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U of I CS 231 - Intro & Base Conversions

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CS 231Intro & Base ConversionsCelal ZiftciAug 26, 2005Announcements Mallard is up. Try to log on and take the first quiz (due Monday, Aug 29) Course webpage is down Mallard link is posted on the newsgroup(https://mallard2.cites.uiuc.edu/CS231/) First homework assignment to be released sometime next weekTips for CS 231 Don’t hesitate to: Ask questions Interrupt when you don’t understand Be comfortable with binary and hex Learn device inputs & outputs Allows abstraction Easier to understand Show your workTips for CS 231 Efficiency is important! Neatness is important! Unreadable solutions are considered wrong Readable ones are neat, short (big favor for both you and the graders) Digits = powers of 10… 100, 10, 1,1/10,1/100,1/1000 ……102, 101,100,10-1,10-2,10-3 …Ex: (36.25)10= 3*10 + 6*1 + 2*1/10 + 5* 1/100 Bits = powers of 2…8, 4,2,1,1/2,1/4,1/8 ……23, 22, 21,20,2-1,2-2,2-3 …Ex: (100100.01)2= 1*32 + 1*4 + 1*1/4Digits vs. bitsBinary to decimal add powers that have a 1(101001)2= 32 + 8 + 1(0.1001)2= 1/2 + 1/16(10110.1011)2= ?16 + 4 + 2 + 1/2 + 1/8 + 1/16= 22.6875Decimal to binary Left of decimal point Repeatedly divide integer part by 2 until you get 0 Read remainders bottom to up22 = (?)22211 R 05R 12R 11R 00R 1Decimal to binary Left of decimal point Repeatedly divide integer part by 2 until you get 0 Read remainders bottom to up22 = (10110)22211 R 05R 12R 11R 00R 1Decimal to binary Right of decimal point Repeatedly multiply fractional part by 2 until you get 1 Read integer portion top to bottom0.8125 = (?)20.81251.62501.250.51.0Decimal to binary Right of decimal point Repeatedly multiply fractional part by 2 until you get 1 Read integer portion top to bottom0.8125 = (0.1101)20.81251.62501.250.51.0Decimal to binary What if there are both left and right of the decimal point? Do them separately and combine22.8125 = (?)2Decimal to binary What if there are both left and right of the decimal point? Do them separately and combine22.8125 = (?)22211 R 05R 12R 11R 00R 10.81251.62501.250.51.0Decimal to binary What if there are both left and right of the decimal point? Do them separately and combine22.8125 = (10110.1101)22211 R 05R 12R 11R 00R 10.81251.62501.250.51.0up downDecimal to binary Toy example3.25 = (?)231 R 10 R 10. 250.501.0Decimal to binary Toy example3.25 = (11.01)231 R 10 R 10. 250.501.0Hexadecimal base Hex digits = powers of 16…256, 16, 1,1/16,1/256 ……162, 161,160,16-1,16-2 … use digits 0-9, A-F A=10, B=11, C=12, D=13, E=14, F=15 often preceded by 0x book subscript notation (24.4)16Ex: (24.4)16= 2*16 + 4*1 + 4*1/16Hexadecimal base Hex (hexadecimal) Hex digit is a group of 4 bits Memorize this table!!dec. hex binary0 0 00001 1 00012 2 00103 3 00114 4 01005 5 01016 6 01107 7 01118 8 10009 9 100110 A 101011 B 101112 C 110013 D 110114 E 111015 F 1111 Hex (hexadecimal) Group from decimal point outward Pad with zeros to get groups of 4(1101101001010.101001)2(0001 1011 0100 1010 . 1010 0100)21 B 4 A . A 4(1101101001010.101001)2= (1B4A.A4)16Binary to Hex Octal digits = powers of 8…64, 8, 1,1/8,1/64 ……82, 81,80,8-1,8-2 … use digits 0-7 sometimes preceded by 0 book subscript notation (24.4)8Ex: (44.2)8= 4*8 + 4*1 + 2*1/8Octal base Octal Octal digits are groups of 3 bits Pad with zerosdec. octal binary0000011001220103301144100551016611077111Octal base Octal Group from decimal point outward Pad with zeros to get groups of 3(1101101001010.101001)2(001 101 101 001 010 . 101 001)21 5 5 1 2 . 5 1(1101101001010.101001)2= (15512.51)8Binary to Octal What about other conversions such as: Octal Æ Hex Decimal Æ Hex … Use other conversions you already know Octal Æ Binary Æ Hex Decimal Æ Binary Æ HexOther conversionsI. Decimal Æ BinaryBinary Æ DecimalII. Binary Æ HexBinary Æ OctalIII. Other conversions Use the conversions you already


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U of I CS 231 - Intro & Base Conversions

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