Lecture VII Part 2: Syntactic Analysis Bottom-up Parsing: LR Parsing. Prof. Bodik CS Berkley University 1
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1 Lecture VII Part 2: Syntactic Analysis Bottom-up Parsing: LR Parsing. Prof. Bodik CS Berkley University 1
2 Bottom-Up Parsing Bottom-up parsing is more general than topdown parsing And just as efficient Builds on ideas in top-down parsing Preferred method in practice Also called LR parsing L means that tokens are read left to right R means that it constructs a rightmost derivation! Prof. Bodik CS 164 Lecture 7-8 2
3 An Introductory xample LR parsers don t need left-factored grammars and can also handle left-recursive grammars Consider the following grammar: ( ) Why is this not LL(1)? Consider the string: ( ) ( ) Prof. Bodik CS 164 Lecture 7-8 3
4 The Idea LR parsing reduces a string to the start symbol by inverting productions: str: input string of terminals repeat Identify β in str such that A β is a production (i.e., str = α β γ) Replace β by A in str (i.e., str becomes α A γ) until str = S Prof. Bodik CS 164 Lecture 7-8 4
5 A Bottom-up Parse in Detail (1) () () ( ) ( ) Prof. Bodik CS 164 Lecture 7-8 5
6 A Bottom-up Parse in Detail (2) () () () () ( ) ( ) Prof. Bodik CS 164 Lecture 7-8 6
7 A Bottom-up Parse in Detail (3) () () () () () () ( ) ( ) Prof. Bodik CS 164 Lecture 7-8 7
8 A Bottom-up Parse in Detail (4) () () () () () () () ( ) ( ) Prof. Bodik CS 164 Lecture 7-8 8
9 A Bottom-up Parse in Detail (5) () () () () () () () () ( ) ( ) Prof. Bodik CS 164 Lecture 7-8 9
10 A Bottom-up Parse in Detail (6) () () () () () () () () A rightmost derivation in reverse ( ) ( ) Prof. Bodik CS 164 Lecture
11 Important Fact #1 Important Fact #1 about bottom-up parsing: An LR parser traces a rightmost derivation in reverse Prof. Bodik CS 164 Lecture
12 Where Do Reductions Happen Important Fact #1 has an eresting consequence: Let αβγ be a step of a bottom-up parse Assume the next reduction is by A β Then γ is a string of terminals! Why? Because αaγ αβγ is a step in a rightmost derivation. Prof. Bodik CS 164 Lecture
13 Notation Idea: Split string o two substrings Right substring (a string of terminals) is as yet unexamined by parser Left substring has terminals and non-terminals The dividing po is marked by a I The I is not part of the string Initially, all input is unexamined: Ix 1 x 2... x n Prof. Bodik CS 164 Lecture
14 Shift-Reduce Parsing Bottom-up parsing uses only two kinds of actions: Shift Reduce Prof. Bodik CS 164 Lecture
15 Shift Shift: Move I one place to the right Shifts a terminal to the left string (I ) ( I ) Prof. Bodik CS 164 Lecture
16 Reduce Reduce: Apply an inverse production at the right end of the left string If ( ) is a production, then ( ( ) I ) ( I ) Prof. Bodik CS 164 Lecture
17 Shift-Reduce xample I () ()$ shift ( ) ( )
18 Shift-Reduce xample I () ()$ shift I () ()$ red. ( ) ( )
19 Shift-Reduce xample I () ()$ shift I () ()$ red. I () ()$ shift 3 times ( ) ( )
20 Shift-Reduce xample I () ()$ shift I () ()$ red. I () ()$ shift 3 times ( I ) ()$ red. ( ) ( )
21 Shift-Reduce xample I () ()$ shift I () ()$ red. I () ()$ shift 3 times ( I ) ()$ ( I ) ()$ red. shift ( ) ( )
22 Shift-Reduce xample I () ()$ shift I () ()$ red. I () ()$ shift 3 times ( I ) ()$ ( I ) ()$ () I ()$ red. shift red. () ( ) ( )
23 Shift-Reduce xample I () ()$ shift I () ()$ red. I () ()$ shift 3 times ( I ) ()$ ( I ) ()$ () I ()$ I ()$ red. shift red. () shift 3 times ( ) ( )
24 Shift-Reduce xample I () ()$ shift I () ()$ red. I () ()$ shift 3 times ( I ) ()$ ( I ) ()$ () I ()$ I ()$ red. shift red. () shift 3 times ( I )$ red. ( ) ( )
25 Shift-Reduce xample I () ()$ shift I () ()$ red. I () ()$ shift 3 times ( I ) ()$ red. ( I ) ()$ shift () I ()$ red. () I ()$ shift 3 times ( I )$ red. ( I )$ shift ( ) ( )
26 Shift-Reduce xample I () ()$ shift I () ()$ red. I () ()$ shift 3 times ( I ) ()$ ( I ) ()$ () I ()$ I ()$ red. shift red. () shift 3 times ( I )$ ( I )$ red. shift () I $ red. () ( ) ( )
27 Shift-Reduce xample I () ()$ shift I () ()$ red. I () ()$ shift 3 times ( I ) ()$ ( I ) ()$ () I ()$ I ()$ red. shift red. () shift 3 times ( I )$ ( I )$ red. shift () I $ I $ red. () accept ( ) ( )
28 The Stack Left string can be implemented by a stack Top of the stack is the I Shift pushes a terminal on the stack Reduce pops 0 or more symbols off of the stack (production rhs) and pushes a nonterminal on the stack (production lhs) Prof. Bodik CS 164 Lecture
29 Key Issue: When to Shift or Reduce? Decide based on the stack s content and the lookahead Idea: use a finite automaton (DFA) to decide when to shift or reduce The DFA input is the stack s content The alphabet consists of terminals and non-terminals We run the DFA on the stack and we examine the resulting state X and the token tok after I If X has a transition labeled tok then shift If X is labeled with A β on tok then reduce Prof. Bodik CS 164 Lecture
30 accept on $ LR(1) Parsing. An xample () on $, 0 1 ( ) 10 6 ( ) 5 11 on $, 9 on ), () on ), I () ()$ shift I () ()$ I () ()$ shift(x3) ( I ) ()$ ( I ) ()$ shift () I ()$ () I ()$ shift (x3) ( I )$ ( I )$ shift () I $ () I $ accept
31 Representing the DFA Parsers represent the DFA as a 2D table Lines correspond to DFA states Columns correspond to terminals and nonterminals Typically columns are split o: Those for terminals: action table Those for non-terminals: goto table Prof. Bodik CS 164 Lecture
32 Representing the DFA. xample The table for a fragment of our DFA: ( ) 7 () on $, 5 on ), ( ) $ 3 s4 4 s5 g6 5 r -> r -> 6 s8 s7 7 r -> () r -> () Prof. Bodik CS 164 Lecture
33 A Hierarchy of Grammar Classes From Andrew Appel, Modern Compiler Implementation in Java Prof. Bodik CS 164 Lecture
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