Systems Architecture Lecture 2 Slides PDF

Systems Architecture IN1006 Number Systems and Their Conversion —Dr H. Asad Positional Number Systems  Numbers can be represented in any base  Humans are used to base 10  The digits used in base 10 are: 0,1,2,3,4,5,6,7,8 and 9  We can analyse nu...

Systems Architecture IN1006 Number Systems and Their Conversion —Dr H. Asad Positional Number Systems  Numbers can be represented in any base  Humans are used to base 10  The digits used in base 10 are: 0,1,2,3,4,5,6,7,8 and 9  We can analyse numbers based on position and powers of 10 Example 3245 = 3 x 103 + Similarly, we can write: Thousand Hundred Ten Units 789 6 7 8 9 6 7896 = 7000 800 + 90 + 6 + 7896 = 8 x 100 + 9 x 10 + 6x1 7 x 1000 7896 = + 8 x102 9 x 101 6 x 100 + +  This can3 be applied to any base, not just 10. 7 x 10 Indicating the base  We should subscript numbers with their base (e.g., decimal numbers with ten and binary numbers with two )  00112 or 0011two or 0011b (which equals 310 or 3d)  1110 or 11ten or 11d (which equals 10112)  ten, d and 10 in the above alternative representations indicate base 10 Positional Numbers and Base (a systematic approach) In any number base, the value of the i th digit d in a number can be written as: d x base i where i starts at position 0 of the number and increases to the left. In 124710 what is represented by 2? The position i of 2 is 2 (as the position starts from right to left and the first position is 0) The digit d is 2 Hence, the value represented by the 2 digit in the second position is 2 X 102 = 200 Binary Numbers  Base 2 is usually used in computers (Binary)  This is because of the natural correspondence between high/low signals and 1/0, respectively  The digits used in this base are 0 and 1  A single binary digit is known as a bit  bit = binary digit  Example, binary numbers with two bits:  00  0 (decimal)  01  1 (decimal)  10  2 (decimal)  11  3 (decimal) Example: Decimal Number vs Binary Number Consider the number 1310. We can write this in the 10-base as: 1310 = 1 x 101 + 3 x 100 (as 1 x 101 + 3 x 100 = 1 x 10 + 3 x 1 = 13) And in the 2-base, following the formula d x base i, as 11012: 11012 = 1 x 23 +1 x 22 + 0 x 21 +1 x 20 = 1 x 810 +1 x 410 + 0 x 210 +1 x 110 Decimal Number to Binary Number conversion  Step 1: Divide the given decimal number by 2 and note down the remainder.  Step 2: Divide the obtained quotient by 2, and note the remainder again.  Step 3: Repeat the above step until you get 0 as the quotient.  Step 4: Write the remainders in such a way that the last remainder is written first, followed by the rest in the reverse order.  Step 5: Result number is the binary value of the given decimal number. Decimal Number to Binary Number conversion Hexadecimal  Long Binary numbers are hard to read!  One Hexadecimal (base 16) digit can represent 4 bits Hexadecimal (cont’d)  Uses the characters 0123456789ABCDEF to represent the numbers 0.. 15 corresponding to the 4-bit binary values 0000.. 1111 1357 9BDFhex 0001 0011 0101 0111 1001 1011 1101 1111two  Octal  Octal is a number representation system with base 8  The octal representation uses the characters: 0 1 2 3 4 5 6 7  Each Octal character corresponds to three bits in binary What is 27ten in Octal? Answer: 1. 27 / 8 24 = 3 x 8 and the remainder is 3 2. 3 / 8  0 and the remainder is 3, we stop 3. Write the remainders from bottom to top: 338  338 = 3 x 81 + 3 x 80 Exercises  Convert 2710 into binary, octal and hexadecimal Hint:  Divide by the base of the target representation  Compose number as sequence of remainders Convert 100111101101 hexadecimal Hint:  transform subsequences of three bits for octal  transform subsequences of four bits for hexadecimal DECIMAL BINARY OCTAL HEXADECIMAL 0 0 0 0 1 1 1 1 2 10 2 2 3 11 3 3 4 100 4 4 5 101 5 5 6 110 6 6 7 111 7 7 8 1000 10 8 9 1001 11 9 10 1010 12 A 11 1011 13 B 12 1100 14 C 13 1101 15 D 14 1110 16 E 15 1111 17 F 16 10000 20 10 17 10001 21 11 18 10010 22 12 19 10011 23 13 20 10100 24 14 21 22 10101 10110 25 26 15 16 100111101101 23 10111 27 17 24 11000 30 18 25 11001 31 19 26 11010 32 1A 27 11011 33 1B 28 11100 34 1C 29 11101 35 1D 30 11110 36 1E 31 11111 37 1F 32 100000 40 20 Systems Architecture IN1006 Signed & Unsigned no representation —Dr H. Asad Bits, Bytes and Words  Bit: (binary digit) the most basic unit of information in a digital computer  e.g. 1101 4 bits (also called a nibble) are used  Byte: a collection of 8 bits  e.g. 11010001  Word: two or more adjacent bytes that are addressed and handled collectively  E.g. 11010001 11110000 is a word with two bytes or 16-bits  The word size represents the data size that is handled more efficiently by a particular architecture  The number of bits we use to represent a number defines the largest number we can store  In contemporary computers, words with 32 bits (i.e., 4 bytes of 8 bits each) or 64 bits (i.e., 8 bytes of 8 bits each) are usual Representing Positive Integers  Binary numbers (32 bits) MSB LSB 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 1  Least significant bit (LSB) is the “smallest” bit in a word, i.e., the bit at position 0 (representing the number: LSB x 20, which may be the decimal 1 or 0)  Most significant bit (MSB) is the “biggest” bit in a word, i.e. the bit at position 31 (representing the number: LSB x 231, which may be the decimal13 number 2,147,483,648 or ???) Systems Architecture IN1006 Floating point and Non-Numeric Data Representation —Dr H. Asad  Endianness  To identify the order in which bytes that constitute words are stored in memory  Endianness defines this (i.e., the order in which bytes are arranged in memory)  When programming at a low level, it is important to know what your machine supports! Endian First Byte Last Byte Mnemonic Example computers Big Most Significant Least Normal way of M68000, Significant writing numbers IBM/360,motorolla Little Least Most Arithmetic X86, Z80,intel Significant Significant calculation order Some computers are bi-endian (ARM, PowerPC) Representing Positive Integers: The simple pattern Question: Assume that we have a hypothetical 3-bit word, how many different patterns of 0s and 1s can we have? Answer: We can represent 23 different 3-bit patterns, i.e., n-bits can represent 2n different numbers from 000 001 010 011 100 101 110 111 0 to 2n - 1 This represents numbers from 0 to 7 (7 = 23 – 1) Question: If we have a 32-bit word, how many different patterns of 0s and 1s can we have? Answer: We can represent 232 different 32-bit patterns. This represents numbers from 0 to (232 –1), i.e., 0 to 4,294,967,296 – 1 (or to 4,294,967,29510) Representing Positive Integers (cont’d) 32-bit word, how many different patterns of 0s and 1s can we have ? 0000 0000 0000 0000 0000 0000 0000 0000two = 0ten 0000 0000 0000 0000 0000 0000 0000 0001two = 1ten 0000 0000 0000 0000 0000 0000 0000 0010two = 2ten 0000 0000 0000 0000 0000 0000 0000 0011two = 3ten 0000 0000 0000 0000 0000 0000 0000 0100two = 4ten … … … 1111 1111 1111 1111 1111 1111 1111 1101two = 4,294,967,293ten 1111 1111 1111 1111 1111 1111 1111 1110two = 4,294,967,294ten 1111 1111 1111 1111 1111 1111 1111 1111two = 4,294,967,295ten Representing Positive Integers (the general case) A 32 binary number of the form: n=x31x30x29…x1x0 can be represented as the bit value times a power of 2 The general formula: (x31 * 231)+(x30 * 230)+(x29 * 229)+…+(x1 * 21)+(x0 * 20) (xi means the i th bit of n)  Example: 0000 0000 0000 0000 0000 0000 0000 0010two = 2ten 0000 0000 0000 0000 0000 0000 0000 0011two = 3ten … … … … … 1111 1111 1111 1111 1111 1111 1111 1110two = 4,294,967,294ten …  Representing Negative Integers  So far have involved only unsigned integer numbers  They cannot represent negative integer numbers  To represent negative and positive integer numbers, we need to have signed integer numbers.  Signed binary integer numbers may be expressed by:  Signed magnitude  One’s complement  Two’s complement Signed-Magnitude Representation  Assign one bit to represent the sign (typically the left most digit is used)  Use the rest of the bits to represent magnitude (value) Example: Assume that 1 represents “-” and 0 represents “+” Then consider 8-bit words: o -1 is 1 000 0001 o +1 is 0 000 0001  What is the range of numbers we can represent with N bits? -(2(N-1) -1) to 2(N-1) -1  This representation has some disadvantages:  Using a special sign bit means that we can represent fewer numbers  Both +0 (e.g. 0000 0000) and -0 (e.g. 1000 0000) are valid, leading to programmer headaches  Arithmetic circuits are complicated by the calculation of sign (as we will Systems Architecture IN1006 Arithmetic Operation Part 1 —Dr H. Asad 1-bit Binary Addition  two 1-bit values gives four cases: carry-out bit: 1 1 0 1 0 1 1 0 0 0 1 1 0 1-bit sum  The simplicity of this system makes it possible for digital circuits to carry out arithmetic operations. Binary Addition: how the “carry out” bit is propagated?  2 bits show all possibilities: (0) (0) (1) (1) (0) (0) (Carry out bit) 0 0 0 1 1 1 a 0 0 0 1 1 0 b …(0) 0 (0) 0 (0) 1 (1 1 (1) 0 (0) 1 ) Binary Addition: an Example  Example: using signed magnitude binary arithmetic, find the sum of 75 and 46. Binary Addition: an Example Example: using signed magnitude binary arithmetic, find the sum of 75 and 46. Solution: 1) convert 75 and 46 to binary, and arrange as a sum, but separate the (positive) sign bits from the magnitude bits. 2) Just as in decimal arithmetic, we find the sum starting with the rightmost bit and work left. 3) In the second bit, we have a “carry” bit, so we note it above the third bit. 4) The third and fourth bits also give us “carries”. 5) Once we have worked our way through all eight bits, we are done. sign bit is separated Binary Addition: Example  Example: Using signed magnitude binary arithmetic, find the sum of 107 and 46. Overflow bit  We see that the carry from the seventh bit overflows and is discarded  BUT then the addition gives us an erroneous result: 107 + 46 = 25. Overflow  It occurs if the result of “add” (or any other arithmetic operation) cannot be represented by the available hardware bits.  While we can’t always prevent overflow, we can always detect overflow.  Note: This is physical overflow.  What happens to the bits that “fall off the left end” is dependant on implementation Binary Addition for Negative Numbers  Example: Using signed magnitude binary arithmetic, find the sum of - 46 and - 25.  Because the signs are the same, all we do is add the numbers and supply the negative sign when we are done. sign bit is separated Binary subtraction (at the basic bit level) four cases: 1 0 1 0 ⎯ 1 1 0 0 0 1* 1 0 * But requires borrowing a digit from a higher position (left bit)  Binary subtraction: Examples with whole numbers  Determine which operand has the largest magnitude  The sign of the result gets the sign of the number that 2 is larger 0 0 0 0 0 0 1 0  The magnitude is obtained by subtracting the smaller 1 0 0 0 0 0 0 1 one from the larger one 0 0 0 0 0 0 0 1  It might require borrowing from the next non zero bit (to the left) or two, three etc. bit to the 1 left (until we find a non zero bit) 2 2  When we borrow from a higher bit to the next 0 0 0 0 0 1 0 0 lower position, we borrow “2” and reduce the bit that we borrowed from by “1” 1 0 0 0 0 0 0 1 Binary subtraction:  Example: Using signed magnitude binary arithmetic, find the sum of 46 and – 25. 4610 - 2510 2110 Systems Architecture IN1006 Arithmetic Operation Part 2 —Dr H. Asad Complement based representation (rationale)  Signed magnitude representation easy for people to understand, but requires complicated computer hardware.  Another disadvantage of signed magnitude is that it allows two different representations for zero: positive zero and negative zero  For these reasons (among others) computer systems employ complement systems for numeric value representation. Two’s complement (what is it?)  Two’s complement is the radix complement of the binary system: the radix complement of a non-zero number N in base r with d digits is r d – N for N ≠0, and 0 for N = 0. Two’s complement (what is it?)  Makes the hardware simple  MSB is regarded as sign bit (0:+ve, 1:-ve) Examples:  Radix complement of 2 in base 10 with 3 digits (or 10’s complement of 2 with 3 digits) 1 1 1 0 2 2 is: 103 – 2 = 1000 – 2 = 998 2 1  2’s complement of 4-bit number 00112 1 0 0 0 00 0 01 01 is: 24 – 00112 = 100002 – 00112 = 11012 0 0 0 0 1 11 00 0 0 0 0 Visualize two’s complement numbers on a 4 bit number To express a value in two’s complement representation: 1) If the number is positive, just convert it to binary and you’re done. 2) If the number is negative, flip bits and then add 1. Visualize two’s complement numbers on a 4 bit number Number 410 is 01002 Number –410 would be represented using 2’s complement as follows Binary number 0100 Inverse 1011 Add 1 1011 +1 2’s complement for – 4 1100 - 410 is 11002 on two’s complement system Two’s complement (some simple examples) Binary value 1s complement 2’s complement Decimal value 3 00000011 00000011 No change as it is a positive number -3 00000011 11111100 11111101 In 8-bit binary, 3 (flip bits) (add 1) Adding 1 gives us -3 in two’s complement Two’s complement (some more complex examples) Two’s complement shortcut 1. Negation: find the negative of a Positive binary number: invert (flip) the bits and add one 2. Transforming a negative decimal number into 2’s complement: write the binary of the magnitude and invert (flip) the bits and add one 3. Transforming a negative two’s complement number into decimal: invert (flip) the bits and add one. Calculate the decimal number from the binary and add (-) Two’s complement shortcut (example) Example: find the negative number of the decimal number 2. 0000 0000 0000 0000 0000 0000 0000 0010 = 2 Invert the bits and add one 1111 1111 1111 1111 1111 1111 1111 1101 + 0000 0000 0000 0000 0000 0000 0000 0001 1111 1111 1111 1111 1111 1111 1111 1110 = -2 Two’s complement shortcut (example) Example: find the decimal number of the two’s complement above (don’t forget the sign -2) invert the bits and add one BUT going in the other direction 0000 0000 0000 0000 0000 0000 0000 0001 + 0000 0000 0000 0000 0000 0000 0000 0001 0000 0000 0000 0000 0000 0000 0000 0010 = 2(adding the sign -> result = -2) Exercise  Convert the two’s complement number into decimal 1111 1111 1111 1111 1111 10102 1. Flip the bits: 0000 0000 0000 0000 0000 0101 2. Add 1: 0000 0000 0000 0000 0000 0001 0000 0000 0000 0000 0000 0110 = 610 3. Check the left-most bit of the original negative number: number is 1 and so it’s a -6 Two’s complement: some noteworthy points  the positive half of the numbers, from 0 to 2,147,483,647ten (231-1), are as before  The pattern 1000…0000two represents the most negative number –2,147,483,648ten (-231)  This is followed by a declining set of negative numbers (1000…0001two) down to (1111…1111two)  The number –2,147,483,648ten (-231) has no corresponding positive number 231 Binary addition and subtraction Subtracting 6ten from 7ten is (7-6)=1: 0000 0000 0000 0000 0000 0000 0000 0111two = - 0000 0000 0000 0000 0000 0000 0000 0110two7 ten = 0000 0000 0000 0000 0000 0000 0000 0001two= 6 ten Or via addition using the two’s complement (7 + = (-6)): (this requires very simple hardware!) 1 ten 0000 0000 0000 0000 0000 0000 0000 0111two = 7ten + 1111 1111 1111 1111 1111 1111 1111 1010two = -6ten = 0000 0000 0000 0000 0000 0000 0000 0001two = 1ten Arithmetic in 2’s Complement  Example:  Using two’s complement binary arithmetic, find the sum of 23 and -9.  We see that there is carry into the sign bit and a carry out. The final result is correct: 23 + (-9) = 14. Overflow in 2’s Complement  It may occur but it does not introduce erroneous results  Rule for detecting signed 2’s complement overflow:  When the “carry in” and the “carry out” of the sign bit differ, overflow has occurred  If the carry into the sign bit equals the carry out of the sign bit, no overflow has occurred.  Two positive numbers are added and the result is negative. Two negative numbers are added and the result is positive. Systems Architecture IN1006 Floating point and Non-Numeric Data Representation —Dr H. Asad Floating Point Numbers So far we have represented positive and negative numbers using N bits How do we represent?  Very large numbers: 9,349,398,989,787,762,244,859,087,678  Very small numbers: 0.0000000000000000000000045691  Rational numbers: 2/3  Irrational numbers: 2 Recall Scientific Notation  Computers use a form of scientific notation for floating-point representation  Numbers written in scientific notation have three components: Single precision FP Representation  Floating point numbers are of the form: S  where: S = sign, F = significand field (mantissa), 1  F  2E E = exponent Single precision FP Representation  These must be packed into a word of bits, in general:  For example: 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 s exponent significand 1 bit 8 bits 23 bits  The size of the exponent field determines the range of values that can be represented.  The size of the significand determines the precision of the representation.  Double precision FP Numbers  Bits allocated to the exponent and the significant:  Increasing the bits for the exponent increases the range.  Increasing the bits for the significand increases the accuracy.  Double precision floating point numbers: 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 s exponent significantt 1 bit 11 bits 20 significand bits (continued) Non numerical data: Characters  ASCII and UNICODE  To use ‘1’s and ‘0’s to represent text and characters ASCII and UNICODE Non-numerical data: Bitmaps Summary 1) Binary 2) Positive and Negative Integers 3) Addition and subtraction 4) Floating Point Numbers 5) Hexadecimal 6) Character Data 7) Bitmaps School of Science & Technology City, University of London Northampton Square London EC1V 0HB United Kingdom T: +44 (0)20 7040 5060 E: [email protected] www.city.ac.uk/department Acknowledgement: Prof George Spanoudakis