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LR(1) Languages An Introduction

LR(1) Languages An Introduction. CMSC 431 Shon Vick. Predictive Parsing Summary. First and Follow sets are used to construct predictive tables For non-terminal A and input t, use a production A  a where t  First( a )

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LR(1) Languages An Introduction

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  1. LR(1) LanguagesAn Introduction CMSC 431 Shon Vick

  2. Predictive Parsing Summary • First and Follow sets are used to construct predictive tables • For non-terminal A and input t, use a production A  a where t  First(a) • For non-terminal A and input t, if e  First(A) and t  Follow(a), then use a production A  a where e First(a) • We’ll see First and Follow sets again . . .

  3. Bottom-Up Parsing • Bottom-up parsing is more general than top-down parsing • And just as efficient • Builds on ideas in top-down parsing • Bottom-up is the preferred method in practice • Concepts today, algorithms next time

  4. An Introductory Example • Bottom-up parsers don’t need left-factored grammars • Hence we can revert to the “natural” grammar for our example: E  T + E | T T  int * T | int | (E) • Consider the string: int * int + int

  5. The Idea Bottom-up parsing reduces a string to the start symbol by inverting productions:

  6. Terminology • Derivation Sequence starting with the start symbol S and proceeding through a sequence of derivation steps • Derivation step A non-terminal on the left side of a derivation is replaced by the right hand side of a production rule in the next step of the derivation A if A is a production, and,  are arbitrary strings of grammar symbols • Sentential Form Any derivation step • Sentence Sentential form with no non-terminals

  7.        Derivation example • Grammar EE + E | E * E | ( E ) | - E | id • Simple derivatiozn E  - E • Derivation of -(id+id) E  -E  -(E)  -(E+E)  -(id+E)  -(id+id) E-(id+id) derives it 1  2  …  n or simply   Select a nonterminal to replace and an alternative at each step of a derivation

  8. Leftmost Derivation • The derivationE  -E  -(E)  -(E+E)  -(id+E)  -(id+id)is leftmost which we designate asE lm -E lm -(E) lm -(E+E) lm -(id+E) lm -(id+id) • A sentential form is a derivation step. • A leftmost derivation step is a left sentential form, for example: • (Denoted*lm for typographical convenience)

  9. Leftmost Derivation • A derivation in which only the leftmost non-terminal in any sentential form is replaced at each step. • Unique derivation for a string most of the time

  10. Rightmost Derivation • The rightmost non-terminal is replaced in the derivation process in each step. • Also referred to as Canonical Derivation • Right sentential form: a sentential form produced via a rightmost derivation • If S *, then  is a sentential form of the CFG • If S *rm, then  is a right sentential form

  11. Right Sentential Form Grammar: S  aABe A Abc | b B  d • Reduce abbcde to S by four steps abbcde aAbcde aAdeaABe S Why not chose d here?

  12. TEST YOURSELF #1 • Question: find the rightmost derivation of the string int * int + int

  13. Observation • Read the productions found by bottom-up parse in reverse (i.e., from bottom to top) • This is a rightmost derivation!

  14. Important Fact #1 Important Fact #1 about bottom-up parsing: A bottom-up parser traces a rightmost derivation in reverse

  15. A Bottom-up Parse E T E T T + int int int *

  16. A Bottom-up Parse in Detail (1) + int int int *

  17. A Bottom-up Parse in Detail (2) T + int int int *

  18. A Bottom-up Parse in Detail (3) T T + int int int *

  19. A Bottom-up Parse in Detail (4) T T T + int int int *

  20. A Bottom-up Parse in Detail (5) T E T T + int int int *

  21. A Bottom-up Parse in Detail (6) E T E T T + int int int *

  22. A Trivial Bottom-Up Parsing Algorithm Let I = input string repeat pick a non-empty substring  of I where X  is a production if no such , backtrack replace one  by X in I until I = “S” (the start symbol) or all possibilities are exhausted

  23. Questions • Does this algorithm terminate? • How fast is the algorithm? • Does the algorithm handle all cases? • How do we choose the substring to reduce at each step?

  24. Where Do Reductions Happen Important Fact #1 has an interesting consequence: • Let  be a step of a bottom-up parse • Assume the next reduction is by X  • Then  is a string of terminals Why? Because X   is a step in a right-most derivation

  25. Notation • Idea: Split string into two substrings • Right substring is as yet unexamined by parsing (a string of terminals) • Left substring has terminals and non-terminals • The dividing point is marked by a | • The |is not part of the string • Initially, all input is unexamined |x1x2 . . . xn

  26. Shift-Reduce Parsing Bottom-up parsing uses only two kinds of actions: Shift Reduce

  27. Shift • Shift: Move | one place to the right • Shifts a terminal to the left string ABC|xyz  ABCx|yz

  28. Reduce • Apply an inverse production at the right end of the left string • If A  xy is a production, then Cbxy|ijk  CbA|ijk

  29. The Example with Reductions Only

  30. The Example with Shift-Reduce Parsing

  31. A Shift-Reduce Parse in Detail (1) + int int int * 

  32. A Shift-Reduce Parse in Detail (2) + int int int * 

  33. A Shift-Reduce Parse in Detail (3) + int int int * 

  34. A Shift-Reduce Parse in Detail (4) + int int int * 

  35. A Shift-Reduce Parse in Detail (5) T + int int int * 

  36. A Shift-Reduce Parse in Detail (6) T T + int int int * 

  37. A Shift-Reduce Parse in Detail (7) T T + int int int * 

  38. A Shift-Reduce Parse in Detail (8) T T + int int int * 

  39. A Shift-Reduce Parse in Detail (9) T T T + int int int * 

  40. A Shift-Reduce Parse in Detail (10) T E T T + int int int * 

  41. A Shift-Reduce Parse in Detail (11) E T E T T + int int int * 

  42. The Stack • Left string can be implemented by a stack • Top of the stack is the | • Shift pushes a terminal on the stack • Reduce pops 0 or more symbols off of the stack (production rhs) and pushes a non-terminal on the stack (production lhs)

  43. Key Issue (will be resolved by algorithms) • How do we decide when to shift or reduce? • Consider step int | * int + int • We could reduce by T  int giving T | * int + int • A fatal mistake: No way to reduce to the start symbol E

  44. Conflicts • Generic shift-reduce strategy: • If there is a handle on top of the stack, reduce • Otherwise, shift • But what if there is a choice? • If it is legal to shift or reduce, there is a shift-reduce conflict • If it is legal to reduce by two different productions, there is a reduce-reduce conflict

  45. Source of Conflicts • Ambiguous grammars always cause conflicts • But beware, so do many non-ambiguous grammars

  46. Conflict Example Consider everybody's favorite ambiguous grammar:

  47. One Shift-Reduce Parse

  48. Another Shift-Reduce Parse

  49. Example Notes • In the second step E * E | + int we can either shift or reduce by E  E * E • Choice determines associativity of + and * • As noted previously, grammar can be rewritten to enforce precedence • Precedence declarations are an alternative

  50. Precedence Declarations Revisited • Precedence declarations cause shift-reduce parsers to resolve conflicts in certain ways • Declaring “* has greater precedence than +” causes parser to reduce at E * E | + int • More precisely, precedence declaration is used to resolve conflict between reducing a * and shifting a +

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