CS308 Compiler Principles

Lexical Analyzer

Fan WuDepartment of Computer Science and Engineering

Shanghai Jiao Tong University

Fall 2012

Compiler Principles

Lexical Analyzer• Lexical Analyzer reads the source program

character by character to produce tokens.– strips out comments and whitespaces– returns a token when the parser asks for– correlates error messages with the source

program

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Compiler Principles

Token• A token is a pair of a token name and an optional

attribute value.– Token name specifies the pattern of the token– Attribute stores the lexeme of the token

• Tokens– Keyword: “begin”, “if”, “else”, …– Identifier: string of letters or digits, starting with a letter– Integer: a non-empty string of digits– Punctuation symbol: “,”, “;”, “(”, “)”, …

• Regular expressions are widely used to specify patterns of the tokens.

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Compiler Principles

Token Example

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Compiler Principles

Terminology of Languages• Alphabet: a finite set of symbols

– ASCII– Unicode

• String: a finite sequence of symbols on an alphabet is the empty string– |s| is the length of string s– Concatenation: xy represents x followed by y – Exponentiation: sn

= s s s .. s ( n times) s0 =

• Language: a set of strings over some fixed alphabet the empty set is a language– The set of well-formed C programs is a language

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Compiler Principles

Operations on Languages• Union: L1 L2 = { s | s L1 or s L2 }

• Concatenation: L1L2 = { s1s2 | s1 L1 and s2 L2 }

• (Kleene) Closure:

• Positive Closure:

0

*

i

iLL

1i

iLL

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Compiler Principles

Example• L1 = {a,b,c,d} L2 = {1,2}

• L1 L2 = {a,b,c,d,1,2}

• L1L2 = {a1,a2,b1,b2,c1,c2,d1,d2}

• L1* = all strings using letters a,b,c,d including

the empty string

• L1+ = all strings using letters a,b,c,d without

the empty string9

Compiler Principles

Regular Expressions• Regular expression is a representation of a

language that can be built from the operators applied to the symbols of some alphabet.

• A regular expression is built up of smaller regular expressions (using defining rules).

• Each regular expression r denotes a language L(r).

• A language denoted by a regular expression is called as a regular set.

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Compiler Principles

Regular Expressions (Rules)Regular expressions over alphabet

Reg. Expr Language it denotes L() = {}a L(a) = {a}(r1) | (r2) L(r1) L(r2)(r1) (r2) L(r1) L(r2)(r)* (L(r))*

(r) L(r)

Extension (r)+

= (r)(r)* (L(r))+

(r)? = (r) | L(r) {} zero or one instance

[a1-an] L(a1|a2|…|an) character class

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Compiler Principles

Regular Expressions (cont.)• We may remove parentheses by using

precedence rules:– * highest– concatenation second highest– | lowest

• (a(b)*)|(c) ab*|c

• Example: = {0,1}– 0|1 => {0,1}– (0|1)(0|1) => {00,01,10,11}– 0*

=> { ,0,00,000,0000,....}– (0|1)*

=> all strings with 0 and 1, including the empty string

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Compiler Principles

Regular Definitions• We can give names to regular expressions, and

use these names as symbols to define other regular expressions.

• A regular definition is a sequence of the definitions of the form:d1 r1 where di is a innovative symbol and

d2 r2 ri is a regular expression over symbols

… in {d1,d2,...,di-1}

dn rn

alphabet previously defined symbols

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Compiler Principles

Regular Definitions Example• Example: Identifiers in Pascal

letter A | B | ... | Z | a | b | ... | z

digit 0 | 1 | ... | 9

id letter (letter | digit ) *

– If we try to write the regular expression representing identifiers without using regular definitions, that regular expression will be complex.

(A|...|Z|a|...|z) ( (A|...|Z|a|...|z) | (0|...|9) ) *

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Compiler Principles

Grammar

Regular Definitions

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Compiler Principles

Transition Diagram• State: represents a condition that could

occur during scanning– start/initial state: – accepting/final state: lexeme found– intermediate state:

• Edge: directs from one state to another, labeled with one or a set of symbols

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Compiler Principles

Transition Diagram for relop

Transition Diagram for ``relop < | > |< = | >= | = | <>’’

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Compiler Principles

Transition-Diagram-Based Lexical Analyzer

Implementation of relop transition diagram18

Compiler Principles

Transition Diagram for Others

A transition diagram for id's and keywords

A transition diagram for unsigned numbers19

Compiler Principles

Practice• Draw the transition diagram for recognizing

the following regular expression

a(a|b)*a

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1 2 3aa

a|b

1 2 3aa

b

b a

Compiler Principles

Finite Automata• A finite automaton is a recognizer that takes a

string, and answers “yes” if the string matches a pattern of a specified language, and “no” otherwise.

• Two kinds:– Nondeterministic finite automaton (NFA)

• no restriction on the labels of their edges– Deterministic finite automaton (DFA)

• exactly one edge with a distinguished symbol goes out of each state

• Both NFA and DFA have the same capability

• We may use NFA or DFA as lexical analyzer

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Compiler Principles

Nondeterministic Finite Automaton (NFA)

• A NFA consists of:– S: a set of states– Σ: a set of input symbols (alphabet)– A transition function: maps state-symbol pairs to sets of

states– s0: a start (initial) state– F: a set of accepting states (final states)

• NFA can be represented by a transition graph• Accepts a string x, if and only if there is a path from

the starting state to one of accepting states such that edge labels along this path spell out x.

• Remarks– The same symbol can label edges from one state to

several different states– An edge may be labeled by ε, the empty string

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Compiler Principles

NFA Example (1)The language recognized by this NFA is (a|b) * a b

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Compiler Principles

NFA Example (2)

NFA accepting aa* |bb*

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Compiler Principles

Implementing an NFAS -closure({s0}) { set all of states can be accessible

from s0 by -transitions }c nextchar()while (c != eof) {

begin S -closure(move(S,c))

c nextcharend

if (SF != ) then { if S contains an accepting state }return “yes”

elsereturn “no”

{ set of all states can be accessible from a state in S by a transition on c}

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Compiler Principles

Deterministic Finite Automaton (DFA)

• A Deterministic Finite Automaton (DFA) is a special form of a NFA.– No state has ε- transition– For each symbol a and state s, there is at

most one a labeled edge leaving s.

The language recognized by this DFA is also (a|b) * a b

start

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Compiler Principles

Implementing a DFAs s0 { start from the initial state }c nextchar { get the next character from the

input string }while (c != eof) do { do until the end of the

string }begin

s move(s,c) { transition function } c nextchar

endif (s in F) then { if s is an accepting state }

return “yes”else

return “no”

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Compiler Principles

NFA vs. DFACompactibility Readability Speed

NFA Good Good Slow

DFA Bad Bad Fast

• DFAs are widely used to build lexical analyzers.

NFA DFAThe language recognized 0*1*

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Compiler Principles

NFA vs. DFACompactibility Readability Speed

NFA Good Good Slow

DFA Bad Bad Fast

• DFAs are widely used to build lexical analyzers.

NFA DFAThe language recognized (a|b) * a b

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Compiler Principles31

(a) 1 2 3 4 5

6 7 8 9

0

0 0 0

0

00

1 1

1

111

1

(b)1 2 3 4 5

a

a aaa

Test Yourself

1) What are the languages presented by the two FAs?

2) For a language only accepting characters from {0,1}, please design a DFA which represents all strings containing three ‘0’s.

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Compiler Principles

Regular Expression NFA• McNaughton-Yamada-Thompson (MYT)

construction– Simple and systematic– Guarantees the resulting NFA will have

exactly one final state, and one start state.– Construction starts from the simplest parts

(alphabet symbols).– For a complex regular expression, sub-

expressions are combined to create its NFA.

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Compiler Principles

MYT Construction• Basic rules: for subexpressions with no

operators– For expression

– For a symbol a in the alphabet

i fstart

i fastart

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Compiler Principles

MYT Construction Cont’d• Inductive rules: for constructing larger

NFAs from the NFAs of subexpressions(Let N(r1) and N(r2) denote NFAs for regular

expressions r1 and r2, respectively)

– For regular expression r1 | r2

i

N(r1)

N(r2)

f

start

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Compiler Principles

MYT Construction Cont’d– For regular expression r1r2

– For regular expression r*

i N(r1) f N(r2)start

N(r)i f

start

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Compiler Principles36

Example: (a|b)*a

a:a

bb:

(a|b):a

b

b

a

(a|b)*:

b

a

a(a|b)*a:

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Compiler Principles

Properties of the Constructed NFA1. N(r) has at most twice as many states as there

are operators and operands in r. – This bound follows from the fact that each step of

the algorithm creates at most two new states.

2. N(r) has one start state and one accepting state. The accepting state has no outgoing transitions, and the start state has no incoming transitions.

3. Each state of N(r) other than the accepting state has either one outgoing transition on a symbol in or two outgoing transitions, both on .

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Compiler Principles

Conversion of an NFA to a DFA• Approach: Subset Construction

– each state of the constructed DFA corresponds to a set of NFA states

• Details①Create transition table Dtran for the DFA②Insert -closure(s0) to Dstates as initial state③Pick a not visited state T in Dstates④For each input symbol a, Create state -closure(move(T, a)), and add it to Dstates and

Dtran⑤Repeat step (3) and (4) until all states in Dstates

are vistited

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Compiler Principles

The Subset Construction

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Compiler Principles

NFA to DFA ExampleNFA for (a|b) * abb

Transition table for DFA Equivalent DFA

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Compiler Principles

Regular Expression DFA• First, augment the given regular expression by

concatenating a special symbol #

r r# augmented regular expression

• Second, create a syntax tree for the augmented regular expression. – All leaves are alphabet symbols (plus # and the

empty string)– All inner nodes are operators

• Third, number each alphabet symbol (plus #) (position numbers)

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Compiler Principles44

Regular Expression DFA Cont’d(a|b)*a (a|b)*a# augmented regular expression

*

|

b

a

#

a1

4

3

2

• each symbol is at a leaf• each symbol is numbered (positions)• inner nodes are operators

Syntax tree of (a|b)*a#

3 F

2

1

b

a

a4

#

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Compiler Principles45

followposThen we define the function followpos for the positions (positions assigned to leaves).

followpos(i) -- the set of positions which can follow the position i in the strings generated by the augmented regular expression.

Example: ( a | b) * a #1 2 3 4

followpos(1) = {1,2,3}followpos(2) = {1,2,3}followpos(3) = {4}followpos(4) = {}

followpos() is just defined for leaves,not defined for inner nodes.

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Compiler Principles

firstpos, lastpos, nullable• To compute followpos, we need three more

functions defined for the nodes (not just for leaves) of the syntax tree.– firstpos(n) -- the set of the positions of the first

symbols of strings generated by the sub-expression rooted by n.

– lastpos(n) -- the set of the positions of the last symbols of strings generated by the sub-expression rooted by n.

– nullable(n) -- true if the empty string is a member of strings generated by the sub-expression rooted by n; false otherwise

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Compiler Principles

Usage of the Functions

*

|

b

a

#

a1

4

3

2

(a|b)*a (a|b)*a# augmented regular expression

Syntax tree of (a|b)*a#

n

m

nullable(n) = falsenullable(m) = true

firstpos(n) = {1, 2, 3}

lastpos(n) = {3}

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Compiler Principles48

Computing nullable, firstpos, lastpos

n nullable(n) firstpos(n) lastpos(n)leaf labeled true

leaf labeled with position i

false {i} {i}

|

c1 c2

nullable(c1) or nullable(c2)

firstpos(c1) firstpos(c2) lastpos(c1) lastpos(c2)

c1 c2

nullable(c1) and nullable(c2)

if (nullable(c1))

firstpos(c1)firstpos(c2)

else firstpos(c1)

if (nullable(c2))

lastpos(c1)lastpos(c2)

else lastpos(c2)

*

c1

true firstpos(c1) lastpos(c1)

Compiler Principles

How to evaluate followpos• Two-rules define the function followpos:

1. If n is concatenation-node with left child c1 and right child c2, and i is a position in lastpos(c1), then all positions in firstpos(c2) are in followpos(i).

2. If n is a star-node, and i is a position in lastpos(n), then all positions in firstpos(n) are in followpos(i).

• If firstpos and lastpos have been computed for each node, followpos of each position can be computed by making one depth-first traversal of the syntax tree.

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Compiler Principles50

Example -- ( a | b) * a #

*

|

b

a

#

a1

4

3

2{1}{1}

{1,2,3}

{3}

{1,2,3}

{1,2}

{1,2}

{2}

{4}

{4}

{4}{3}

{3}{1,2}

{1,2}

{2}

red – firstposblue – lastpos

Then we can calculate followpos

followpos(1) = {1,2,3}followpos(2) = {1,2,3}followpos(3) = {4}followpos(4) = {}

• After we calculate follow positions, we are ready to create DFA for the regular expression.

Compiler Principles

Algorithm (RE DFA)1. Create the syntax tree of (r) #2. Calculate nullable, firstpos, lastpos, followpos3. Put firstpos(root) into the states of DFA as an unmarked state.4. while (there is an unmarked state S in the states of DFA) do

– mark S– for each input symbol a do

• let s1,...,sn are positions in S and symbols in those positions are a• S’ followpos(s1) ... followpos(sn)• Dtran[S,a] S’• if (S’ is not in the states of DFA)

– put S’ into the states of DFA as an unmarked state.

• the start state of DFA is firstpos(root)• the accepting states of DFA are all states containing the position of #

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Compiler Principles

Example -- ( a | b) * a #

followpos(1)={1,2,3} followpos(2)={1,2,3} followpos(3)={4} followpos(4)={}

S1=firstpos(root)={1,2,3}

mark S1

a: followpos(1) followpos(3)={1,2,3,4}=S2 Dtran[S1,a]=S2

b: followpos(2)={1,2,3}=S1 Dtran[S1,b]=S1

mark S2

a: followpos(1) followpos(3)={1,2,3,4}=S2 Dtran[S2,a]=S2

b: followpos(2)={1,2,3}=S1 Dtran[S2,b]=S1

start state: S1

accepting states: {S2}

1 2 3 4

S1 S2

ab

b

a

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Compiler Principles53

Example -- ( a | ) b c* # 1 2 3 4

followpos(1)={2} followpos(2)={3,4} followpos(3)={3,4} followpos(4)={}

S1=firstpos(root)={1,2}

mark S1

a: followpos(1)={2}=S2 Dtran[S1,a]=S2

b: followpos(2)={3,4}=S3 Dtran[S1,b]=S3

mark S2

b: followpos(2)={3,4}=S3 Dtran[S2,b]=S3

mark S3

c: followpos(3)={3,4}=S3 Dtran[S3,c]=S3

start state: S1

accepting states: {S3}

S3

S2

S1

c

ab

b

Compiler Principles

Minimizing Number of DFA States• For any regular language, there is always a unique

minimum state DFA, which can be constructed from any DFA of the language.

• Algorithm:– Partition the set of states into two groups:

• G1 : set of accepting states• G2 : set of non-accepting states

– For each new group G• partition G into subgroups such that states s1 and s2 are in the

same group ifffor all input symbols a, states s1 and s2 have transitions to states in the same group.

– Start state of the minimized DFA is the group containing the start state of the original DFA.

– Accepting states of the minimized DFA are the groups containing the accepting states of the original DFA.

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Compiler Principles55

Minimizing DFA – Example (1)

b a

a

a

b

b

3

2

1

G1 = {2}G2 = {1,3}

G2 cannot be partitioned becauseDtran[1,a]=2 Dtran[1,b]=3Dtran[3,a]=2 Dtran[3,b]=3

So, the minimized DFA (with minimum states) is

1 2

a

a

b

b

Compiler Principles56

Minimizing DFA – Example (2)

Groups: {1,2,3} {4}

a b 1->2 1->32->2 2->33->4 3->3

{1,2} {3}no more partitioning

Minimized DFA

b

b

b

a

a

a

a

b 4

3

2

1

3

1

2b

a

a

a

b

b

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Compiler Principles57

Architecture of A Lexical Analyzer

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Compiler Principles

An NFA for Lex program

• Create an NFA for each regular expression

• Combine all the NFAs into one

• Introduce a new start state• Connect it with ε-transitions to the start states of the NFAs

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Compiler Principles

Pattern Matching with NFA① The lexical analyzer read

in input calculates the set of states it is in at each symbol.

② Eventually, it reach a point with no next state.

③ It looks backwards in the sequence of sets of states, until it finds a set including one or more accepting states.

④ It picks the one associated with the earliest pattern in the list from the Lex program.

⑤ It performs the associated action of the pattern.

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Compiler Principles

Pattern Matching with NFA -- Example

Input: aaba

Report pattern: a*b+

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Compiler Principles

Pattern Matching with DFA① Convert the NFA for all the

patterns into an equivalent DFA. For each DFA state with more than one accepting NFA states, choose the pattern, who is defined earliest, the output of the DFA state.

② Simulate the DFA until there is no next state.

③ Trace back to the nearest accepting DFA state, and perform the associated action.

Input: abba

0137 247 58 68

Report pattern abb

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Compiler Principles

Summary• How lexical analysers work

– Convert REs to NFA

– Convert NFA to DFA

– Minimise DFA

– Use the minimised DFA to recognise tokens in the input

– Use priorities, longest matching rule

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Compiler Principles

Homework• Exercise 3.7.1 (c)

• Exercise 3.7.3 (c)

• Exercise 3.9.4 (a)

• Due date: Sept. 29, 2012 (Saturday)

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