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Lecture 2 Data Representation 2. Integers. Readings. Stalling, Computer Architecture & Organization, pp 291-301. Also summarised in Appendix A. A Bit of History. Numeral – a single symbol that represents a quantity or number.

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## Lecture 2 Data Representation 2

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**Lecture 2Data Representation 2**Integers**Readings**• Stalling, Computer Architecture & Organization, pp 291-301. Also summarised in Appendix A.**A Bit of History**• Numeral – a single symbol that represents a quantity or number. • Number system – a way of assigning meaning to combinations of numerals. • Base 10 (denary) number system used by most cultures (10 fingers). • Mayas, Celts, Aztecs – base 20. • Sumerians and Babylonians – base 60.**Numbers**123.45 = 100 + 20 + 3 + 0.4 + 0.05 = 1*102 + 2*101 + 3*100 + 4*10-1 + 5*10-2 This is a base 10 number (denary) system. It has 10 symbols: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.**Numbers (continued)**• There are other number systems (bases). • The 'value' of a number is determined by positional notation. • In general, def.xyzb = • d*b2 + e*b1 + f*b0 + x*b-1+ y*b-2+ z*b-3 • where b is the base.**Binary Number System**We said the computer is a binary system, i.e. base 2. Thus there are two symbols: 0, 1 110011 = 1*25 + 1*24 + 0*23 + 0*22 + 1*21 + 1*20 = 32 + 16 + 0 + 0 + 2 + 1 = 51 110.011 = 1*22 + 1*21 + 0*20 + 0*2-1 + 1*2-2 + 1*2-3 = 4 + 2 + 0 + 0 + 1/4 + 1/8 = 6.375**Decimal to Binary Conversion**• Convert 38.6875 to binary. • For the whole number part, repeatedly divide by 2. Use the remainders from bottom to top.**Decimal to Binary Conversion (continued)**Dividend Quotient Remainder 38 19 0 LSB 19 9 1 9 4 1 4 2 0 2 1 0 1 0 1 MSB**Decimal to Binary Conversion (continued)**• For the fractional part, multiply by 2. • Split the result into an integer part and a fractional part. • Continue multiplying the fractional part until it is zero or you get tired (run out of space). • Use the integer part from top to bottom.**Decimal to Binary Conversion (continued)**Multiplicand Integer Fractional Part Part 0.6875 1 .375 LSB 0.375 0 .750 0.75 1 .500 0.50 1 .000**Decimal to Binary Conversion (continued)**From earlier slide: 3810 = 1001102 From previous slide: 0.687510 = 10112 Combining: 38.687510 = 100110.10112**Repeating & Irrational Numbers**• If the denominator has factors other than 2 or 5 it will produce a recurring decimal fraction, e.g. 1/3 = 0.3333 … • Irrational numbers, such as √2, and transcendental numbers, such as p, also never terminate. • Thus most decimal fractions will produce recurring binary fractions. • Truncate (or round) these to fit in available storage.**Binary to Decimal Conversion**POWER 2P 2-P 0 1 1 2 .5 2 4 .25 3 8 .125 4 16 .0625 5 32 .03125 6 64 7 128 8 256 9 512 10 1,024**Binary to Decimal Conversion (continued)**POWER 2P 11 2,048 12 4,096 13 8,192 14 16,384 15 32,768 16 65,536 20 1,048,576 30 1,073,741,824 31 2,147,483,648 32 4,294,967,296**Binary to Decimal Conversion (continued)**Work outwards from the binary point: 100110.1011 = 2 + 4 + 32 + .5 + .125 + .0625 = 38.6875**Octal Number System**• A base 8 number system. • Uses: 0, 1, 2, 3, 4, 5, 6, 7. • Takes 3 bits. • To convert from binary, group the bits in threes outwards from the binary point.**Octal Number System**000 0 001 1 010 2 011 3 100 4 101 5 110 6 111 7 101 110 001 011 56138**Hexadecimal Number System**• A base 16 number system. • Uses 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F. • Takes 4 bits. • Often called hex for short. • Group bits in fours from the binary point.**Hexadecimal Number System (continued)**0000 0 1000 8 0001 1 1001 9 0010 2 1010 A 0011 3 1011 B 0100 4 1100 C 0101 5 1101 D 0110 6 1110 E 0111 7 1111 F 100 1101 1001 1111 4D9F16**decimal**binary octal hex Other Conversions Conversions between decimal and octal or hexadecimal are usually done via a binary representation, i.e. :**Introduction to Negative Integers**• So far we have concentrated on positive integers. We have to devise some scheme to represent negative numbers. • 4 possible ways: • Excess notation. • Sign and magnitude. • One's complement. • Two's complement.**Excess Notation**• Shift the numbers by about half of the range of representation. • For example, assume we have 3 bits to represent integers. That allows for 8 (23) possible values. Let's shift by 4 which is called excess-4 notation.**Excess Notation (continued)**Actual Stored Number Number -4 000 -3 001 -2 010 -1 011 0 100 1 101 2 110 3 111**stored**number actual number excess notation = + actual number stored number excess notation = - Excess Notation Algorithms Use the following equations to convert between the actual number and the stored number.**Excess NotationExample 1**A computer stores decimal integer values in 8 bits, using excess-128 notation. How would the decimal value -60 be stored? stored number = -60 + 128 = 68 = 64 + 4 = 10001002 or 0100 0100 in a byte**Excess NotationExample 2**The binary number, 0101 1101, is in excess-128 binary notation. What is the number in decimal? 0101 1101 = 64 + 16 + 8 + 4 + 1 = 93 in excess-128 actual number = 93 - 128 = -35**Excess NotationSummary**Negation is complicated. Addition and subtraction is easy. One value for zero. Range of integers is (2n-1 - 1) to - (2n-1).**Signed Magnitude**Assign the most significant bit as a sign bit instead of a power of two. 0 - positive 1 - negative Using 4 bits: 2 0010 -2 1010 For a byte: 84 0101 0100 -84 1101 0100**Signed Magnitude Summary**Negation is easy. Addition and subtraction is complicated. Two values for zero. Range of integers is (2n-1 - 1) to - (2n-1 - 1).**One's Complement**If positive, do nothing. If negative, complement (reverse) each bit. 84 0101 0100 -84 1010 1011**One's ComplementSummary**Negation is easy. Addition and subtraction is easier. Two values for zero. Range of integers is (2n-1 - 1) to - (2n-1 - 1).**Two's Complement**If positive, do nothing. If negative, complement (reverse) each bit, and then add 1 to the result.**Two's Complement Example**Convert –84 to two's complement: 84 = 0101 0100 0101 0100 original 1010 1011 each bit complemented + 1 add one 1010 1100 -84 in two's complement**Two's Complement to Decimal**If the sign bit is 0, convert the binary number to decimal. If the sign bit is 1, subtract 1 from the binary number, complement (reverse) each bit, convert the binary number to decimal, and put a minus sign in front.**Two's Complement to Decimal Example**Convert the 8-bit number 1110 1010 in two's complement to decimal. 1110 1010 - 1 1110 1001 Complement each bit: 0001 0110 0001 0110 = 16 + 4 + 2 = 22 Answer = -22**Two's Complement Summary**Negation is relatively easy. Addition and subtraction is easy. One value for zero. Range of integers is (2n-1 - 1) to - (2n-1).**4-Bit Summary**Decimal SM 1's 2's +7 0111 0111 0111 +6 0110 0110 0110 +5 0101 0101 0101 +4 0100 0100 0100 +3 0011 0011 0011 +2 0010 0010 0010 +1 0001 0001 0001 0 0000 0000 0000 -0 1000 1111**4-Bit Summary**Decimal SM 1's 2's -1 1001 1110 1111 -2 1010 1101 1110 -3 1011 1100 1101 -4 1100 1011 1100 -5 1101 1010 1011 -6 1110 1001 1010 -7 1111 1000 1001 -8 1000**Arithmetic with Negative Numbers (1)**• Addition and subtraction in signed magnitude representation requires consideration of the signs of both numbers and consideration of their absolute value. Leads to complex rules (complex logic). • Complement forms take advantage of the continuum of representation.**Arithmetic with Negative Numbers (2)**• How to choose between 1's and 2's complement? • 1's complement easier to form. • 2's complement has only one 0 which makes logic for zero comparison easier. • Most machines use 2's complement form.**Two's Complement Example 1**Subtract 5 from 18 in 8-bit, 2's complement. 18 = 0001 0010 -5 = 1111 1011 10000 1101 Ignoring the high-order (9th) bit gives: 0000 1101 = 13**Two's Complement Example 2**Subtract 5 from -18 in 8-bit, 2's complement. -18 = 1110 1110 -5 = 1111 1011 11110 1001 Ignoring the high-order (9th) bit gives: 1110 1001 = -23**Two's Complement Overflow**• Over/Underflow occurs if any of the following situations occur (assuming that both A and B are positive): • A + B negative. • A – (-B) negative. • -A – B positive. • -A + -B positive.**Next Lecture**Real Numbers (floats and doubles)

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