Number Representations and Computer Arithmetic 120301 Computing with

Number Representations and Computer Arithmetic 12/03/01

“Computing” with Computers n Back in the “old days” (WW II time), the word “computer” meant this • “Computers” were used to compute things such as trajectory tables, etc. – redundant “computers” for reliability and speed • Quite effective! CS 21 a 9/23/02 – new computers were tested against human computers © Luis F. G. Sarmenta and John Paul Vergara, Ateneo de Manila University Odds and Ends Slide 12/03/01

“Computing” with Computers n n n Of course, we don’t do things that way anymore! But HOW do (electronic) computers compute? The trick is to use BINARY numbers n n CS 21 a 9/23/02 use 1’s and 0’s corresponds on current/voltage being ON or OFF easy to implement resilient against noise © Luis F. G. Sarmenta and John Paul Vergara, Ateneo de Manila University Odds and Ends Slide 12/03/01

Java types: int and double n n n int range: -2, 147, 483, 648 to 2, 147, 483, 647 double range: 4. 94 e-324 to 1. 80 e+308 These ranges depend on: n n CS 21 a 9/23/02 Storage size Internal data representation © Luis F. G. Sarmenta and John Paul Vergara, Ateneo de Manila University Odds and Ends Slide 12/03/01

Review: Decimal Numbers n Integer Representation n number is sum of DIGIT * “place value” d 7 d 6 d 5 d 4 d 3 d 2 d 1 d 0 Range 0 to 10 n - 1 n 107 106 105 104 103 102 101 100 379210 = 3 103 + 7 102 + 9 = 3000 + 700 + 90 + 2 101 + 2 100 Adding two decimal numbers n add by “place value”, one digit at a time 3792 + 531 ? ? ? CS 21 a 9/23/02 0 0 3 7 9 2 1 3 7 9 2 + 0 5 3 1 0 4 3 2 3 “carry 1” because 9+3 = 12 © Luis F. G. Sarmenta and John Paul Vergara, Ateneo de Manila University Odds and Ends Slide 12/03/01

Binary Numbers • Humans can naturally count up to 10 values, • But computers can count only up to 2 values (OFF and ON, or 0 and 1) n (Unsigned) Binary Integer Representation n “base” of place values is 2, not 10 b 7 b 6 b 5 b 4 b 3 b 2 b 1 b 0 Range 0 to 2 n - 1 0 1 1 0 0 27 26 25 24 23 22 21 20 011001002 = 26 + 25 + 22 = 64 + 32 + 4 = 10010 aka “ 0 b 01100100” CS 21 a 9/23/02 © Luis F. G. Sarmenta and John Paul Vergara, Ateneo de Manila University Odds and Ends Slide 12/03/01

Converting from Binary to Decimal n n n VERY USEFUL trick for a CS/MIS person … Memorize powers of 2 from 20 up to 210 Lets you approximate any power of 2 n n Example 1: what is 216? n n “ 1 KB” is actually 1, 024 bytes, not 1000 bytes “ 1 MB” is 1 K*1 KB = 220 bytes = 1, 048, 576 bytes = approximately 1 million bytes “ 1 GB” is 1 K * 1 MB = 230 bytes = (approx 1 billion) 26 * 210 = 64 * 1024 = 64 K = 65, 536 Example 2: n n CS 21 a 9/23/02 20 21 22 23 24 25 26 27 28 29 210 = = = 1 2 4 8 16 32 64 128 256 512 1, 024 or “ 1 K” The Pentium processor does integer math with 32 -bit numbers. What’s the highest unsigned number it can handle (approximately)? range = 2 n-1 = 232 - 1 232 = 22 * 230 = 4 * 1 G = approximately 4 billion = actually 4*1024*1024 = 4, 294, 967, 296 (minus 1) © Luis F. G. Sarmenta and John Paul Vergara, Ateneo de Manila University Odds and Ends Slide 12/03/01

Converting from Binary to Decimal n More practice 0 b 00000010 0 b 00001110 0 b 00110001 0 b 10110001 CS 21 a 9/23/02 © Luis F. G. Sarmenta and John Paul Vergara, Ateneo de Manila University 20 21 22 23 24 25 26 27 28 29 210 = = = 1 2 4 8 16 32 64 128 256 512 1, 024 or “ 1 K” Odds and Ends Slide 12/03/01

From Decimal to Binary n n General rule: Divide and write remainder from right to left Why this works n n n remainder gives bit 0 odd numbers have a 1 in bit 0 Note that this rule works in converting decimal to any base n e. g. , HEX numbers (more later) 3792 / 2 = 1896 rem 0 1896 / 2 = 948 rem 0 948 / 2 = 474 rem 0 474 / 2 = 237 rem 0 237 / 2 = 118 rem 1 118 / 2 = 59 rem 0 59 / 2 = 29 rem 1 29 / 2 = 14 rem 1 14 / 2 = 7 rem 0 7 / 2 = 3 rem 1 3 / 2 = 1 rem 1 1 / 2 = 0 rem 1 0 b 111011010000 = 211+210+29+27+26+24 = 3792 CS 21 a 9/23/02 © Luis F. G. Sarmenta and John Paul Vergara, Ateneo de Manila University Odds and Ends Slide 12/03/01

Binary Arithmetic n Adding two decimal numbers 3792 + 531 ? ? ? n 0 4 3 2 3 “carry 1” because 9+3 = 12 Adding two binary numbers n CS 21 a 9/23/02 1 3 7 9 2 + 0 5 3 1 same, but “ 1 + 1 = 10” 14 + 7 1 1 1 0 + 0 1 1 1 21 1 0 1 “carry 1” because 1+1 = 10 © Luis F. G. Sarmenta and John Paul Vergara, Ateneo de Manila University Odds and Ends Slide 12/03/01

Binary Arithmetic n n In general: Add up to 3 bits at a time per place value n n n 1 1 1 0 0 1 1 1 0 + 0 1 1 1 sum bits carry-out bits 0 1 1 1 0 Output 2 bits at a time n n CS 21 a 9/23/02 A and B “carry in” carry-in bits A bits B bits sum bit for that place value “carry out” bit (becomes carry-in of next bit) © Luis F. G. Sarmenta and John Paul Vergara, Ateneo de Manila University Odds and Ends Slide 12/03/01

So what about negative values? n n In math, it is easy to represent negative values Just put a negative sign (-) prefix In computers, we extend the binary notation in order to support signed values We can use the following 3 methods: n n n CS 21 a 9/23/02 Sign and magnitude (using the Most Significant Bit or MSB) 1’s complement 2’s complement © Luis F. G. Sarmenta and John Paul Vergara, Ateneo de Manila University Odds and Ends Slide 12/03/01

Signed Integers n Sign-and-Magnitude n n n 210 0 010 0 000 -210 1 010 -010 1 000 flip bits easy to do subtraction STILL has 2 “zeros” 210 -210 0 010 1 101 010 -010 0 000 1 111 210 -210 0 010 1 110 010 -010 0 000 1’s complement n n MSB represents sign bit (0 for positive, 1 for negative) has 2 “zeroes” 2’s complement n n flip bits, then add 1 easy subtraction (just negate, then add) has only 1 zero Another interpretation: n CS 21 a 9/23/02 add place values as before, except that MSB is negative n-1)John Paul Vergara, F. G. value Sarmenta (i. e. , MSB©is. Luis place is 2 and Ateneo de Manila University Odds and Ends Slide 12/03/01

Binary Subtraction n 1’s complement subtraction: flip the negative 5 - 3 = 2 0 1 0101 0011 flip 1100 -3 in 1’s complement form 3 - 5 = -2 0011 0101 flip 1010 +1 1 0 0 0 1 1 0 0 1 0 Add the carry overflow 2 0 0 1 1 +1 0 1 -2 -5 in 1’s complement form CS 21 a 9/23/02 © Luis F. G. Sarmenta and John Paul Vergara, Ateneo de Manila University Odds and Ends Slide 12/03/01

Binary Subtraction n 2’s complement subtraction: flip then add 1 5 - 3 = 2 0101 0011 flip 1100 +1 1101 -3 in 2’s complement form 3 - 5 = -2 0011 0101 flip 1010 +1 1011 -5 in 2’s complement form CS 21 a 9/23/02 0 1 +1 1 0 0 1 0 2 ignore overflow 0 0 1 1 +1 0 1 1 1 0 0001 flip 0010 +1 -2 (flip+1 also gives positive of negative number) © Luis F. G. Sarmenta and John Paul Vergara, Ateneo de Manila University Odds and Ends Slide 12/03/01

Range of binary numbers Java (and most computer platforms today) use 2’s comp. n Hence the range for the different int types n byte (8 bits): ? n short (16 bits): ? n int (32 bits): -2, 147, 483, 648 to 2, 147, 483, 647 n long (64 bits): ? n CS 21 a 9/23/02 © Luis F. G. Sarmenta and John Paul Vergara, Ateneo de Manila University Odds and Ends Slide 12/03/01

Hexadecimal (Hex) Numbers n n n aka “ 0 x 64” Base 16 Each hex digit goes 6416 = 6*1610 + 4 = from 0 -9, then A-F Hex is a convenient shortform for binary b 7 b 6 b 5 b 4 b 3 b 2 b 1 4 bits = 1 hex digit 0 1 1 0 0 1 0 (aka nibble) 27 26 25 24 23 22 21 1 byte = 8 bits = 2 hex digits Another useful trick: n n b 0 0 20 memorize binary of 0 to F Addition and conversion to/from decimal is similar n CS 21 a 9/23/02 10010 except use base 16 instead of 2 © Luis F. G. Sarmenta and John Paul Vergara, Ateneo de Manila University Odds and Ends Slide 12/03/01

Some Exercises n n What’s the range of an n-bit sign-mag integer? How about 1’s comp? 2’s comp? Convert the ff signed 8 -bit values to decimal n n Convert the ff to 16 -bit binary and hex numbers: n n n CS 21 a 9/23/02 413, 39, 1045, -3, -124, -134 Add the following pairs n n 0 b 0101, 0 b 1110111, 0 x 14, 0 x 41 0 b 00001011 + 0 b 00100100, 0 x 3 F + 0 x 2 F If 0 x 7 F and 0 x 32 are signed 8 -bit integers, what’s wrong with adding them and storing the result in a byte? Puzzle: How can I use my 10 fingers to count up to 1000? © Luis F. G. Sarmenta and John Paul Vergara, Ateneo de Manila University Odds and Ends Slide 12/03/01