# Introduction to Parsing. Lecture 5

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1 Introduction to Parsing Lecture 5 1

2 Outline Regular languages revisited Parser overview Context-free grammars (CFG s) Derivations Ambiguity 2

3 Languages and Automata Formal languages are very important in CS specially in programming languages Regular languages The weakest formal languages widely used Many applications (as we ve seen) We will also study context-free languages, tree languages 3

4 Beyond Regular Languages Difficulty with regular languages is that many languages are not regular Some are very important They can t be expressed using Rs and FAs x. Strings of balanced parentheses are not regular: Note this is fairly representative of lots of programming constructs {() i i i 0} Note: given as set not R 4

5 Beyond Regular Languages x. Nested arithmetic expressions ((1+2) * 3) x. Nested if then else statements if if if fi fi then fi then then if here acts like ( in previous example Note that even if language doesn t have the fi like Cool, it is usually implied 5

6 An xample To Help Understand the Limitations Consider the following DFA 1 0 x: What does it recognize? 0 Note: doesn t have any way of knowing length of input string 6

7 Beyond Regular Languages In general: Nesting constructs cannot be handled by regular expressions Raises the questions: What can be expressed? Why are Rs insufficient for recognizing arbitrary nesting constructs? 7

8 What Can Regular Languages xpress? Languages requiring counting modulo a fixed integer.g., parity Intuition: A finite automaton that runs long enough must repeat states Finite automaton can t remember # of times it has visited a particular state 8

9 The Functionality of the Parser Input: sequence of tokens from lexer Output: parse tree of the program (But some parsers never produce a parse tree...) 9

10 xample Cool if x = y then 1 else 2 fi Parser input (from lexical analyzer) IF ID = ID THN INT LS INT FI Parser output IF-THN-LS = INT INT ID ID 10

11 xample Note: nesting structure has been made explicit by tree Also the three components of the if then else Predicate Then branch lse branch IF-THN-LS = INT INT ID ID 11

12 Comparison with Lexical Analysis Phase Input Output Lexer Parser String of characters String of tokens String of tokens Parse tree 12

13 Couple of things As mentioned, sometimes parse tree is only implicit More on this later Many compilers do build full parse tree, many do not There are compilers that combine lexer and parser phases into one phase verything done by the parser Parsing technology powerful enough to express lexical analysis in addition to parsing But most compilers use two phases, because Rs are such a good match for lexical analysis 13

14 The Role of the Parser Not all strings of tokens are programs parser must distinguish between valid and invalid strings of tokens And give error messages for the invalid ones We need A language for describing valid strings of tokens An algorithm for distinguishing valid from invalid strings of tokens 14

15 Context-Free Grammars Programming language constructs have recursive structure An XPR in Cool can be if XPR then XPR else XPR fi while XPR loop XPR pool Note: Recursively composed of other expressions Context-free grammars are a natural notation for this recursive structure 15

16 What is a Context-Free Grammar (CFG)? A CFG consists of A set of terminals T A set of non-terminals N A start symbol S (a non-terminal) A set of productions X Y 1 Y 2!Y n where X N and Y T N { ε} i 16

17 Notational Conventions In these lecture notes Non-terminals are written upper-case Terminals are written lower-case The start symbol is the left-hand side of the first production This is standard for CFGs 17

18 xamples of CFGs S ( S ) S ε 18

19 xamples of CFGs S ( S ) S ε What are the parts of the grammar: N =? T =? Start =? 19

20 xamples of CFGs S ( S ) S ε What are the parts of the grammar: N = { S } T = { (, ) } Start = S (the only nonterminal) 20

21 xamples of CFGs S ( S ) S ε What are the parts of the grammar: N = { S } T = { (, ) } Start = S (the only nonterminal) Productions? 21

22 xamples of CFGs A fragment of Cool: XPR if XPR then XPR else XPR fi while XPR loop XPR pool id 22

23 xamples of CFGs (cont.) Simple arithmetic expressions: + ( ) id 23

24 The Language of a CFG Read productions as rules: X Y 1!Y n Means X can be replaced by Y 1!Y n That is, in general, the right hand side can replace the left hand side. 24

25 Key Idea 1. Begin with a string consisting of the start symbol S 2. Replace any non-terminal X in the string by a the right-hand side of some production X Y 1!Y n 3. Repeat (2) until there are no non-terminals in the string So note, the string is changing over time 25

26 The Language of a CFG (Cont.) More formally, write X 1! X i! X n X 1! X i 1 Y 1!Y m X i+1! X n if there is a production X i Y 1!Y m and say that the left hand side derives the right, or can derive the right hand side, etc. This is one step of a context-free derivation. 26

27 The Language of a CFG (Cont.) Write if X 1! X n Y 1!Y m in 0 or more steps X 1! X n!! Y 1!Y m We say the left hand side rewrites in zero or more steps to the right hand side 27

28 So, in general When we write X 0 X n it is shorthand for saying that there is some sequence of individual productions (rules) that get us from X 0 to X n in zero or more steps 28

29 The Language of a CFG Let G be a context-free grammar with start symbol S. Then the language, L(G), of G is: # & \$ a 1 a n S a 1 a n and every a i is a terminal' % ( 29

30 Terminals Terminals are so-called because there are no rules for replacing them Once generated, terminals are permanent feature of the string Terminals ought to be tokens of the language 30

31 Recall earlier xample L(G) is the language of CFG G Strings of balanced parentheses {() i i i 0} Two grammars: S S ( S) ε OR S ( S) ε 31

32 Cool xample A fragment of COOL: XPR if XPR then XPR else XPR fi while XPR loop XPR pool id Recall: Non-terminals are written upper-case Terminals are written lower-case Also, could have written as three productions 32

33 Cool xample (Cont.) Some elements of the language (why?) id if id then id else id fi while id loop id pool if while id loop id pool then id else id if if id then id else id fi then id else id fi 33

34 Arithmetic xample Simple arithmetic expressions: + () id Some elements of the language: id id + id (id) id id (id) id id (id) 34

35 Notes The idea of a CFG is a big step. But: Membership in a language is yes or no ; also need parse tree of the input Must handle errors gracefully Need an implementation of CFG s (e.g., bison) 35

36 More Notes Form of the grammar is important Many grammars generate the same language Tools are sensitive to the grammar Note: Tools for regular languages (e.g., flex) are sensitive to the form of the regular expression, but this is rarely a problem in practice 36

37 Derivations and Parse Trees A derivation is a sequence of productions S!!! A derivation can be drawn as a tree Start symbol is the tree s root For a production add children to node X X Y 1!Y n X Y 1!Y n Y 1 Y n 37

38 Derivation xample Grammar + () id String id id + id We wish to parse the string 38

39 Derivation xample (Cont.) + + id id id + id + id + id + * id id id parse tree (of the input string) 39

40 Derivation in Detail (1) 40

41 Derivation in Detail (2)

42 Derivation in Detail (3) * 42

43 Derivation in Detail (4) * id + id 43

44 Derivation in Detail (5) id + * id id + id id 44

45 Derivation in Detail (6) id + * id id id id + id + id id id 45

46 Some Interesting Things About Parse Trees A parse tree has Terminals at the leaves Non-terminals at the interior nodes An in-order traversal of the leaves is the original input Let s go back and take a look The parse tree shows the association of operations, the input string does not Note * binds more tightly than + because * is a subtree of the parse tree 46

47 An Interesting Question How did I know to pick this particular parse tree for the derivation? It turns out that there is more than one 47

48 Left-most and Right-most Derivations The example we did is a left-most derivation At each step, replace the left-most non-terminal There is an equivalent notion of a right-most derivation + + id + id id + id id + id 48

49 Left-most and Right-most Derivations The example we did is a left-most derivation At each step, replace the left-most non-terminal There is an equivalent notion of a right-most derivation + +id + id id + id id id + id 49

50 Right-most Derivation in Detail (1) 50

51 Right-most Derivation in Detail (2)

52 Right-most Derivation in Detail (3) id id 52

53 Right-most Derivation in Detail (4) + + +id * id + id 53

54 Right-most Derivation in Detail (5) + + +id + id * id id + id id 54

55 Right-most Derivation in Detail (6) + +id + + id * id id + id id id + id id id 55

56 Derivations and Parse Trees Note that right-most and left-most derivations have the same parse tree In this case And this is not an accident The difference is the order in which branches are added Finally, there could be other parse trees that arise from neither left-most or right-most derivation But we are most interested in left-most and rightmost 56

57 Summary of Derivations We are not just interested in whether s is in L(G) We need a parse tree for s A derivation defines a parse tree But one parse tree may have many derivations Left-most and right-most derivations are important in parser implementation 57

58 Ambiguity Grammar + () id String id id + id 58

59 Ambiguity (Cont.) This string has two parse trees + * * id id + id id id id 59

60 Ambiguity (Cont.) A grammar is ambiguous if it has more than one parse tree for some string quivalently, there is more than one right-most or left-most derivation for some string Ambiguity is BAD Leaves meaning of some programs ill-defined 60

61 Dealing with Ambiguity There are several ways to handle ambiguity Most direct method is to rewrite grammar unambiguously + ' ' id ʹ id () ʹ () ' nforces precedence of * over + 61

62 Ambiguity in Arithmetic xpressions Recall the grammar + * ( ) int The string int * int + int has two parse trees: + * * int int + int int int int 62

63 Ambiguity: The Dangling lse Consider the grammar if then if then else OTHR This grammar is also ambiguous 63

64 The Dangling lse: xample The expression if 1 then if 2 then 3 else 4 has two parse trees if if 1 if 4 1 if Typically we want the second form 64

65 The Dangling lse: A Fix else matches the closest unmatched then We can describe this in the grammar MIF /* all then are matched */ UIF /* some then is unmatched */ MIF if then MIF else MIF OTHR UIF if then if then MIF else UIF Describes the same set of strings 65

66 The Dangling lse: xample Revisited The expression if 1 then if 2 then 3 else 4 if if 1 if 1 if A valid parse tree (for a UIF) Not valid because the then expression is not a MIF 66

67 Ambiguity No general techniques for handling ambiguity Impossible to convert automatically an ambiguous grammar to an unambiguous one Used with care, ambiguity can simplify the grammar Sometimes allows more natural definitions We need disambiguation mechanisms 67

68 Precedence and Associativity Declarations Instead of rewriting the grammar Use the more natural (ambiguous) grammar Along with disambiguating declarations Most tools allow precedence and associativity declarations to disambiguate grammars xamples 68

69 Associativity Declarations Consider the grammar + int Ambiguous: two parse trees of int + int + int int int + int int int int Left associativity declaration: %left + 69

70 Precedence Declarations Consider the grammar + * int And the string int + int * int * + + int int * int int int Precedence declarations: %left + %left * int 70

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