Compiler Compiler ConstructionConstruction

Compiler Compiler ConstructionConstruction

Sohail Aslam

Lecture 19

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LL(1) Table ConstructionLL(1) Table ConstructionLL(1) Table ConstructionLL(1) Table ConstructionFor each production A →

1. for each terminal a in FIRST(), add A → to M[A,a].

2. If is in FIRST(), add A → to M[A,b] for each terminal b in FOLLOW(A).

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LL(1) Table ConstructionLL(1) Table ConstructionLL(1) Table ConstructionLL(1) Table ConstructionFor each production A →

1. for each terminal a in FIRST(), add A → to M[A,a].

2. If is in FIRST(), add A → to M[A,b] for each terminal b in FOLLOW(A).

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LL(1) Table ConstructionLL(1) Table ConstructionLL(1) Table ConstructionLL(1) Table ConstructionFor each production A →

1. for each terminal a in FIRST(), add A → to M[A,a].

2. If is in FIRST(), add A → to M[A,b] for each terminal b in FOLLOW(A).

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LL(1) Table ConstructionLL(1) Table ConstructionLL(1) Table ConstructionLL(1) Table Construction

(2 cont.)If is in FIRST(), and $ is in FOLLOW(A), add A → to M[A,$].

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LL(1) Table ConstructionLL(1) Table ConstructionLL(1) Table ConstructionLL(1) Table Construction

Make each undefined entry of M be error.

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LL(1) Table ConstructionLL(1) Table ConstructionLL(1) Table ConstructionLL(1) Table Construction

Let us apply the algorithm to the expression grammar.

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LL(1) Table ConstructionLL(1) Table ConstructionLL(1) Table ConstructionLL(1) Table Construction

Since FIRST(TE ') = FIRST(T ) = { (, id },production E →TE' cause M[E,(] and

M[E,id] to get E →TE'

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id + * ( ) $

E E →TE' E →TE'

E' E' → +TE'

E' → E' →

T T →FT' T →FT'T' T' → T

→*FT'T' → T' →

F F → id F →(E )

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LL(1) Table ConstructionLL(1) Table ConstructionLL(1) Table ConstructionLL(1) Table Construction

Production E' → +TE' causes

M[E',+] to get E' →+TE'

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id + * ( ) $

E E →TE' E →TE'

E' E' → +TE'

E' → E' →

T T →FT' T →FT'T' T' → T

→*FT'T' → T' →

F F → id F →(E )

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LL(1) Table ConstructionLL(1) Table ConstructionLL(1) Table ConstructionLL(1) Table Construction

Production E' → causes M[E',)] and M[E',$] to get E' → since

FOLLOW (E' ) = { ), $ }

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id + * ( ) $

E E →TE' E →TE'

E' E' → +TE'

E' → E' →

T T →FT' T →FT'T' T' → T

→*FT'T' → T' →

F F → id F →(E )

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LL(1) Table ConstructionLL(1) Table ConstructionLL(1) Table ConstructionLL(1) Table Construction

And so on.

The final parsing table produced is:

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LL(1) Parsing TableLL(1) Parsing TableLL(1) Parsing TableLL(1) Parsing Tableid + * ( ) $

E E →TE' E →TE'

E' E' → +TE'

E' → E' →

T T →FT' T →FT'T' T' → T

→*FT'T' → T' →

F F → id F →(E )

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Left FactoringLeft FactoringLeft FactoringLeft Factoring

Consider the grammar

E → T + E | TT → int | int T | (E)

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Left FactoringLeft FactoringLeft FactoringLeft Factoring

E → T + E | TT → int | int T | (E)

Impossible to predict because for T, two productions start with int

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Left FactoringLeft FactoringLeft FactoringLeft Factoring

E → T + E | TT → int | int T | (E)

For E, it is not clear how to predict

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Left FactoringLeft FactoringLeft FactoringLeft Factoring

A grammar must be left factored before use for predictive parsing

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Left FactoringLeft FactoringLeft FactoringLeft Factoring

If , replace all productionsA → 1 | 2 | .... | n |

withA → Z |

Z → 1| 2| .... | n

where Z is new nonterminal

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Left FactoringLeft FactoringLeft FactoringLeft Factoring

If , replace all productionsA → 1 | 2 | .... | n |

withA → Z |

Z → 1| 2| .... | n

where Z is new nonterminal

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Left FactoringLeft FactoringLeft FactoringLeft FactoringA graphical explanation

A 2

1

3

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Left FactoringLeft FactoringLeft FactoringLeft Factoring

becomes

Z 2

1

3

A

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Left FactoringLeft FactoringLeft FactoringLeft Factoring

Consider following fragment of expression grammarFactor → id

| id [ ExprList ] | id ( ExprList )

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Left FactoringLeft FactoringLeft FactoringLeft FactoringFactor → id

| id [ ExprList ]| id ( ExprList )

FIRST(rhs1) = {id}FIRST(rhs2) = {id}FIRST(rhs3) = {id}

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Left FactoringLeft FactoringLeft FactoringLeft Factoring

after left factoring

Factor → id ArgsArgs → [ ExprList ]

| ( ExprList ) |

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Left FactoringLeft FactoringLeft FactoringLeft FactoringFactor → id ArgsArgs → [ ExprList ]

| ( ExprList ) |

FIRST(rhs1) = {id}FIRST(rhs2) = { [ }FIRST(rhs3) = { ( }

FIRST(rhs4) = FOLLOW(Factor)

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Left FactoringLeft FactoringLeft FactoringLeft Factoring

[

(

Factor id ExprList

ExprList

]

)id

id

No basis for choice

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Left FactoringLeft FactoringLeft FactoringLeft Factoring

[

(

Factor id ExprList

ExprList

]

)

correct choice possible

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Left FactoringLeft FactoringLeft FactoringLeft FactoringQuestion

By eliminating left recursion and left factoring, can we transform an arbitrary CFG to a form where it meets the LL(1) condition (and can be parsed predictively with a single token lookahead?)

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Left FactoringLeft FactoringLeft FactoringLeft Factoring

AnswerGiven a CFG that does not meet the LL(1) condition, it is undecidable whether or not an equivalent LL(1) grammar exists