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Lexical Analysis III Recognizing Tokens Lecture 4 CS 4318/5331 Apan Qasem Texas State University Spring 2015.

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Presentation on theme: "Lexical Analysis III Recognizing Tokens Lecture 4 CS 4318/5331 Apan Qasem Texas State University Spring 2015."— Presentation transcript:

1 Lexical Analysis III Recognizing Tokens Lecture 4 CS 4318/5331 Apan Qasem Texas State University Spring 2015

2 Announcements Assg 1 due this Friday at 11:59 PM Test instances on github No lecture at RRC this week

3 Lexical Analysis int main() { int i for (i = 0; i < MAX; i++) printf(“Hello World”); } Scanner What do we do if we encounter a missing semi-colon? Nothing!

4 Lexical Analysis int main() { int i; for (i = 0; i < MAX; i++) abcprintf(“Hello World”); } Scanner What do we do if we encounter an undefined function name? Nothing!

5 Lexical Analysis int main() { int i; for (i = 0; i < MAX; i++) abcprintf(“Hello World”); } Scanner What do we do if we encounter an undefined function name? Nothing!

6 Lexical Analysis intmain(){inti;for(i=0;i<MAX;i++)printf(“Hello World”);} Scanner Legal C program? Passes Scanner? No Yes

7 Lexical Analysis intmain(){inti;for(i=0;i<MAX;i++)printf(“Hello World”);} Scanner Legal C program? Passes Scanner? No Yes

8 Lexical Analysis int main() { int %$*&i; for (i = 0; i < MAX; i++) printf(“Hello World”); } Scanner What’s an illegal C program at the scanner phase? Very Few! C/C++ has become too large!

9 Breaking Down Lexical Analysis Further … 1.Specify patterns for tokens Look at language description and identify the types of tokens needed for the language usually trivial Use regular expressions to specify a pattern for each token patterns for some tokens are trivial 2.Recognize patterns in the input stream and generate tokens for the parser

10 Recognizing Tokens We can specify the regular expression while for the while keyword in C How do we recognize it if we see it in the input stream? Essentially a pattern-matching algorithm

11 Code for Recognizing while if (nextchar() == ‘w’) if (nextchar() == ‘h’) if (nextchar() == ‘i’) if (nextchar() == ‘l’) if (nextchar() == ‘e’) return KEYWORD_WHILE; else // do something else // do something else // do something else // do something else // do something This approach works for more complex REs as well while (nextchar() == ‘a’ || …) Need to decide what to do for strings like when Need to account for strings like whileabc Need to account for strings like abcwhile Can we generate this code automatically?

12 Code for Recognizing while if (nextchar() == ‘w’) if (nextchar() == ‘h’) if (nextchar() == ‘i’) if (nextchar() == ‘l’) if (nextchar() == ‘e’) return KEYWORD_WHILE; else // do something else // do something else // do something else // do something else // do something Each ‘if clause’ represents a state The state is determined solely based on what we have seen so far in the input stream No need to go back and rescan input At each state we make a decision to move to a new state based on the next input symbol This is exactly the idea behind (deterministic) finite state machines

13 Recognizing Tokens General idea Consume a character from the input stream Based on the value of the character move to a new state If the character just consumed produces a valid token and no more characters to consume then DONE leads to a valid token, move to a valid state produces an invalid token go to error state and finish Repeat above recognizes one token

14 Recognizing Tokens Need to construct a recognizer based on regular expressions A recognizer for a regular expression is a machine that recognizes the language described by the RE Given an input string constructed from the alphabet, the recognizer will Say “yes” if the string is in the language (ACCEPT) Say “no” if the string is not in the language (REJECT) Implications Must produce a yes or no answer on every input Cannot say yes when the string is not in the language (false positives)

15 RE and DFA For every RE there is a recognizer that recognizes the corresponding RL If you build it … it will be recognizable! The recognizers are called deterministic finite automata (DFAs) Kleene’s Theorem (1952)

16 Deterministic Finite Automata Formal mathematical construct Abstract state machines that can recognize regular languages A set of states with transitions defined on each input symbol on every state Formal definition in Text (Section 2.2.1) Convenient to reason about DFAs using state transition diagrams

17 DFA Diagram s0s2s1s3 int E initial state input error state final state error states sometimes implicit only one initial state can have multiple final states i n t

18 Acceptance Criteria for DFAs A DFA accepts a string if and only if the DFA ends up in a final state after consuming all input symbols Implications A DFA built to recognize int will _______ intmain A DFA built to recognize intmain will _______ int reject Easy fix if we want the machine to recognize int AND intmain

19 DFA Example : if s0 s1 if s2

20 DFA Example: int | if s0 s1 if s3 s2 n s4 t

21 DFA for if | int s0s1 i f s3 s4 n s2 i t s5 Non-determinism

22 DFA Example : Integers Σ = {0-9} Digit : 0|1|2|3|… |9 Integer : 0 | (1|2|3|… |9)(Digit)* s0s2 E s1 1-9 0-9 0

23 REs and DFAs every RL has a DFA that recognizes it and every DFA has a corresponding RL there are algorithms that allow us to convert an RE to a DFA and vice versa we can automate scanning! to convert REs to DFAs we need to first look at non-deterministic finite automata (NFA)

24 Non-determinism DFAs do not allow non-determinism Must have a transition defined on every state on every possible input symbol Cannot move to a new state without consuming an input symbol Cannot have multiple transitions on the same input symbol

25 NFA DFAs with transitions To run NFAs, start at the initial state and guess the right transition at each step Always guess correctly If some sequence of correct guesses leads to a final state then accept Sounds dubious But works!

26 NFA for if | int s0s0 s1s1 i f s3s3 s4s4 n s2s2 i t s5s5 NFA, multiple transitions on i in state s 0

27 NFA and DFA Although NFAs allow non-determinism it has been shown that NFAs and DFAs are equivalent! Scott and Rabin (1959) DFAs are just specialized forms of NFAs NFAs and DFAs both recognize the same set of languages Can simulate a DFA with an NFA Can construct corresponding DFAs for any NFA Implication For every RE there is also an NFA Relatively easy to construct an NFA from an RE

28 RE to NFA : Empty String 1. is a regular expression that denotes { }, the set that contains the empty string s0s1

29 RE to NFA : Symbol 2. For each, a is a regular expression denoting {a}, the set containing the string a. s0s1 a

30 RE to NFA : Union 3. r | s is an RE denoting L(r) U L(s) e.g., RE = a | b L(RE) = {a, b} s0s1 b s0s1 a s3 a s2s4 b s5 s0

31 RE to NFA : Concatenation 4. rs is an RE denoting L(r)L(s) e.g., RE = ab L(RE) = {ab} s0s1 b s0s1 a s3s0 ab s2

32 RE to NFA : Closure 5. r* is an RE denoting L(r)* e.g., RE = a* L(RE) = {, a, aa, aaa, aaaa, …} s1s3s0 a s2 s0s1 a

33 RE to NFA The algorithm for converting REs to NFAs is known as Thompson’s construction Repeated application of the five conversion rules! Named after Ken Thompson (1968)

34 Example : NFA for a(b|c)* Work inside parentheses b|c s0s1 c s0s1 b s0 s5

35 Example : NFA for a(b|c)* Work inside parentheses b|c s2s4 c s1s3 b s0 s5 Adjust final states Rename states

36 Example : NFA for a (b|c)* Step 3: * (closure) (b | c)* s1s3 b s2s4 c s5s0s5s0

37 Example : NFA for a (b|c)* Step 3: * (closure) (b | c)* s2s4 b s3s5 c s6s0s7s1

38 Example : NFA for a (b|c)* Step 4: concatenation s4s5 b s6s7 c s8 s1 s9 s3s2 s0 a

39 Cycle of Construction RE Minimized DFA NFA Code Thompson’s Construction Subset Construction Hopcroft’s Algorithm


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