1. Compiler Construction Sohail Aslam Lecture 19

2. 2 LL(1) Table Construction For 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).

3. 3 LL(1) Table Construction For 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).

4. 4 LL(1) Table Construction For 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).

5. 5 LL(1) Table Construction (2 cont.) If  is in FIRST(  ), and $ is in FOLLOW(A), add A →  to M [ A,$ ].

6. 6 LL(1) Table Construction Make each undefined entry of M be error.

7. 7 LL(1) Table Construction Let us apply the algorithm to the expression grammar.

8. 8 LL(1) Table Construction Since FIRST(TE ') = FIRST(T ) = { (, id }, production E → TE' cause M [ E,( ] and M[E,id] to get E → TE'

9. 9 id +*()$ EE → TE' E'E' → +TE' E' →  TT → FT' T' T' →  T → *FT' T' →  FF → idF → (E )

10. 10 LL(1) Table Construction Production E' → +TE' causes M[E',+] to get E' → +TE'

11. 11 id +*()$ EE → TE' E'E' → +TE' E' →  TT → FT' T' T' →  T → *FT' T' →  FF → idF → (E )

12. 12 LL(1) Table Construction Production E' →  causes M[E',)] and M[E',$] to get E' →  since FOLLOW (E' ) = { ), $ }

13. 13 id +*()$ EE → TE' E'E' → +TE' E' →  TT → FT' T' T' →  T → *FT' T' →  FF → idF → (E )

14. 14 LL(1) Table Construction And so on. The final parsing table produced is:

15. 15 LL(1) Parsing Table id +*()$ EE → TE' E'E' → +TE' E' →  TT → FT' T' T' →  T → *FT' T' →  FF → idF → (E )

16. 16 Left Factoring Consider the grammar E → T + E | T T → int | int  T | (E)

17. 17 Left Factoring E → T + E | T T → int | int  T | (E) Impossible to predict because for T, two productions start with int

18. 18 Left Factoring E → T + E | T T → int | int  T | (E) For E, it is not clear how to predict

19. 19 Left Factoring A grammar must be left factored before use for predictive parsing

20. 20 Left Factoring If   , replace all productions A →  1 |  2 |.... |  n |  with A →  Z |  Z →  1 |  2 |.... |  n where Z is new nonterminal

21. 21 Left Factoring If   , replace all productions A →  1 |  2 |.... |  n |  with A →  Z |  Z →  1 |  2 |.... |  n where Z is new nonterminal

22. 22 Left Factoring A graphical explanation A  2  1  3

23. 23 Left Factoring becomes ZZ 22 11 33 A

24. 24 Left Factoring Consider following fragment of expression grammar Factor → id | id [ ExprList ] | id ( ExprList )

25. 25 Left Factoring Factor → id | id [ ExprList ] | id ( ExprList ) FIRST(rhs 1 ) = {id} FIRST(rhs 2 ) = {id} FIRST(rhs 3 ) = {id}

26. 26 Left Factoring after left factoring Factor → id Args Args → [ ExprList ] | ( ExprList ) | 

27. 27 Left Factoring Factor → id Args Args → [ ExprList ] | ( ExprList ) |  FIRST(rhs 1 ) = {id} FIRST(rhs 2 ) = { [ } FIRST(rhs 3 ) = { ( } FIRST(rhs 4 ) = FOLLOW(Factor)

28. 28 Left Factoring [ ( Factor id ExprList ] ) id No basis for choice

29. 29 Left Factoring [  ( Factor id ExprList ] ) correct choice possible

30. 30 Left Factoring Question 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?)

31. 31 Left Factoring Answer Given a CFG that does not meet the LL(1) condition, it is undecidable whether or not an equivalent LL(1) grammar exists