Wednesday, September 6, 2017

Transcription

1 Wednesday, September 6, 2017 Topics for today Arithmetic operations and status bits Logical operators Introduction to bigger bases Encoding characters Coding in general Status bits We saw last time that the CPU gives us simple feedback on an operation it performs. This is true both for arithmetic operations and logical operations (see later). Input 1 Input 2 Operation Select N Z V C Result N - the result was negative Z - the result was zero C - there was a carry out from the most significant stage V - there was overflow (=error) Comp 162 Notes Page 1 of 13 September 6, 2017

2 Here are example 5-bit additions showing the operands, result and the values of each of the 4 status bits Operand Operand Result (1)00000 (1)11110 (1) N Z V C We will see later (chapter 6) how we can use these status bits to make flow-of-control decisions inside a program. Logical operations (section 3.3) In general a CPU will have an "Arithmetic and Logic Unit" (ALU). Arithmetic operations: addition, subtraction (maybe multiplication and division) Logical operations: AND, OR, XOR and shift. Logical operations are fast compared with arithmetic operations because they are typically bitwise operations in which the bit positions of a result can be computed independently in parallel. In contrast, an ADD operation may have to wait for a carry to propagate the length of the operands (consider ). Truth tables (using 0 for false and 1 for true) A B A and B A or B A xor B The "Exclusive-OR" (XOR) is so named because it excludes from being true the case where both inputs are true. Examples AND OR XOR Comp 162 Notes Page 2 of 13 September 6, 2017

3 Q. How many other functions of two variables are there (not all useful)? A. 13. If there are N variables, there are 2 N rows in the truth table. A single column can thus be filled in 2^2 N ways each representing a function. With N=2 above there are 2 4 = 16 functions of which we show 3. Here are a couple of applications of XOR (1) Exchanging values We can exchange the values in two variables without using a third. For example, if the variables are A and B, the sequence of operations is A = A xor B B = A xor B A = A xor B Trace A B Initial A = A xor B B = A xor B A = A xor B (2) Encryption To encrypt a message, we XOR it with a key of the same length. To decrypt it we XOR it with the same key see following example Example Message Key Encrypted message Encrypted message Key Message Comp 162 Notes Page 3 of 13 September 6, 2017

4 Without the key, the encrypted message is unbreakable 1. Using a particular key only once yields a strong system particularly if the bits in the key are truly random. Some combinations of logical and math operations Here are some interesting results of combining arithmetic with logical operations (1) x and (x-1) : result is x with rightmost 1 removed Example (84) and (83) rightmost 1 of 84 goes away (2) x and (-x) result is rightmost 1 of x ( = highest power of 2 that divides x) 2 Example (84) and (-84) result: the rightmost 1 of 84 4 is the highest power of 2 that divides 84 (3) x or (-x) smear rightmost 1 of x to the left Example (84) or (-84) the rightmost 1 of 84 copied left (4) x exor (-x) remove and smear rightmost 1 to left Example (84) exor (-84) removed and copied left 1 See story about carrier pigeon at It is thought that a one-time pad might have been used. Without the correct page, the message is undecipherable. 2 In Pep/9 assembly code this operation requires only three instructions: LDWA X,d ; Register A contains X NEGA ; Register A contains -X ANDA X,d ; Register A contains (X logicaland X) Comp 162 Notes Page 4 of 13 September 6, 2017

5 (5) x or (x-1) smear rightmost 1 to right Example (84) or (83) the rightmost 1 smeared right (6) x exor (x-1) extract then smear rightmost 1 to right Example (84) exor (83) the rightmost 1 smeared right Shift operations Shift operations are also logical operations (though arithmetic shift can be also be regarded as an arithmetic operation). A typical shift operation takes a single operand and moves all the bits one place left or right. Suppose we start with a 6-bit object containing a b c d e f The following table shows the results of six different types of shift operations. Type of Shift Left Right Arithmetic b c d e f 0 a a b c d e Logical b c d e f 0 0 a b c d e Rotate/Circular b c d e f a f a b c d e Arithmetic: Shift left is equivalent to multiplication by 2 (unless it overflows). Shift right is equivalent to truncated division by 2. For example, rightshift(7) = 3. Note the sign extension the sign bit does not change so if we shift a negative number rightwards, it is still negative. Comp 162 Notes Page 5 of 13 September 6, 2017

6 Pep/9 has arithmetic shift instructions which we can use to implement the following pseudocode algorithm that computes X=A*B for A >=0. X=0 while (A 0) { if (A%2=1) X=X+B A /= 2 B *= 2; } A%2=1 logicaland(a,1) and test result A /= 2 rightshift(a) B *= 2 leftshift(b) Logical: Both left and right shifts brings in zero bits at the appropriate end Rotate/Circular: May or may not involve the carry bit depending on implementation. If not, a bit that falls off one end is introduced at the other end. In the Pep/9 implementation of Rotate, the Carry bit is in the loop: C bit... Packing and unpacking data Logical operations used in combination are useful in packing and unpacking data like putting several items into a lunchbox. For example, we can pack several independent data items into a 16- bit word. The following shows how five items of student data can be stored in this way Field Size (bits) Depicted Values Student type 1 a 0= undergraduate, 1=graduate Level 2 bb If undergraduate: freshman, senior If graduate: 2ndBS/MS/MBA/PhD Major 5 ccccc Encode up to 32 majors GPA 3 ddd 8 GPA bands: , Enrolled units 5 eeeee Up to 31 units Comp 162 Notes Page 6 of 13 September 6, 2017

7 Thus represents a junior undergraduate Major = 3 (History perhaps) GPA in range 3.00 to 3.49 enrolled in 15 units. Packing the data Suppose the five values to be stored are in appropriately named variables. We can build up the packed data representation in a variable (R) with a sequence of shift and add operations. Initialize R R = R = leftshift(1) + undergrad (0) R = R = leftshift(2) + junior (10) R = R = leftshift(5)+ major (00011) R = R = leftshift(3)+gpa (110) R = R = leftshift(5)+units (01111) R = Unpacking the data Suppose we have student data packed into a variable S and we wish to extract the GPA field (ddd). abbcccccdddeeeee we can accomplish that by first using an AND operation to clear all the other bits to zero then using a shift operation to move the ddd field down to the right end If the data is in S. Then S = S AND ' ' S = rightshift(5) S is now ddd00000 S is now ddd Bigger bases (section 3.4) Binary numbers are tedious to write and it is easy to make mistakes. Look at the binary constant in the AND operation above; it is easy to get the number of zeros wrong. Using a bigger base will result in shorter numbers but we have seen that conversions between binary and decimal can be time-consuming. Comp 162 Notes Page 7 of 13 September 6, 2017

8 Bases 8 and 16 are often more convenient than decimal because each is about as compact as decimal and conversion to and from binary involves simply mapping a group of bits onto a single digit. In octal the groups are 3 bits and in hexadecimal the groups have 4 bits. Octal (Base 8, digits 0..7) The constant that we wrote as would be = in octal. We group digits in threes outwards from where the binary point would be. So a fractional binary number Would be (0) = in octal. (add an extra zero to make a group of three) The problem with octal is that the size of data objects in a computer is typically a multiple of 4 not 3. For example, we have 8-bit bytes, 32-bit and 64-bit registers and so on. This leads us to prefer hexadecimal Hexadecimal (Base 16, use 0..9 and A..F) We group bits in fours so there are 16 possible combinations. In addition to the decimal digits we use the first 6 letters of the alphabet (it usually doesn t matter whether we use upper or lower case). The main problem with hexadecimal is remembering which letter represents which bit pattern is A 1011 is B 1100 is C 1101 is D 1110 is E 1111 is F To convert binary to hexadecimal, group digits in fours outwards from where the binary point would be. For example = = 00E0 in hex = = A35E As in our octal example, if we have a fraction we may have to pad out the fraction bits Comp 162 Notes Page 8 of 13 September 6, 2017

9 = (0) = DC.CA If the number of bits to the left of the point is not a multiple of 4, we pad them also but need to know if the number is signed or unsigned. For instance if signed (pad out with sign bit) = ED if unsigned (pad out with zero) = 2D.1 When a programming language permits both decimal and hexadecimal constants it usually tags hex constants in some way. For example, the number 1234 (base 16) might appear in a program as 0x1234 or #1234h Pep/9 has a instruction (HEXO) that outputs an integer in hexadecimal but no corresponding input instruction. We know that integers can be represented in many ways. If we see a byte and we know it represents an integer, we would need to know the encoding method used in order to know what number it represents. The following table shows the value of in various mappings. Number system Value unsigned s complement s complement -67 Sign and magnitude -61 Excess magnitude 67 If we don t even know it is an integer, there are other things it might be. Characters The ASCII coding scheme is shown in Figure It shows how 128 characters (some printable characters, some control characters) are mapped onto 7-bit codewords. Note that the letters are assigned to consecutive codes and the digits are assigned to consecutive codes. This makes sorting of alphanumeric data much simpler. Comp 162 Notes Page 9 of 13 September 6, 2017

10 Also note that the codewords for an upper case letter and the corresponding lower case letter differ in only one bit position, e.g. H = h = ^ Case bit These features of the ASCII code makes it possible to write: * a simple check to see if a letter is upper case or lower case * instructions to change the case of a letter * case-insensitive matching ( so artichoke = ArtiCHoKe ). Unicode ASCII and similar 7-bit and 8-bit codes are being superceded by Unicode ( which is a 16-bit code. The space requirements are thus twice as big but we can encode a much greater variety of alphabets (Chinese, Korean, Cyrillic, Japanese, Arabic etc). Here are some examples پ Ж א ĝ There are many technical issues to be worked out when dealing with different languages. An example is determining how strings are to be sorted what is the collating order of characters? In Swedish z < ö whereas in German ö < z Coding in general If you need to encode an item (e.g., day of week, student s major, color of a car) how many bits do you need? It the item has M possible different values you would need ceiling(log2 (M)) bits. So Day of week = 3 bits Day of month = 5 bits Month in year = 4 bits and so on. Comp 162 Notes Page 10 of 13 September 6, 2017

11 Generally we don't need to know the exact value of log2(m), just be familiar with powers of 2. For example if M is 200, we know that 2 7 is 128 and 2 8 is 256 so we would need 8 bits to represent values. Example. How could we encode the time of day (to the nearest second)? Method 1: Method 2: We could represent it by the number of seconds after midnight. This would be a number in the range *60*60 = This is a number between 2 16 (65536) and 2 17 (131072) so we would need 17 bits. We could represent the time by a 3-field object (hours-minutes-seconds). Hours (0..23) would need 5 bits, minutes (0..60) would need 6 bits and seconds (0..60) would need 6 bits. Total 17 bits same as Method 1. (This might suggest that the information content of the time to the nearest second is 17 bits.) The different representations, even though same size, have different advantages/ disadvantages. For example, using Method 1 to encode times, it is easy to find the number of seconds between two times. Using Method 2 it is easier to output a time in the HH:MM:SS format. Similarly to represent a date within a year requires 9 bits either as days-since-jan-1st or as Month (4 bits):day (5 bits). Alternatives are not always the same size. Consider the following variation on a problem in the text book Q: You own a (small) paint store and wish to attach a binary code to each paint can. How many bits does each label need if There are three sizes: 5 gallon, 1 gallon, half-gallon There are five colors: black, white, red, green, blue There are two types of paint: indoor, outdoor? A: If we code the fields separately we need 6 bits (2 for size, 3 for color, 1 for type). However, there are only 30 combinations so we could devise a 5-bit code Reading: Chapter 3 except for Sections 3.5 and 3.6. Comp 162 Notes Page 11 of 13 September 6, 2017

12 Review Questions 1. What do the 2 s complement binary representations of N and N have in common? For example, look at 26 and -26 and at 16 and What is the result of (a) exor-ing a number with itself, (b) exor-ing a number with zero (c) exor-ing a number with -1 (all 1s)? 3. Express the answers to the following 16-bit operations as hexadecimal numbers (a) 0x1234 AND 0x1277 (b) 0x7742 OR 0x1003 (c) 0x1750 EXOR 0xABCD 4. A student, given an 8-bit number to convert to octal, grouped the bits in threes from the lefthand end instead of the right-hand end and came up with 513 (a) What was the binary number? (b) What is the correct octal? 5. A student, given an 11-bit number to convert to hexadecimal, grouped the bits in fours from the left-hand end instead of the right-hand end and came up with AF1 (a) What was the binary number? (b) What is the correct octal? 6. Suppose byte X contains the ASCII code of a letter. What logical operation will isolate the case bit. 7. The Unix designers chose 32-bit signed binary integers to represent time. Suppose instead they had used the same number of bits to code the following fields separately Month, day, hour, minute, seconds, am/pm How many bits could be used for the year? When will this encoding fail? Comp 162 Notes Page 12 of 13 September 6, 2017

13 Review Answers 1. From right to left they are identical up to and including the first 1 digit 2. (a) zero (b) the original number (c) the ones complement of the number 3. (a) 0x1234 (b) 0x7743 (c) 0xBC9D 4. (a) (b) (a) (b) AND with (2016). Result will be if upper case and if lower case. 7. The number of bits to encode the named fields separately is 26 as follows: Month Day Hour Minutes Seconds AM/PM This leaves 6 bits for the year, so the largest year possible is = Comp 162 Notes Page 13 of 13 September 6, 2017