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## Lexical Analysis

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**Lexical Analysis**• Recognize tokens and ignore white spaces, comments • Error reporting • Model using regular expressions • Recognize using Finite State Automata Generates token stream**Lexical Analysis**• Sentences consist of string of tokens (a syntactic category) for example number, identifier, keyword, string • Sequences of characters in a token is lexeme for example 100.01, counter, const, “How are you?” • Rule of description is pattern for example letter(letter/digit)* • Discard whatever does not contribute to parsing like white spaces (blanks, tabs, newlines) and comments • construct constants: convert numbers to token num and pass number as its attribute for example integer 31 becomes <num, 31> • recognize keyword and identifiers for example counter = counter + incrementbecomes id = id + id /*check if id is a keyword*/**Interface to other phases**Read characters Token • Push back is required due to lookahead for example > = and > • It is implemented through a buffer • Keep input in a buffer • Move pointers over the input Lexical Analyzer Syntax Analyzer Input Push back Extra characters Ask for token**Approaches to implementation**• Use assembly language Most efficient but most difficult to implement • Use high level languages like C Efficient but difficult to implement • Use tools like lex, flex Easy to implement but not as efficient as the first two cases**Construct a lexical analyzer**• Allow white spaces, numbers and arithmetic operators in an expression • Return tokens and attributes to the syntax analyzer • A global variable tokenval is set to the value of the number • Design requires that • A finite set of tokens be defined • Describe strings belonging to each token**#include <stdio.h>**• #include <ctype.h> • int lineno = 1; • int tokenval = NONE; • int lex() { • int t; • while (1) { • t = getchar (); • if (t ==‘ ‘ || t == ‘\t’); • else if (t == ‘\n’) lineno = lineno + 1; • else if (isdigit (t) ) { • tokenval = t – ‘0’ ; • t = getchar (); • while (isdigit(t)) { • tokenval = tokenval * 10 + t – ‘0’ ; • t = getchar(); • } • ungetc(t,stdin); • return num; • } • else { tokenval = NONE; return t; } • } • }**Problems**• Scans text character by character • Look ahead character determines what kind of token to read and when the current token ends • First character cannot determine what kind of token we are going to read**Symbol Table**• Stores information for subsequent phases • Interface to the symbol table • Insert(s,t): save lexeme s and token t and return pointer • Lookup(s): return index of entry for lexeme s or 0 if s is not found Implementation of symbol table • Fixed amount of space to store lexemes. Not advisable as it waste space. • Store lexemes in a separate array. Each lexeme is separated by eos. Symbol table has pointers to lexemes.**Fixed space for lexemes**Other attributes Other attributes …… eos eos lexeme1 lexeme2 lexeme3 Usually 32 bytes Usually 4 bytes**How to handle keywords?**• Consider token DIV and MOD with lexemes div and mod. • Initialize symbol table with insert( “div” , DIV ) and insert( “mod” , MOD). • Any subsequent lookup returns a nonzero value, therefore, cannot be used as identifier.**Difficulties in design of lexical analyzers**• Is it as simple as it sounds? • Lexemes in a fixed position. Fix format vs. free format languages • Handling of blanks • in Pascal blanks separate identifiers • in Fortran blanks are important only in literal strings for example variable counter is same as count er • Another example DO 10 I = 1.25 DO10I=1.25 DO 10 I = 1,25 DO10I=1,25**The first line is variable assignment**DO10I=1.25 • second line is beginning of a Do loop • Reading from left to right one can not distinguish between the two until the “;” or “.” is reached • Fortran white space and fixed format rules came into force due to punch cards and errors in punching**PL/1 Problems**• Keywords are not reserved in PL/1 if then then then = else else else = then if if then then = then + 1 • PL/1 declarations Declare(arg1,arg2,arg3,…….,argn) • Can not tell whether Declare is a keyword or array reference until after “)” • Requires arbitrary lookahead and very large buffers. Worse, the buffers may have to be reloaded.**Problem continues even today!!**• C++ template syntax: Foo<Bar> • C++ stream syntax: cin >> var; • Nested templates: Foo<Bar<Bazz>> • Can these problems be resolved by lexical analyzers alone?**How to specify tokens?**• How to describe tokens 2.e0 20.e-01 2.000 • How to break text into token if (x==0) a = x << 1; iff (x==0) a = x < 1; • How to break input into token efficiently • Tokens may have similar prefixes • Each character should be looked at only once**How to describe tokens?**• Programming language tokens can be described by regular languages • Regular languages • Are easy to understand • There is a well understood and useful theory • They have efficient implementation • Regular languages have been discussed in great detail in the “Theory of Computation” course**Operations on languages**• L U M = {s | s is in L or s is in M} • LM = {st | s is in L and t is in M} • L* = Union of Li such that 0 ≤ i ≤ ∞ Where L0 = є and Li = L i-1 L**Example**• Let L = {a, b, .., z} and D = {0, 1, 2, … 9} then • LUD is set of letters and digits • LD is set of strings consisting of a letter followed by a digit • L* is a set of all strings of letters including є • L(LUD)* is set of all strings of letters and digits beginning with a letter • D+ is the set of strings of one or more digits**Notation**• Let Σ be a set of characters. A language over Σ is a set of strings of characters belonging to Σ • A regular expression r denotes a language L(r) • Rules that define the regular expressions over Σ • Є is a regular expression that denotes {є} the set containing the empty string • If a is a symbol in Σ then a is a regular expression that denotes {a}**If r and s are regular expressions denoting the languages**L(r) and L(s) then • (r)|(s) is a regular expression denoting L(r) U L(s) • (r)(s) is a regular expression denoting L(r)L(s) • (r)* is a regular expression denoting (L(r))* • (r) is a regular expression denoting L(r)**Let Σ = {a, b}**• The regular expression a|b denotes the set {a, b} • The regular expression (a|b)(a|b) denotes {aa, ab, ba, bb} • The regular expression a* denotes the set of all strings {є, a, aa, aaa, …} • The regular expression (a|b)* denotes the set of all strings containing є and all strings of a’s and b’s • The regular expression a|a*b denotes the set containing the string a and all strings consisting of zero or more a’s followed by a b**Precedence and associativity**• *, concatenation, and | are left associative • * has the highest precedence • Concatenation has the second highest precedence • | has the lowest precedence**How to specify tokens**• Regular definitions • Let ri be a regular expression and di be a distinct name • Regular definition is a sequence of definitions of the form d1 r1 d2 r2 ….. dn rn • Where each ri is a regular expression over Σ U {d1, d2, …, di-1}**Examples**• My fax number 91-(512)-259-7586 • Σ = digits U {-, (, ) } • Country digit+ • Area ‘(‘ digit+ ‘)’ • Exchange digit+ • Phone digit+ • Number country ‘-’ area ‘-’ exchange ‘-’ phone digit2 digit3 digit3 digit4**Examples …**• My email address ska@iitk.ac.in • Σ = letter U {@, . } • Letter a| b| …| z| A| B| …| Z • Name letter+ • Address name ‘@’ name ‘.’ name ‘.’ name**Examples …**• Identifier letter a| b| …|z| A| B| …| Z digit 0| 1| …| 9 identifier letter(letter|digit)* • Unsigned number in Pascal digit 0| 1| …|9 digits digit+ fraction ’.’ digits | є exponent (E ( ‘+’ | ‘-’ | є) digits) | є number digits fraction exponent**Regular expressions in specifications**• Regular expressions describe many useful languages • Regular expressions are only specifications; implementation is still required • Given a string s and a regular expression R, does s Є L(R) ? • Solution to this problem is the basis of the lexical analyzers • However, just the yes/no answer is not important • Goal: Partition the input into tokens**Construct R matching all lexemes of all tokens**• R = R1 + R2 + R3 + ….. • Let input be x1…xn • for 1 ≤ i ≤ n check x1…xiЄ L(R) • x1…xiЄ L(R) x1…xiЄ L(Rj) for some j • Write a regular expression for lexemes of each token • number digit+ • identifier letter(letter|digit)+ • smallest such j is token class of x1…xi • Remove x1…xi from input; go to (3)**The algorithm gives priority to tokens listed earlier**• Treats “if” as keyword and not identifier • How much input is used? What if • x1…xiЄ L(R) • x1…xjЄ L(R) • Pick up the longest possible string in L(R) • The principle of “maximal munch” • Regular expressions provide a concise and useful notation for string patterns • Good algorithms require single pass over the input**else**x = 0 elsex = 0 How to break up text • Elsex=0 • Regular expressions alone are not enough • Normally longest match wins • Ties are resolved by prioritizing tokens • Lexical definitions consist of regular definitions, priority rules and maximal munch principle**Finite Automata**• Regular expression are declarative specifications • Finite automata is implementation • A finite automata consists of • An input alphabet belonging to Σ • A set of states S • A set of transitions statei statej • A set of final states F • A start state n • Transition s1 s2 is read: in state s1 on input a go to state s2 • If end of input is reached in a final state then accept • Otherwise, reject input a**a**i j Pictorial notation • A state • A final state • Transition • Transition from state i to state j on input a**How to recognize tokens**• Consider relop < | <= | = | <> | >= | > id letter(letter|digit)* num digit+ (‘.’ digit+)? (E(‘+’|’-’)? digit+)? delim blank | tab | newline ws delim+ • Construct an analyzer that will return <token, attribute> pairs**Transition diagram for relops**token is relop, lexeme is >= > = * token is relop, lexeme is > other * token is relop, lexeme is < < other token is relop, lexeme is <> > token is relop, lexeme is <= = = token is relop, lexeme is = = token is relop, lexeme is >= > token is relop, lexeme is > * other**Transition diagram for identifier**letter * letter other digit Transition diagram for white spaces delim * delim other**Transition diagram for unsigned numbers**digit digit digit . * E + digit digit digit others - E digit digit digit . * digit digit others digit digit * others**The lexeme for a given token must be the longest possible**• Assume input to be 12.34E56 • Starting in the third diagram the accept state will be reached after 12 • Therefore, the matching should always start with the first transition diagram • If failure occurs in one transition diagram then retract the forward pointer to the start state and activate the next diagram • If failure occurs in all diagrams then a lexical error has occurred**Implementation of transition diagrams**Token nexttoken() { while(1) { switch (state) { …… case 10: c=nextchar(); if(isletter(c)) state=10; elseif (isdigit(c)) state=10; else state=11; break; …… } } }**.*** E + digit digit digit others - E digit others others Another transition diagram for unsigned numbers digit digit digit A more complex transition diagram is difficult to implement and may give rise to errors during coding**Lexical analyzer generator**• Input to the generator • List of regular expressions in priority order • Associated actions for each of regular expression (generates kind of token and other book keeping information) • Output of the generator • Program that reads input character stream and breaks that into tokens • Reports lexical errors (unexpected characters)**LEX: A lexical analyzer generator**lex.yy.c C code for Lexical analyzer Token specifications C Compiler LEX Object code Lexical analyzer Input program tokens Refer to LEX User’s Manual**How does LEX work?**• Regular expressions describe the languages that can be recognized by finite automata • Translate each token regular expression into a non deterministic finite automaton (NFA) • Convert the NFA into equivalent DFA • Minimize DFA to reduce number of states • Emit code driven by DFA tables