Winter 2012-2013 Compiler Principles Syntax Analysis (Parsing) – Part 1

Winter 2012-2013Compiler Principles

Syntax Analysis (Parsing) – Part 1Mayer Goldberg and Roman Manevich

Ben-Gurion University

2

BooksCompilersPrinciples, Techniques, and ToolsAlfred V. Aho, Ravi Sethi, Jeffrey D. Ullman

Advanced Compiler Design and ImplementationSteven Muchnik

Modern Compiler DesignD. Grune, H. Bal, C. Jacobs, K. Langendoen

Modern Compiler Implementation in JavaAndrew W. Appel

3

Today

• Understand role of syntax analysis• Context-free grammars– Basic definitions– Ambiguities

• Top-down parsing– Predictive parsing

• Next time: bottom-up parsing method

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The bigger picture

• Compilers include different kinds of program analyses each further constrains the set of legal programs– Lexical constraints

– Syntax constraints

– Semantic constraints

– “Logical” constraints(Verifying Compiler grand challenge)

Program consists of legal tokens

Program included in a given context-free language

Type checking, legal inheritance graph, variables initialized before used

Memory safety: null dereference, array-out-of-bounds access, data races, assertion violation

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Role of syntax analysis

• Recover structure from stream of tokens– Parse tree / abstract syntax tree

• Error reporting (recovery)• Other possible tasks

– Syntax directed translation (one pass compilers)– Create symbol table– Create pretty-printed version of the program, e.g., Auto

Formatting function in Eclipse

High-levelLanguage

(scheme)

Executable

Code

LexicalAnalysis

Syntax Analysis

Parsing

AST SymbolTableetc.

Inter.Rep.(IR)

CodeGeneration

From tokens to abstract syntax trees5 + (7 * x)

) id * num ( + num

Lexical Analyzer

program text

token stream

Parser

Grammar:E id E numE E + EE E * EE ( E ) +

num

num id

*

Abstract Syntax Tree

validsyntaxerror

6

Regular expressionsFinite automata

Context-free grammarsPush-down automata

Example grammar

shorthand for Statement

shorthand for Expression

shorthand for List(of expressions)

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S S ; SS id := E S print (L)E idE numE E + EL EL L, E

CFG terminology

8

Symbols: Terminals (tokens): ; := ( ) id num printNon-terminals: S E L

Start non-terminal: SConvention: the non-terminal appearingin the first derivation rule

Grammar productions (rules)N α


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Language of a CFG

• A sentence ω is in L(G) (valid program) if– There exists a corresponding derivation– There exists a corresponding parse tree

Derivations • Show that a sentence ω is in a grammar G– Start with the start symbol– Repeatedly replace one of the non-terminals by a

right-hand side of a production– Stop when the sentence contains only terminals

• Given a sentence αNβ and rule NµαNβ => αµβ

• ω is in L(G) if S =>* ω– Rightmost derivation– Leftmost derivation

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Leftmost derivation

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S

=> S ; S

=> id := E ; S

=> id := num ; S

=> id := num ; id := E

=> id := num ; id := E + E

=> id := num ; id := num + E

=> id := num ; id := num + num

a := 56 ; b := 7 + 3

id := num ; id := num + num


Rightmost derivation

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S

=> S ; S

=> S ; id := E

=> S ; id := E + E

=> S ; id := E + num

=> S ; id := num + num

=> id := E ; id := num + num

=> id := num ; id := num + num

a := 56 ; b := 7 + 3

id := num ; id := num + num


Parse trees• Tree nodes are symbols, children ordered left-to-right• Each internal node is non-terminal and its children

correspond to one of its productions

N µ1 … µk

• Root is start non-terminal• Leaves are tokens• Yield of parse tree: left-to-right walk over leaves

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µ1 µk

N

…

Parse tree example

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S S ; SS id := E S print (L)E idE numE E + EL EL L, E id := num ; id := num num+

Draw parse tree for expression

Parse tree example

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id := num ; id := num num+

E E E E

S E

S

S

Order-independent representation


(S(S(Ea)E := (E56)E)S ; (S(Eb)E := (E(E7)E + (E3)E)E)S)S

E

Equivalently add parentheses labeled by non-terminal names

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Capabilities and limitations of CFGs• CFGs naturally express– Hierarchical structure

• A program is a list of classes,A Class is a list of definition,A definition is either…

– Beginning-end type of constraints• Balanced parentheses S (S)S | ε

• Cannot express– Correlations between unbounded strings (identifiers)– Variables are declared before use: ω S ω– Handled by semantic analysis

p. 173

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Sometimes there are two parse trees

Leftmost derivationEE + Enum + Enum + E + Enum + num + Enum + num + num

num(1)

E

E E

+

E E

+num(2) num(3)

Rightmost derivationEE + EE + numE + E + numE + num + numnum + num + num

+ num(3)+num(1) num(2)

Arithmetic expressions:E id E numE E + EE E * EE ( E )

1 + 2 + 3

E

E E

E

E

1 + (2 + 3) (1 + 2) + 3

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Is ambiguity a problem?

Leftmost derivationEE + Enum + Enum + E + Enum + num + Enum + num + num

num(1)

E

E E

+

E E

+num(2) num(3)

Rightmost derivationEE + EE + numE + E + numE + num + numnum + num + num

+ num(3)+num(1) num(2)


1 + 2 + 3

E

E E

E

E

= 6 = 6

1 + (2 + 3) (1 + 2) + 3Depends on semantics

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Problematic ambiguity example

Leftmost derivationEE + Enum + Enum + E * Enum + num * Enum + num * num

num(1)

E

E E

+

E E

*num(2) num(3)

Rightmost derivationEE * EE * numE + E * numE + num * numnum + num * num

* num(3)+num(1) num(2)


1 + 2 * 3

This is what we usually want: * has precedence over +

E

E E

E

E

= 7 = 9

1 + (2 * 3) (1 + 2) * 3

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Ambiguous grammars• A grammar is ambiguous if there exists a sentence

for which there are– Two different leftmost derivations– Two different rightmost derivations– Two different parse trees

• Property of grammars, not languages• Some languages are inherently ambiguous – no

unambiguous grammars exist• No algorithm to detect whether arbitrary

grammar is ambiguous

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Drawbacks of ambiguous grammars

• Ambiguous semantics• Parsing complexity• May affect other phases• Solutions– Transform grammar into non-ambiguous– Handle as part of parsing method• Using special form of “precedence”• Wait for bottom-up parsing lecture

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Transforming ambiguous grammars to non-ambiguous by layering

Ambiguous grammarE E + EE E * EE id E numE ( E )

Unambiguous grammarE E + TE TT T * FT FF idF numF ( E )

Layer 1

Layer 2

Layer 3

Let’s derive 1 + 2 * 3

Each layer takes care of one way of composing sub-strings to form a string:1: by +2: by *3: atoms

Transformed grammar: * precedes +

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Unambiguous grammarE E + TE TT T * FT FF idF numF ( E )

Derivation E=> E + T=> T + T=> F + T=> 1 + T=> 1 + T * F=> 1 + F * F=> 1 + 2 * F=> 1 + 2 * 3

+ * 321

F F F

T

TE

T

E

Parse tree

Transformed grammar: + precedes *

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Unambiguous grammarE E * TE TT T + FT FF idF numF ( E )

+ * 321

Derivation E=> E * T=> T * T=> T + F * T=> F + F * T=> 1 + F * T=> 1 + 2 * T=> 1 + 2 * F=> 1 + 2 * 3

F F F

T

T

E

T

E

Parse tree

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Another example for layering

Ambiguous grammarP ε | P P | ( P )

Unambiguous grammarS P S | εP ( S )

Takes care of “concatenation”

Takes care of nesting

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“dangling-else” exampleAmbiguous grammar S if E then S S | if E then S else S | other

if

S

Sthen

thenif elseE S S

E

E1

E2 S1 S2

if

S

Sthen

thenif

else

E S

SE

E1

E2 S1

S2

if E1 then (if E2 then S1 else S2) if E1 then (if E2 then S1) else S2

This is what we usually want: match else to closest

unmatched then

if E1 then if E2 then S1 else S2

p. 174

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“dangling-else” example

if

S

Sthen

thenif else

Ambiguous grammar S if E then S S | if E then S else S | other

E S S

E

E1

E2 S1 S2

if

S

Sthen

thenif

else

E S

SE

E1

E2 S1

S2

if E1 then (if E2 then S1 else S2) if E1 then (if E2 then S1) else S2

Unambiguous grammar S M | UM if E then M else M | otherU if E then S | if E then M else U

if E1 then if E2 then S1 else S2

Matched statements

Unmatched statements

p. 174

Broad kinds of parsers • Parsers for arbitrary grammars

– Earley’s method, CYK method O(n3)– Not used in practice

• Top-Down– Construct parse tree in a top-down matter– Find the leftmost derivation– Predictive: for every non-terminal and k-tokens predict the next

production LL(k)– Preorder tree traversal

• Bottom-Up– Construct parse tree in a bottom-up manner– Find the rightmost derivation in a reverse order– For every potential right hand side and k-tokens decide when a

production is found LR(k)– Postorder tree traversal

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Top-down vs. bottom-up

• Top-down parsing– Beginning with the start symbol, try to guess the

productions to apply to end up at the user's program

• Bottom-up parsing– Beginning with the user's program, try to apply

productions in reverse to convert the program back into the start symbol

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Top-down parsingUnambiguous grammarE E * TE TT T + FT FF idF numF ( E )

+ * 321

F F F

T

T

E

T

E

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We need this rule to get the *

+ * 321

E

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+ * 321

E

T

E

33


+ * 321

F

T

E

T

E

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+ * 321

F F

T

T

E

T

E

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+ * 321

F F F

T

T

E

T

E

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+ * 321

F F F

T

T

E

T

E

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Bottom-up parsingUnambiguous grammarE E * TE TT T + FT FF idF numF ( E )

+ * 321

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+ * 321

F

39


+ * 321

F F

40


+ * 321

F F

T

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+ * 321

F F

T

F

42


+ * 321

F F

T

F

T

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+ * 321

F F

T

F

T

T

44


+ * 321

F F

T

F

T

T

E

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+ * 321

F F

T

F

T

T

E

E

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Challenges in top-down parsing• Top-down parsing begins with virtually no• information– Begins with just the start symbol, which matches every program

• How can we know which productions to apply?• In general, we can‘t– There are some grammars for which the best we can do is guess

and backtrack if we're wrong• If we have to guess, how do we do it?– Parsing as a search algorithm– Too expensive in theory (exponential worst-case time) and

practice

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Predictive parsing• Given a grammar G and a word w attempt to derive w

using G• Idea– Apply production to leftmost nonterminal– Pick production rule based on next input token

• General grammar– More than one option for choosing the next production

based on a token• Restricted grammars (LL)– Know exactly which single rule to apply– May require some lookahead to decide

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Boolean expressions example

not ( not true or false )

E => not E => not ( E OP E ) =>not ( not E OP E ) =>not ( not LIT OP E ) =>not ( not true OP E ) =>not ( not true or E ) =>not ( not true or LIT ) =>not ( not true or false )

not E

E

( E OP E )

not LIT or LIT

true false

production to apply known from next token

E LIT | (E OP E) | not ELIT true | falseOP and | or | xor

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Recursive descent parsing

• Define a function for every nonterminal• Every function work as follows– Find applicable production rule– Terminal function checks match with next input

token– Nonterminal function calls (recursively) other

functions• If there are several applicable productions for

a nonterminal, use lookahead

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Matching tokens

• Variable current holds the current input token

match(token t) { if (current == t) current = next_token() else error}


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Functions for nonterminals

E() { if (current {TRUE, FALSE}) // E LIT LIT(); else if (current == LPAREN) // E ( E OP E ) match(LPAREN); E(); OP(); E(); match(RPAREN); else if (current == NOT) // E not E match(NOT); E(); else error;}

LIT() { if (current == TRUE) match(TRUE); else if (current == FALSE) match(FALSE); else error;}


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Implementation via recursion

E → LIT | ( E OP E ) | not ELIT → true | falseOP → and | or | xor

E() {if (current {TRUE, FALSE}) LIT();else if (current == LPAREN) match(LPARENT);

E(); OP(); E();match(RPAREN);

else if (current == NOT) match(NOT); E();else error;

}

LIT() {if (current == TRUE) match(TRUE);else if (current == FALSE) match(FALSE);else error;

}

OP() {if (current == AND) match(AND);else if (current == OR) match(OR);else if (current == XOR) match(XOR);else error;

}

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Adding semantic actions

• Can add an action to perform on each production rule

• Can build the parse tree– Every function returns an object of type Node– Every Node maintains a list of children– Function calls can add new children

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Building the parse tree

Node E() { result = new Node(); result.name = “E”; if (current {TRUE, FALSE}) // E LIT result.addChild(LIT()); else if (current == LPAREN) // E ( E OP E ) result.addChild(match(LPAREN)); result.addChild(E()); result.addChild(OP()); result.addChild(E()); result.addChild(match(RPAREN)); else if (current == NOT) // E not E result.addChild(match(NOT)); result.addChild(E()); else error; return result;}

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Recursive descent

• How do you pick the right A-production?• Generally – try them all and use

backtracking• In our case – use lookahead

void A() { choose an A-production, A X1X2…Xk; for (i=1; i≤ k; i++) { if (Xi is a nonterminal) call procedure Xi(); elseif (Xi == current) advance input; else

report error; }}

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• The function for indexed_elem will never be tried… – What happens for input of the form ID[expr]

term ID | indexed_elemindexed_elem ID [ expr ]

Problem 1: productions with common prefix

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Problem 2: null productions

int S() { return A() && match(token(‘a’)) && match(token(‘b’));}int A() { return match(token(‘a’)) || 1;}

S A a bA a |

What happens for input “ab”? What happens if you flip order of alternatives and try “aab”?

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Problem 3: left recursion

int E() { return E() && match(token(‘-’)) && term();}

E E - term | term

What happens with this procedure? Recursive descent parsers cannot handle left-recursive grammars

p. 127

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FIRST sets• For every production rule Aα

– FIRST(α) = all terminals that α can start with – Every token that can appear as first in α under some derivation for α

• In our Boolean expressions example– FIRST( LIT ) = { true, false }– FIRST( ( E OP E ) ) = { ‘(‘ }– FIRST( not E ) = { not }

• No intersection between FIRST sets => can always pick a single rule

• If the FIRST sets intersect, may need longer lookahead– LL(k) = class of grammars in which production rule can be determined

using a lookahead of k tokens– LL(1) is an important and useful class

60

Computing FIRST sets

• Assume no null productions A 1. Initially, for all nonterminals A, set

FIRST(A) = { t | A tω for some ω }2. Repeat the following until no changes occur:

for each nonterminal A for each production A Bω set FIRST(A) = FIRST(A) FIRST(B)∪

• This is known a fixed-point

61

FIRST sets computation example

STMT if EXPR then STMT | while EXPR do STMT | EXPR ;EXPR TERM -> id | zero? TERM | not EXPR | ++ id | -- idTERM id | constant

TERM EXPR STMT

62

1. Initialization


TERM EXPR STMTidconstant

zero?Not++--

ifwhile

63

2. Iterate 1



zero?Not++--

ifwhile

zero?Not++--

64

2. Iterate 2



zero?Not++--

ifwhile

idconstant

zero?Not++--

65

2. Iterate 3 – fixed-point



zero?Not++--

ifwhile

idconstant

zero?Not++--

idconstant

66

FOLLOW sets• What do we do with nullable () productions?– AB C D B C – Use what comes afterwards to predict the right

production• For every production rule Aα – FOLLOW(A) = set of tokens that can immediately

follow A

• Can predict the alternative Ak for a non-terminal N when the lookahead token is in the set– FIRST(Ak) (if Ak is nullable then FOLLOW(N))

p. 189

67

LL(k) grammars• A grammar is in the class LL(K) when it can be

derived via:– Top-down derivation– Scanning the input from left to right (L)– Producing the leftmost derivation (L)– With lookahead of k tokens (k)– For every two productions Aα and Aβ we have

FIRST(α) ∩ FIRST(β) = {}and FIRST(A) ∩ FOLLOW(A) = {}

• A language is said to be LL(k) when it has an LL(k) grammar

68

Back to problem 1

• FIRST(term) = { ID }• FIRST(indexed_elem) = { ID }

• FIRST/FIRST conflict


69

Solution: left factoring

• Rewrite the grammar to be in LL(1)

Intuition: just like factoring x*y + x*z into x*(y+z)


term ID after_IDAfter_ID [ expr ] |

70

S if E then S else S | if E then S | T

S if E then S S’ | TS’ else S |

Left factoring – another example

71

Back to problem 2

• FIRST(S) = { a } FOLLOW(S) = { } • FIRST(A) = { a } FOLLOW(A) = { a }

• FIRST/FOLLOW conflict

S A a bA a |

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Solution: substitution

S A a bA a |

S a a b | a b

Substitute A in S

S a after_A after_A a b | b

Left factoring

73

Back to problem 3

• Left recursion cannot be handled with a bounded lookahead

• What can we do?

E E - term | term

74

Left recursion removal

• L(G1) = β, βα, βαα, βααα, …

• L(G2) = same

N Nα | β N βN’ N’ αN’ |

G1 G2

E E - term | term E term TE | termTE - term TE |

For our 3rd example:

p. 130

Can be done algorithmically.Problem: grammar becomes mangled beyond recognition

75

LL(k) Parsers

• Recursive Descent– Manual construction– Uses recursion

• Wanted– A parser that can be generated automatically– Does not use recursion

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• Pushdown automaton uses– Prediction stack– Input stream– Transition table• nonterminals x tokens -> production alternative• Entry indexed by nonterminal N and token t contains

the alternative of N that must be predicated when current input starts with t

LL(k) parsing via pushdown automata

77

LL(k) parsing via pushdown automata

• Two possible moves– Prediction

• When top of stack is nonterminal N, pop N, lookup table[N,t]. If table[N,t] is not empty, push table[N,t] on prediction stack, otherwise – syntax error

– Match• When top of prediction stack is a terminal T, must be equal to

next input token t. If (t == T), pop T and consume t. If (t ≠ T) syntax error

• Parsing terminates when prediction stack is empty– If input is empty at that point, success. Otherwise,

syntax error

78

Model of non-recursivepredictive parser

Predictive Parsing program

Parsing Table

X

Y

Z

$

Stack

$ b + a

Output

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( ) not true false and or xor $E 2 3 1 1

LIT 4 5OP 6 7 8

(1) E → LIT(2) E → ( E OP E ) (3) E → not E(4) LIT → true(5) LIT → false(6) OP → and(7) OP → or(8) OP → xor

Non

term

inal

s

Input tokens

Which rule should be used

Example transition table

80

a b c

A A aAb A c

A aAb | caacbb$

Input suffix Stack content Move

aacbb$ A$ predict(A,a) = A aAbaacbb$ aAb$ match(a,a)

acbb$ Ab$ predict(A,a) = A aAbacbb$ aAbb$ match(a,a)

cbb$ Abb$ predict(A,c) = A ccbb$ cbb$ match(c,c)

bb$ bb$ match(b,b)

b$ b$ match(b,b)

$ $ match($,$) – success

Running parser example

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a b c

A A aAb A c

A aAb | cabcbb$


abcbb$ A$ predict(A,a) = A aAbabcbb$ aAb$ match(a,a)

bcbb$ Ab$ predict(A,b) = ERROR

Illegal input example

82

Error handling and recovery

x = a * (p+q * ( -b * (r-s);

Where should we report the error?

The valid prefix property

Recovery is tricky Heuristics for dropping tokens, skipping to

semicolon, etc.

83

Error handling in LL parsers

• Now what?– Predict b S anyway “missing token b inserted in line XXX”

S a c | b Sc$

a b c

S S a c S b S


c$ S$ predict(S,c) = ERROR

84

Error handling in LL parsers

• Result: infinite loop

S a c | b Sc$

a b c

S S a c S b S


bc$ S$ predict(b,c) = S bSbc$ bS$ match(b,b)

c$ S$ Looks familiar?

85

Error handling

• Requires more systematic treatment• Enrichment– Acceptable-set method– Not part of course material

86

Summary• Parsing– Top-down or bottom-up

• Top-down parsing– Recursive descent– LL(k) grammars– LL(k) parsing with pushdown automata

• LL(K) parsers– Cannot deal with left recursion– Left-recursion removal might result with complicated

grammar

87

See you next time

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Winter 2012-2013 Compiler Principles Syntax Analysis (Parsing) – Part 1

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