addition Representing Negatives Two’s complement notation and addition Overflow problem Today’s lecture covers sections 1.4, 1.5, and 1.6 in Brookshear text

Lecture 2: 8/29/2002CS149D Fall All computers use the binary number system (base 2) basic nature of electronic circuits ON/OFFTRUE/FALSE current flow/does not flow Machine alphabet has two letters “0”, “1” Each letter is a binary digit “bit”. Byte is 8 bits. Introduction

Lecture 2: 8/29/2002CS149D Fall Numbers can be represented in any base (humans use base 10) Symbols for a number system of base B are 0, 1, 2, …, B –1 decimal (base 10) 0, 1, 2,.., 9binary (base 2) 0, 1 notation “number B ” (375 in decimal is written , 1011 in binary is written ) Value of i th digit d is “d * B i” where i starts from 0 and increases from right to left 2 1 0ipositional notation 3 7 5d 5 * 10 0 =5 7 * 10 1 =70 3 * 10 2 =300 Three hundred and seventy five Number Systems

Lecture 2: 8/29/2002CS149D Fall Conversion from binary to decimal Convert to decimal = (1 * 2 0 ) + (1 * 2 1 ) + (0 * 2 2 ) + (1 *2 3 ) = = i d This process can be used for conversion from any number system to decimal (TRY convert to decimal)

Lecture 2: 8/29/2002CS149D Fall Conversion from decimal to binary Convert to binary Step 1: divide value by 2 and record remainder Step 2: as long as quotient not zero, continue to divide the newest quotient by 2 and record the remainder Step 3: when obtain a zero as quotient, binary representation consists of remainders listed from right to left in order OperationQuotientremainder 13 by by by by =

Lecture 2: 8/29/2002CS149D Fall Other Number Systems Octal (base 8) Symbols (0, 1, 2, 3, 4, 5, 6, 7) Working with too long binary numbers is a problem Hexadecimal (base 16) Symbols (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F) Byte = 8 bits = 2 hex digits ( 1 hex digit is 4 bits) B5 16 = ? 2 B 16 is is B5 16 =

Lecture 2: 8/29/2002CS149D Fall Conversion from binary to hex Convert to hex Divide binary number into 4 bits groups from right to left E 16 34E

Lecture 2: 8/29/2002CS149D Fall DecimalBinaryHexadecimal A B C D E F 2 0 = 12 7 = = 22 8 = = 42 9 = = = = = = = = = 8190

Lecture 2: 8/29/2002CS149D Fall Binary Addition Binary addition table = = = = 0 and a carry of

Lecture 2: 8/29/2002CS149D Fall Negative numbers 1/4 Sign and magnitude Left hand bit is used to determine the sign of the number Left hand bit 0 = positive number Left hand bit 1 = negative number 0010 = = -2 Using 4 bits with sign and magnitude, largest positive number represented is 7 (-7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7) 15 values overall can be represented

Lecture 2: 8/29/2002CS149D Fall Negative numbers 2/4 Two’s Complement +2 added to –2 in sign magnitude using regular binary addition does not equal 0 (make sure of that!) First bit indicates a negative value but also bears a position value For a positive number, the two’s complement representation is itself For a negative number, complement positive value, then add 1 3 in two’s complement is in two’s complement is Invert bits011 becomes 100 Add = 101 Invert bits 0 becomes 1 1 Becomes 0

Lecture 2: 8/29/2002CS149D Fall What is 1010 in two’s complement? It is a negative number since left hand bit is a 1 Invert bits1010 becomes 0101 Add = 0110 (+6) Then the original number is -6 Negative numbers 3/4

Lecture 2: 8/29/2002CS149D Fall Negative Numbers 4/4 Two’s complement addition Copyright 2003 Pearson Education, Inc. Subtraction performed the same as addition 7-5 is the same as 7+ (-5) 1011 = -(0100+1) = -(0101) = -5

Lecture 2: 8/29/2002CS149D Fall Overflow Problem Try adding in 4-bit two’s complement notation Overflow. Result is Negative value Actually –(0110+1) = -(0111) = -7 There is a limit to the size of the values that can be represented according to how many bits are used. Normally integers are represented in 32-bit patterns.