# Parsing #1. Leonidas Fegaras. CSE 5317/4305 L3: Parsing #1 1

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1 Parsing #1 Leonidas Fegaras CSE 5317/4305 L3: Parsing #1 1

2 Parser source file get next character scanner get token parser AST token A parser recognizes sequences of tokens according to some grammar and generates Abstract Syntax Trees (ASTs) A context-free grammar (CFG) has a finite set of terminals (tokens) a finite set of nonterminals from which one is the start symbol and a finite set of productions of the form: A ::= X1 X2... Xn where A is a nonterminal and each Xi is either a terminal or nonterminal symbol CSE 5317/4305 L3: Parsing #1 2

3 Example Expressions: E ::= E + T E - T T T ::= T * F T / F F F ::= num id Nonterminals: E T F Start symbol: E Terminals: + - * / id num Example: x+2*y... or equivalently: E ::= E + T E ::= E - T E ::= T T ::= T * F T ::= T / F T ::= F F ::= num F ::= id CSE 5317/4305 L3: Parsing #1 3

4 Derivations Notation: terminals: t, s,... nonterminals: A, B,... symbol (terminal or nonterminal): X, Y,... sequence of symbols: a, b,... Given a production: A ::= X 1 X 2... X n the form aab => ax 1 X 2... X n b is called a derivation eg, using the production T ::= T * F we get T / F x => T * F / F x Leftmost derivation: when you always expand the leftmost nonterminal in the sequence Rightmost derivation:... rightmost nonterminal CSE 5317/4305 L3: Parsing #1 4

5 Top-down Parsing It starts from the start symbol of the grammar and applies derivations until the entire input string is derived Example that matches the input sequence id(x) + num(2) * id(y) E => E + T use E ::= E + T => E + T * F use T ::= T * F => T + T * F use E ::= T => T + F * F use T ::= F => T + num * F use F ::= num => F + num * F use T ::= F => id + num * F use F ::= id => id + num * id use F ::= id You may have more than one choice at each derivation step: my have multiple nonterminals in each sequence for each nonterminal in the sequence, may have many rules to choose from Wrong predictions will cause backtracking need predictive parsing that never backtracks CSE 5317/4305 L3: Parsing #1 5

6 Bottom-up Parsing It starts from the input string and uses derivations in the opposite directions (from right to left) until you derive the start symbol Previous example: id(x) + num(2) * id(y) <= id(x) + num(2) * F use F ::= id <= id(x) + F * F use F ::= num <= id(x) + T * F use T ::= F <= id(x) + T use T ::= T * F <= F + T use F ::= id <= T + T use T ::= F <= E + T use E ::= T <= E use E ::= E + T At each derivation step, need to recognize a handle (the sequence of symbols that matches the right-hand-side of a production) CSE 5317/4305 L3: Parsing #1 6

7 Parse Tree Given the derivations used in the top-down/bottom-up parsing of an input sequence, a parse tree has the start symbol as the root E the terminals of the input sequence as leafs for each production A ::= X 1 X 2... X n used in a derivation, a node A with children X 1 X 2... X n E T T T F F F id(x) + num(2) * id(y) E => E + T => E + T * F => T + T * F => T + F * F => T + num * F => F + num * F => id + num * F => id + num * id CSE 5317/4305 L3: Parsing #1 7

8 Playing with Associativity What about this grammar? E ::= T + E T - E T T ::= F * T F / T F F ::= num id Right associative Now x+y+z is equivalent to x+(y+z) E T E F T E F T F id(x) + id(y) + id(z) CSE 5317/4305 L3: Parsing #1 8

9 Ambiguous Grammars What about this grammar? E ::= E + E E - E E * E E / E num id E E E E E id(x) * id(y) + id(z) E E E E E id(x) * id(y) + id(z) Operators + - * / have the same precedence! It is ambiguous: has more than one parse tree for the same input sequence (depending which derivations are applied each time) CSE 5317/4305 L3: Parsing #1 9

10 Predictive Parsing The goal is to construct a top-down parser that never backtracks Always leftmost derivations left recursion is bad! We must transform a grammar in two ways: eliminate left recursion perform left factoring These rules eliminate most common causes for backtracking although they do not guarantee a completely backtrack-free parsing CSE 5317/4305 L3: Parsing #1 10

11 Left Recursion Elimination For example, the grammar A ::= A a b recognizes the regular expression ba*. But a top-down parser may have hard time to decide which rule to use Need to get rid of left recursion: A ::= b A' A' ::= a A' ie, A' parses the RE a*. The second rule is recursive, but not left recursive CSE 5317/4305 L3: Parsing #1 11

12 Left Recursion Elimination (cont.) For each nonterminal X, we partition the productions for X into two groups: one that contains the left recursive productions the other with the rest That is: X ::= X a 1... X ::= X a n X ::= b 1... X ::= b m where a and b are symbol sequences. Then we eliminate the left recursion by rewriting these rules into: X ::= b 1 X' X' ::= a 1 X' X' ::= a X ::= b m X' n X' X' ::= CSE 5317/4305 L3: Parsing #1 12

13 Example E ::= E + T E - T T T ::= T * F T / F F F ::= num id E ::= T E' E' ::= + T E' - T E' T ::= F T' T' ::= * F T' / F T' F ::= num id CSE 5317/4305 L3: Parsing #1 13

14 Example A grammar that recognizes regular expressions: R ::= R R R bar R R * ( R ) char After left recursion elimination: R ::= ( R ) R' char R' R' ::= R R' bar R R' * R' CSE 5317/4305 L3: Parsing #1 14

15 Left Factoring Factors out common prefixes: X ::= a b 1... X ::= a b n becomes: X ::= a X' X' ::= b 1... X' ::= b n Example: E ::= T + E T - E T E ::= T E' E' ::= + E - E CSE 5317/4305 L3: Parsing #1 15

16 Recursive Descent Parsing E ::= T E' E' ::= + T E' - T E' T ::= F T' T' ::= * F T' / F T' F ::= num id static void E () { T(); Eprime(); } static void Eprime () { if (current_token == PLUS) { read_next_token(); T(); Eprime(); } else if (current_token == MINUS) { read_next_token(); T(); Eprime(); }; } static void T () { F(); Tprime(); } static void Tprime() { if (current_token == TIMES) { read_next_token(); F(); Tprime(); } else if (current_token == DIV) { read_next_token(); F(); Tprime(); }; } static void F () { if (current_token == NUM current_token == ID) read_next_token(); else error(); } CSE 5317/4305 L3: Parsing #1 16

17 Predictive Parsing Using a Table The symbol sequence from a derivation is stored in a stack (first symbol on top) if the top of the stack is a terminal, it should match the current token from the input if the top of the stack is a nonterminal X and the current input token is t, we get a rule for the parse table: M[X,t] the rule is used as a derivation to replace X in the stack with the right-hand symbols push(s); read_next_token(); repeat X = pop(); if (X is a terminal or '\$') if (X == current_token) read_next_token(); else error(); else if (M[X,current_token] == "X ::= Y1 Y2... Yk") { push(yk);... push(y1); } else error(); until X == '\$'; CSE 5317/4305 L3: Parsing #1 17

18 Parsing Table Example 1) E ::= T E' \$ 2) E' ::= + T E' 3) - T E' 4) 5) T ::= F T' 6) T' ::= * F T' 7) / F T' 8) 9) F ::= num 10) id num id + - * / \$ E 1 1 E' T 5 5 T' F 9 10 CSE 5317/4305 L3: Parsing #1 18

19 Example: Parsing x-2*y\$ top Stack current_token Rule E x M[E,id] = 1 (using E ::= T E' \$) \$ E' T x M[T,id] = 5 (using T ::= F T') \$ E' T' F x M[F,id] = 10 (using F ::= id) \$ E' T' id x read_next_token \$ E' T' - M[T',-] = 8 (using T' ::= ) \$ E' - M[E',-] = 3 (using E' ::= - T E') \$ E' T - - read_next_token \$ E' T 2 M[T,num] = 5 (using T ::= F T') \$ E' T' F 2 M[F,num] = 9 (using F ::= num) \$ E' T' num 2 read_next_token \$ E' T' * M[T',*] = 6 (using T' ::= * F T') \$ E' T' F * * read_next_token \$ E' T' F y M[F,id] = 10 (using F ::= id) \$ E' T' id y read_next_token \$ E' T' \$ M[T',\$] = 8 (using T' ::= ) \$ E' \$ M[E',\$] = 4 (using E' ::= ) \$ \$ stop (accept) CSE 5317/4305 L3: Parsing #1 19

20 Constructing the Parsing Table FIRST[a] is the set of terminals t that result after a number of derivations on the symbol sequence a ie, a =>... => tb for some symbol sequence b FIRST[ta]={t} eg, FIRST[3+E]={3} FIRST[X]=FIRST[a 1 ] FIRST[a n ] for each production X ::= a i FIRST[Xa]=FIRST[X] but if X has an empty derivation then FIRST[Xa]=FIRST[X] FIRST[a] FOLLOW[X] is the set of all terminals that follow X in any legal derivation find all productions Z ::= a X b in which X appears at the RHS; then FIRST[b] must be included in FOLLOW[X] if b has an empty derivation, FOLLOW[Z] must be included in FOLLOW[X] CSE 5317/4305 L3: Parsing #1 20

21 Example 1) E ::= T E' \$ 2) E' ::= + T E' 3) - T E' 4) 5) T ::= F T' 6) T' ::= * F T' 7) / F T' 8) 9) F ::= num 10) id FIRST FOLLOW E {num,id} {} E' {+,-} {\$} T {num,id} {+,-,\$} T' {*,/} {+,-,\$} F {num,id} {+,-,*,/,\$} CSE 5317/4305 L3: Parsing #1 21

22 Constructing the Parsing Table (cont.) For each rule X ::= a do: for each t in FIRST[a], add X ::= a to M[X,t] if a can be reduced to the empty sequence, then for each t in FOLLOW[X], add X ::= a to M[X,t] 1) E ::= T E' \$ 2) E' ::= + T E' 3) - T E' 4) 5) T ::= F T' 6) T' ::= * F T' 7) / F T' 8) 9) F ::= num 10) id FIRST FOLLOW E {num,id} {} E' {+,-} {\$} T {num,id} {+,-,\$} T' {*,/} {+,-,\$} F {num,id} {+,-,*,/,\$} num id + - * / \$ E 1 1 E' T 5 5 T' F 9 10 CSE 5317/4305 L3: Parsing #1 22

23 Another Example G ::= S \$ S ::= ( L ) a L ::= L, S S 0) G := S \$ 1) S ::= ( L ) 2) S ::= a 3) L ::= S L' 4) L' ::=, S L' 5) L' ::= ( ) a, \$ G 0 0 S 1 2 L 3 3 L' 5 4 CSE 5317/4305 L3: Parsing #1 23

24 LL(1) A grammar is called LL(1) if each element of the parsing table of the grammar has at most one production element the first L in LL(1) means that we read the input from left to right the second L means that it uses left-most derivations only the number 1 means that we need to look one token ahead from the input CSE 5317/4305 L3: Parsing #1 24

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