CS606 Short Notes Lecture 19

Lecture 19

LL(1) Table Construction

Here now is the algorithm to construct a predictive parsing table.

1. For each production A ? α
2. for each terminal a in FIRST(α), add A ? α to M[A,a].
3. If ε is in FIRST(α), add A ? α to M[A ,b] for each terminal b in FOLLOW(A ). If ε is in FIRST(α), and $ is in FOLLOW(A ), add A ? α to M[A ,$].
4. Make each undefined entry of M be error.

Let us apply the algorithm to the expression grammar. Since FIRST(TE ') = FIRST(T ) = { (, id }, the production E ? TE' cause M[E,(] and M[E,id] to get E ? TE'. The production E' ? +TE' causes M[E',+] to get E' ? +TE'. The production E' ? ε causes M[E',)] and M[E',$] to get E' ? ε since FOLLOW (E' ) = { ), $ }. And so on. The final parsing table produced is:

id E E ? TE' E ? TE'

E'

E' ? +TE' E' ? ε E' ? ε

T T ? FT' T ? FT'

T'

T' ? ε T ? *FT' T' ? ε T' ? ε F F ? id F ? (E )

Left Factoring

Consider the grammar

E ? T + E | T T ? int | int∗ T | (E)

It is impossible to predict because for T, two productions start with int. For E, it is not clear how to predict; the two productions start with the non-terminal T. A grammar must be left factored before use for predictive parsing. The procedure to left-factor a grammar is as follows: If α ≠ ε, replace all productions

A ? αβ1 | αβ2 | .... | αβn | γ with A ? α Z | γ Z ? β 1| β2| .... | β n

where Z is a new non-terminal

A graphical explanation:

Example: consider following fragment of expression grammar

Factor ?

id id [ ExprList ] id ( ExprList )

After left factoring, the grammar becomes

Factor ? id Args Args ? [ ExprList ] | ( ExprList ) | ε

Given a CFG that does not meet the LL(1) condition, it is undecidable whether or not an equivalent LL(1) grammar exists. A αβ2 αβ1

αβ3 β 2 β 1

β 3 αZ A