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LL(1) Parsing

LL(1) Parsing. Programming Language Concepts Lecture 7. Prepared by Manuel E. Bermúdez, Ph.D. Associate Professor University of Florida. Example. Build the LL(1) parse table for the following grammar. S → begin SL end {begin} → id := E; {id} SL → SL S {begin,id} → S {begin,id}

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LL(1) Parsing

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  1. LL(1) Parsing Programming Language Concepts Lecture 7 Prepared by Manuel E. Bermúdez, Ph.D. Associate Professor University of Florida

  2. Example • Build the LL(1) parse table for the following grammar. S →begin SL end {begin} → id := E; {id} SL → SL S {begin,id} →S {begin,id} E → E+T {(, id} → T {(, id} T → P*T {(, id} →P {(, id} P → (E) {(} → id {id} * * * * - not LL(1)

  3. Example (cont’d) • Lemma: Left recursion always produces a non-LL(1) grammar (e.g., SL, E above) • Proof: Consider A → A First () or Follow (A) →  First () Follow (A)

  4. Problems with our Grammar • SL is left recursive. • E is left recursive. • T → P * T both begin with the same→ P sequence of symbols (P).

  5. Solution to Problem 3 • Change: T → P * T { (, id } → P { (, id } • to: T → P X { (, id } X →* T { * } → { +, ; , ) } Follow(X) • Follow(T) due to T → P X • Follow(E)due to E → E+T , E → T = { +, ;, ) } due to E →E+T, S→id := E ; and P → (E) Disjoint!

  6. Solution to Problem 3 (cont’d) • In general, change A → 1 → 2 . . . → n to A → X X →1 . . . →n Hopefully all the’s begin with different symbols

  7. Solution to Problems 1 and 2 • We want (…((( T + T) + T) + T)…) • Instead, (T) (+T) (+T) … (+T) Change: E → E + T { (, id } → T { (, id } To: E → T Y { (, id } Y → + T Y { + } → { ; , ) } • Follow(Y) Follow(E) • = { ; , ) } No longer contains ‘+’, because we eliminated the production E → E + T

  8. Solution to Problems 1 and 2 (cont’d) • In general, Change:A → A1 A →  1 . . . . . . → An → m to:A → 1 X X → 1X . . . . . . →m X→nX →

  9. Solution to Problems 1 and 2 (cont’d) • In our example, Change:SL → SL S { begin, id } → S { begin, id } To:SL → S Z { begin, id } Z →S Z { begin, id } → { end }

  10. Modified Grammar S → begin SL end {begin} → id := E ; {id} SL → S Z {begin,id} Z →S Z {begin,id} → {end} E → T Y (,id} Y → + T Y {+} → {;,)} T → P X {(,id} X → * T {*} → {;,+,)} P → (E) {(} → id {id} Disjoint. Grammar is LL(1)

  11. Recursive Descent Parsing • Top-down parsing strategy, suitable for LL(1) grammars. • One procedure per nonterminal. • Contents of stack embedded in recursive call sequence. • Each procedure “commits” to one production, based on the next input symbol, and the select sets. • Good technique for hand-written parsers.

  12. For our Modified, LL(1) Grammar proc S; {S → begin SL end → id := E; } case Next_Token of T_begin : Read(T_begin); SL; Read (T_end); T_id : Read(T_id); Read (T_:=); E; Read (T_;); otherwise Error; end end; “Read (T_X)” verifies that the upcoming token is X, and consumes it. “Next_Token” is the upcoming token.

  13. For our Modified, LL(1) Grammar (cont’d) proc SL; {SL → SZ} S; Z; end; proc E; {E → TY} T; Y; end; Technically, should have insisted that Next Token be either T_begin or T_id, but S will do that anyway. Checking early would aid error recovery. // Ditto for T_( and T_id.

  14. For our Modified, LL(1) Grammar (cont’d) proc Z;{Z → SZ → } case Next Token of T_begin, T_id: S;Z; T_end: ; otherwise Error; end end;

  15. For our Modified, LL(1) Grammar (cont’d) proc Y; {Y → +TY → } if Next Token = T_+ then Read (T_+) T; Y; end; proc T; {T → PX} P; X end; Could have used a case statement Could have checked for T_( and T_id.

  16. For our Modified, LL(1) Grammar (cont’d) proc X;{X → *T → } if Next Token = T_* then Read (T_*); T; end;

  17. For our Modified, LL(1) Grammar (cont’d) proc P; {P →(E) → id } case Next Token of T_(: Read (T_(); E; Read (T_)); T_id: Read (T_id); otherwise Error; end end;

  18. LL(1) Parsing Programming Language Concepts Lecture 7 Prepared by Manuel E. Bermúdez, Ph.D. Associate Professor University of Florida

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