1. Lexical Analysis
Cheng-Chia Chen

2. Outline The goal and niche of lexical analysis in a compiler
Lexical tokens
Regular expressions (RE)
Use regular expressions in lexical specification
Finite automata (FA)
DFA and NFA
from RE to NFA
from NFA to DFA
from DFA to optimized DFA
Lexical-analyzer generators

3. 1. The goal and niche of lexical analysis
Source
Tokens
(token stream)
(char stream)
Parsing
Interm.
Language
Code
Gen.
Machine
Code
Optimization
Goal of lexical analysis:
breaking the input into individual words or “tokens”

4. Lexical Analysis What do we want to do? Example: if (i == j) else
Z = 0;
else
Z = 1;
The input is just a sequence of characters:
\tif (i == j)\n\t\tz = 0;\n\telse\n\t\tz = 1;
Goal: Partition input string into substrings
And determine the categories (token types) to which the substrings belong

5. 2. Lexical Tokens What’s a token ? Token attributes
Normal token and special tokens
Example of tokens and special tokens.

6. What’s a token?
a sequence of characters that can be treated as a unit in the grammar of a PL. Output of lexical analysis is a stream of tokens
Tokens are partitioned into categories called token types. ex:
In English:
book, students, like, help, strong,… : token
- noun, verb, adjective, … : token type
In a programming language:
student, var34, 345, if, class, “abc” … : token
ID, Integer, IF, WHILE, Whitespace, … : token type
Parser relies on the token type instead of token distinctions to analyze:
var32 and var1 are treated the same,
var32(ID), 32(Integer) and if(IF) are treated differently.

7. Token attributes token type :
category of the token; used by syntax analysis.
ex: identifier, integer, string, if, plus, …
token value :
semantic value used in semantic analysis.
ex: [integer, 26], [string, “26”]
token lexeme (member, text):
textual content of a token
[while, “while”], [identifier, “var23”], [plus, “+”],
[integer, “26”],…
positional information:
file + start/end line/position of the textual content in the source program.

8. Notes on Token attributes
Token types affect syntax analysis
Token values affect semantic analysis
lexeme and positional information affect error handling
Only token type information must be supplied by the lexical analyzer for syntax analysis.
Any program performing lexical analysis is called a scanner (lexer, lexical analyzer).

9. Aspects of Token types
Language view: A token type is the set of all lexemes of all its token instances.
ID = {a, ab, … } – {if, do,…}.
Integer = { 123, 456, …}
IF = {if}, WHILE={while};
STRING={“abc”, “if”, “WHILE”,…}
Pattern (regular expression): a rule defining the language of all instances of a token type.
WHILE: w h i l e
ID: letter (letters | digits )*
ArithOp: + | - | * | /

10. Lexical Analyzer: Implementation
An implementation must do two things:
Recognize substrings corresponding to lexemes of tokens
Determine token attributes
type is necessary
value depends on the type/application,
lexeme/positional information depends on applications (eg: debug or not).

11. Example input lines: \tif (i == j)\n\t\tz = 0;\n\telse\n\t\tz = 1;
Token-lexeme pairs returned by the lexer:
[Whitespace, “\t”]
[if, - ]
[OpenPar, “(“]
[Identifier, “i”]
[Relation, “==“]
[Identifier, “j”] …

12. Normal Tokens and special Tokens
Kinds of tokens
normal tokens: needed for later syntax analysis and must be passed to parser.
special tokens
skipped tokens (or nontoken):
do not contribute to parsing,
discarded by the scanner.
Examples: Whitespace, Comments
why need them ?
Question: What happens if we remove all whitespace and all comments prior to scanning?

13. Lexical Analysis in FORTRAN
FORTRAN rule: Whitespace is insignificant
E.g., VAR1 is the same as VA R1
Footnote: FORTRAN whitespace rule motivated by inaccuracy of punch card operators

14. A terrible design! Example
Consider
DO 5 I = 1,25
DO 5 I = 1.25
The first is DO 5 I = 1 , 25
The second is DO5I = 1.25
Reading left-to-right, cannot tell if DO5I is a variable or DO stmt. until after “,” is reached

15. Lexical Analysis in FORTRAN. Lookahead.
Two important points:
The goal is to partition the string. This is implemented by reading left-to-right, recognizing one token at a time
“Lookahead” may be required to decide where one token ends and the next token begins
Even our simple example has lookahead issues
i vs. if
= vs. ==

16. Some token types of a typical PL
Examples ID
foo n14 last
NUM (literal)
REAL(literal)
e67 1.5e-10 IF if
COMMA ,
NOTEQ !=
LPAREN (
RPAREN )

17. Some Special Tokens 1,5 are skipped. 2,3 need preprocess,
4 need to be expanded.
1. comment
/* … */
// …
2. preprocessor directive
#include <stdio.h>
3. macro definition
#define NUMS 5,6
4. macro use
NUMS
5.blank,tabs,newlines
\t \n

18. 3. Regular expressions and Regular Languages

19. The geography of lexical tokens
ID: var1, last5,…
NUM
23 56
0 000
special tokens : \t \n /* … */
IF:if
LPAREN (
REAL
12.35
2.4 e –10 …
RPAREN )
the set of all strings

20. Issues Definition problem:
how to define (formally specify) the set of strings(tokens) belonging to a token type ?
=> regular expressions
(Recognition problem)
How to determine which set (token type) an input string belongs to?
=> DFA!

21. Languages Def. Let S be a set of symbols (or characters).
A language over S is a set of strings of characters drawn from S
(S is called the alphabet )

22. Examples of Languages Alphabet = English characters
Language = English words
Not every string on English characters is an English word
likes, school,…
beee,yykk,…
Alphabet = ASCII
Language = C programs
Note: ASCII character set is different from English character set

23. Regular Expressions
A language (metaLanguage) for representing (or defining) languages(sets of words)
Definition: If S is an alphabet. The set of regular expression(RegExpr) over S is defined recursively as follows:
(Atomic RegExpr) : 1. any symbol c   is a RegExpr.
2. e (empty string) is a RegExpr.
(Compound RegExpr): if A and B are RegExpr, then so are
3. (A | B) (alternation)
4. (A  B) (concatenation)
5. A* (repetition)

24. Semantics (Meaning) of regular expressions
For each regular expression A, we use L(A) to express the language defined by A.
I.e. L is the function:
L: RegExpr(S)  the set of Languages over S
with
L(A) = the language denoted by RegExpr A
The meaning of RegExpr can be made clear by explicitly defining L.

25. Atomic Regular Expressions
1. Single symbol: c
L(c) = { c } (for any c  )
2. Epsilon (empty string): e
L(e) = {e}

26. Compound Regular Expressions
3. alternation ( or union or choice)
L( (A | B) ) = { s | s  L(A) or s  L(B) }
4. Concatenation: AB (where A and B are reg. exp.)
L((A  B)) =L(A)  L(B)
=def { a  b | a  L(A) and b  L(B) }
Note:
Parentheses enclosing (A|B) and (AB) can be omitted if there is no worries of confusion.
MN (set concatenation) and a  b (string concatenation) will be abbreviated to AB and ab, respectively.
AA and L(A)  L(A) are abbreviated as A2 and L(A)2, respectively.

27. Examples if | then | else  { if, then, else}
0 | 1 | … |  { 0, 1, …, 9 }
(0 | 1) (0 | 1)  { 00, 01, 10, 11 }

28. More Compound Regular Expressions
5. repetition ( or Iteration): A*
L(A*) = { e }  L(A)  L(A)L(A)  L(A)3  …
Examples:
0* : {e, 0, 00, 000, …}
10* : strings starting with 1 and followed by 0’s.
(0|1)* 0 : Binary even numbers.
(a|b)*aa(a|b)*: strings of a’s and b’s containing consecutive a’s.
b*(abb*)*(a|e) : strings of a’s and b’s with no consecutive a’s.

29. Example: Keyword
Keyword: else or if or begin …
else | if | begin | …

30. Example: Integers Integer: a non-empty string of digits
( 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 ) ( 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 )*
problem: reuse complicated expression
improvement: define intermediate reg. expr.
digit = 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9
number = digit digit*
Abbreviation: A+ = A A*

31. Regular Definitions Names for regular expressions d1 = r1 d2 = r2 ...
dn = rn
where ri over alphabet È {d1, d2, ..., d i-1}
note: Recursion is not allowed.

32. Example
Identifier: strings of letters or digits, starting with a letter
digit = 0 | 1 | ... | 9
letter = A | … | Z | a | … | z
identifier = letter (letter | digit) *
Is (letter* | digit*) the same as (letter | digit)* ?

33. Whitespace: a non-empty sequence of blanks, newlines, CRNL and tabs
Example: Whitespace
Whitespace: a non-empty sequence of blanks, newlines, CRNL and tabs
WS = (\ | \t | \n | \r\n )+

34. Example: Email Addresses
Consider
 = letters [ { }
name = letter+
address = name name (‘.’ name)*

35. Notational Shorthands
One or more instances
r+ = r r*
r* = (r+ | e)
Zero or one instance
r? = (r | e)
Character classes
[abc] = a | b | c [a-z] = a | b | ... | z
[ac-f] = a | c | d | e | f = S
[^ac-f] = S – [ac-f] [^.] = e

36. Summary Regular expressions describe many useful languages
Regular definition is a language specification
We still need an implementation
problem: Given a string s and a rexp R, is

37. 4. Use Regular expressions in lexical specification

38. Goal
Specifying lexical structure using regular expressions

39. Regular Expressions in Lexical Specification
Last lecture: the specification of all lexemes in a token type using regular expression.
But we want a specification of all lexemes of all token types in a programming language.
Which may enable us to partition the input into lexemes
We will adapt regular expressions to this goal

40. Regular Expressions => Lexical Spec. (1)
Select a set of token types
Number, Keyword, Identifier, ...
Write a rexp for the lexemes of each token type
Number = digit+
Keyword = if | else | …
Identifier = letter (letter | digit)*
LParen = ‘(‘ …

41. Regular Expressions => Lexical Spec. (2)
Construct R, matching all lexemes for all tokens
R = Keyword | Identifier | Number | …
= R | R | R …
Facts: If s  L(R) then s is a lexeme
Furthermore s  L(Ri) for some “i”
This “i” determines the token type that is reported

42. Regular Expressions => Lexical Spec. (3)
4. Let the current input be x1…xn
(x1 ... xn are symbols in the language alphabet)
For 1  i  n check
x1…xi  L(R) ?
5. It must be the case that there is one i such that
x1…xi  L(Rj) for some j
6. Remove t = x1…xi from input
if t is normal token, then pass it to the parser
// else it is whitespace or comments, just skip it!
7.go to (4)

43. Ambiguities (1) Rule: Pick the longest possible substring
There are ambiguities in the algorithm
How much input is used? What if
x1…xi  L(R) and also
x1…xk  L(R) for some i  k.
Rule: Pick the longest possible substring
The longest match principle !!

44. Ambiguities (2) Rule: use rule listed first (j iff j < k)
Which token is used? What if
x1…xi  L(Rj) and also
x1…xi  L(Rk)
Rule: use rule listed first (j iff j < k)
Earlier rule first!
Example:
R1 = Keyword and R2 = Identifier
“if” matches both.
Treats “if” as a keyword not an identifier

45. Error Handling What if No rule matches any prefix of input ?
Problem: Can’t just get stuck …
Solution:
Write a rule matching all “bad” strings
Put it last
Lexer tools allow the writing of:
R = R1 | ... | Rn | Error
Token Error matches if nothing else matches

46. Summary
Regular expressions provide a concise notation for string patterns
Use in lexical analysis requires small extensions
To resolve ambiguities
To handle errors
Efficient algorithms exist (next)
Require only single pass over the input
Few operations per character (table lookup)

47. 5. Finite Automata Regular expressions = specification
Finite automata = implementation
A finite automaton consists of
An input alphabet 
A finite set of states S
A start state n
A set of accepting states F  S
A set of transitions state input state
If the automata is for recognizing a token type , then this type should be associated with the machine.

48. In state s1 on input “a” go to state s2
Finite Automata
Transition
s1 a s2
Is read
In state s1 on input “a” go to state s2
If end of input (or no transition possible)
If in accepting state => accept
Otherwise => reject

49. Finite Automata State Transition Graphs
The start state
An accepting state T
[ T is the tokenType ] a
A transition

50. A Simple Example
A finite automaton that accepts only “1” 1

51. Another Simple Example
A finite automaton accepting any number of 1’s followed by a single 0
Alphabet: {0,1} 1
accepted input: 1*0

52. And Another Example Alphabet {0,1} What language does this recognize?1 1
accepted inputs: to be answered later!

53. And Another Example Alphabet still { 0, 1 }
The operation of the automaton is not completely defined by the input
On input “11” the automaton could be in either state 1

54. Epsilon Moves Another kind of transition: -moves B
Machine can move from state A to state B without reading input

55. Deterministic and Nondeterministic Automata
Deterministic Finite Automata (DFA)
One transition per input per state
No -moves
Nondeterministic Finite Automata (NFA)
Can have multiple transitions for one input in a given state
Can have -moves
Finite automata can have only a finite number of states.

56. Execution of Finite Automata
A DFA can take only one path through the state graph
Completely determined by input
NFAs can choose
Whether to make -moves
Which of multiple transitions for a single input to take

57. Acceptance of NFAs An NFA can get into multiple states Input:1
Input: 1 1
Rule: NFA accepts if it can get in a final state

58. Acceptance of a Finite Automata
A FA (DFA or NFA) accepts an input string s iff there is some path in the transition diagram from the start state to some final state such that the edge labels along this path spell out s

59. NFA vs. DFA (1)
NFAs and DFAs recognize the same set of languages (regular languages)
DFAs are easier to implement
NFA are easier for specification and
converting lexical specification to FA

60. NFA vs. DFA (2)
For a given language the NFA can be simpler than the DFA 1
NFA 1
DFA
DFA can be exponentially larger than NFA

61. Operations on NFA states
e-closure(s): set of NFA states reachable from NFA state s on e-transitions alone
e-closure(S): set of NFA states reachable from some NFA state s in S on e-transitions alone
move(S, c): set of NFA states to which there is a transition on input symbol c from some NFA state s in S
notes:
e-closure(S) = Us ∈ S e-closure(s);
e-closure(s) = e-closure({s});
e-closure(S) = ?

62. Computing e-closure Input. An NFA and a set of NFA states S.
Output. E = e-closure(S).
begin
push all states in S onto stack; T := S;
while stack is not empty do begin
pop t, the top element, off of stack;
for each state u with an edge from t to u labeled e do
if u is not in T do begin
add u to T; push u onto stack
end
end;
return T
end.

63. Simulating an NFA (for recognizing a token)
Input. An input string ended with eof and an NFA with start state s0 and final states F.
Output. The answer “yes” if accepts, “no” otherwise.
begin
S := e-closure({s0});
c := next_symbol();
while c != eof do begin
S := e-closure(move(S, c));
end;
if S Ç F != Æ then return “yes”
else return “no”
end.

64. Simulating an NFA (for recognizing a sequence of tokens)
Input. An input string ended with eof and an NFA with start state s0 and a set F of final states, each marked by a token type.
Output: a token sequence (possibly ended with error ).

65. L0: length = 0 ; buf = new ArrayList(); type = -1;
S := e-closure({s0}); c := next_symbol();
while( c != eof ) {
S := e-closure(move(S, c)); F1 = S Ç F ;
if (F1 != Æ ) { // c goes to final states!
buf.add(c ); length= buf.size(); typr = minTypeOf(F1); }
if ( S == Æ ) { // cannot make c-transition!!
if (length == 0 ) { output( error-token) ; exit() }
else { output token(type, buf[0:length-1] );
puch back buf[length:-] and c into the input;
goto L0; }
else { pos++; buf.add(c); c := next_symbol(); } }
if(length = buf.length() ) {
if (length > 0) output token(type, buf ) ; }
else { output token(type, buf(0, length-1);
output error-token ; }.

66. Regular Expressions to Finite Automata
High-level sketch
NFA
DFA
Regular
expressions
Optimized DFA
Lexical
Specification
Table-driven
Implementation of DFA

67. Regular Expressions to NFA (1)
For each kind of rexp, define an NFA
Notation: NFA for rexp A A
For  
For input a a

68. Regular Expressions to NFA (2)
For AB A B  x y
min(x,y)
For A | B

69. Regular Expressions to NFA (3)
For A*  A  

70. Example of RegExp -> NFA conversion
Consider the regular expression
(1|0)*1
The NFA is  A H 1 C E  B G 1 I J D F 

71. NFA to DFA

72. Regular Expressions to Finite Automata
High-level sketch
NFA
DFA
Regular
expressions
Optimized DFA
Lexical
Specification
Table-driven
Implementation of DFA

73. RegExp -> NFA : an Examlpe
Consider the regular expression
(1+0)*1
The NFA is  A H 1 C E  B G 1 I J D F 

74. NFA to DFA. The Trick Simulate the NFA Each state of DFA
= a non-empty subset of states of the NFA
Start state
= the set of NFA states reachable through -moves from NFA start states
Add a transition S a S’ to DFA iff
S’ is the set of NFA states reachable from any state in S after seeing the input a
considering -moves as well

75. NFA -> DFA Example  1   1      1 1 1 C E A B G H I J D FG H I J   D F  
FGABCDHI 1
ABCDHI 1 1
EJGABCDHI

76. NFA to DFA. Remark An NFA may be in many states at any time
How many different states ?
If there are N states, the NFA must be in some subset of those N states
How many non-empty subsets are there?
2N - 1 = finitely many

77. From an NFA to a DFA Subset construction Algorithm. Input. An NFA N.
Output. A DFA D with states S and transition table mv.
begin
add e-closure(s0) as an unmarked state to S;
while there is an unmarked state T in S do begin
mark T; let TokenType(T) = min{ type( s) | s ∈ T ∩ F }.
for each input symbol a do begin
U := e-closure(move(T, a));
if U is not in S then
add U as an unmarked state to S;
mv[T, a] := U
end end end.

78. Implementation A DFA can be implemented by a 2D table T
One dimension is “states”
Other dimension is “input symbols”
For every transition Si a Sk define mv[i,a] = k
DFA “execution”
If in state Si and input a, read mv[i,a] (= k) and move to state Sk
Very efficient

79. Table Implementation of a DFAT 1 S 1 1 U 1 S T U

80. Simulation of a DFA
Input. An input string ended with eof and a DFA with start state s0 and final states F.
Output. The answer “yes” if accepts, “no” otherwise.
begin
s := s0;
c := next_symbol();
while c <> eof do begin
s := mv(s, c);
c := next_symbol()
end;
if s is in F then return “yes”
else return “no”
end.

81. Simulation of a DFA (for recognizing token sequence )
Input. An input string ended with eof and a DFA with start state s0 and a set F of final states, each with a type.
Output. a sequence of tokens possibly ended with error token.
begin
length = 0; buf = new ArrayList(); type = -1;
s := s0;
c := next_symbol();
while( c <> eof ) {
s := mv(s, c);
if( s is not error state ) {
if(s is final) { length= buf.size() + 1; type = type(s); }
buf.add(c) ; }

82. if( s is error-state ) {
if(length == 0 ) { output error-token ; exit() }
else { output token(type, buf[0:length-1]) ;
push back buf[length:-] and c into input ;
length = 0; type = -1; } }
c := next_symbol()
if(s is not final ) { output error-token ;}
end.

83. Implementation (Cont.)
NFA -> DFA conversion is at the heart of tools such as flex
But, DFAs can be huge
DFA => optimized DFA : try to decrease the number of states.
not always helpful!
In practice, flex-like tools trade off speed for space in the choice of NFA and DFA representations

84. Time-Space Tradeoffs RE to NFA, simulate NFA
time: O(|r| |x|) , space: O(|r|)
RE to NFA, NFA to DFA, simulate DFA
time: O(|x|), space: O(2|r| )
Lazy transition evaluation
transitions are computed as needed at run time;
computed transitions are stored in cache for later use

85. DFA to optimized DFA

86. Motivations Problems:
1. Given a DFA M with k states, is it possible to find an equivalent DFA M’ (I.e., L(M) = L(M’)) with state number fewer than k ?
2. Given a regular language A, how to find a machine with minimum number of states ?
Ex: A = L((a+b)*aba(a+b)*) can be accepted by the following NFA:
By applying the subset
construction, we can construct
a DFA M2 with 24=16 states,
of which only 6 are accessible from the initial state {s}. s t u v a b
a,b

87. Inaccessible states
A state p  Q is said to be inaccessible (or unreachable) [from the initial state] if there exists no path from from the initial state to it. If a state is not inaccessible, it is accessible.
Inaccessible states can be removed from the DFA without affecting the behavior of the machine.
Problem: Given a DFA (or NFA), how to find all inaccessible states ?

88. Finding all accessible states: (like e-closure)
Input. An FA (DFA or NFA)
Output. the set of all accessible states
begin
push all start states onto stack;
Add all start states into A;
while stack is not empty do begin
pop t, the top element, off of stack;
for each state u with an edge from t to u do
if u is not in A do begin
add u to A; push u onto stack
end
end;
return A end.

89. Minimization process Minimization process for a DFA:
1. Remove all inaccessible states
2. Collapse all equivalent states
What does it mean that two states are equivalent?
both states have the same observable behaviors.i.e.,
there is no way to distinguish their difference, or
more formally, we say p and q are not equivalent(or distinguishable) iff there is a string x S* s.t. exactly one of D(p,x) and D(q,x) is a final state,
where D(p,x) is the ending state of the path from p with x as the input.
Equivalents sates can be merged to form a simpler machine.

90. Example: 1 2 4 3 a
a,b b 5 5
a,b
1,2
3,4

91. Quotient Construction
M=(Q,S, d,s,F): a DFA.
: a relation on Q defined by:
p  q <=>for all x  S* D(p,x)  F iff D(q,x)  F
Property:  is an equivalence relation.
Hence it partitions Q into equivalence classes
[p] = {q  Q | p  q} for p  Q. and the quotient set
Q/ = {[p] | p  Q}.
Every p  Q belongs to exactly one class [p] and
p  q iff [p]=[q].
Define the quotient machine M/ = <Q’,S, d’,s’,F’> where
Q’=Q/ ; s’=[s]; F’={[p] | p  F}; and
d’([p],a)=[d(p,a)] for all p  Q and a  S.

92. Minimization algorithm
input: a DFA
output: a optimized DFA
1. Write down a table of all pairs {p,q}, initially unmarked.
2. mark {p,q} if p ∈ F and q  F or vice versa.
3. Repeat until no more change:
3.1 if ∃ unmarked pair {p,q} s.t. {move(p,a), move(q,a)} is marked for some a ∈ S, then mark {p,q}.
4. When done, p  q iff {p,q} is not marked.
5. merge all equivalent states into one class and return the resulting machine
Note:For recognizing multiple token types, 2 need change to
2’ mark {p,q} if type(p) ≠ type(q) [assume non final state has the same type ]

93. An Example:
The DFA: a b
>0 1 2 1F 3 4 2F 5 5F

94. Initial Table 1 - 2 3 4 5

95. After step 2 1 M 2 - 3 4 5

96. After first pass of step 31 M 2 - 3 4 5

97. 2nd pass of step 3.
The result : 1  2 and 3  4. 1 M 2 - 3 M2 4 5 M1