Slr parser
- 1. SLR PARSER Steps: 1. create augment grammar 2. generate kernel items 3. find closure 4. compute goto() 5. construct parsing table 6. parse the string Let us consider grammar: S->a S->(L) L->S L->L,S
- 2. • Step :1 Create augment grammar S is start symbol in the given grammar Augment grammar is S’-> S • Step :2 Generate kernel items Introduce dot in RHS of the production`` S’-> .S • Step :3 Find closure (Rule: A -> α.Xβ i.e if there is nonterminal next to dot then Include X production)) S’-> .S S-> . a S->.(L) I0
- 3. Step :4 compute goto() S’-> S . I1 Goto (I0,a) S-> a . I2 Goto (I0,( ) S’-> .S S-> . a S->.(L) I0 S-> (.L) L->.S L>.L,S S->.a S->.(L) I3 Goto (I3,L ) S-> (L .) L->L . ,S I4 Goto (I3,S ) L->S . I5 Goto (I3,a ) S->a . I2 Goto (I3, ( ) S-> (. L ) L->.S L>.L,S S->.a S->.(L) I3 Goto (I0,S)
- 4. Goto (I4, ) ) S-> (L) . I6 Goto (I4, , ) L->L , . S S-> . a S->.(L) I7 Goto (I7,S ) L->L , S . I8 Goto (I7,a ) S-> a . I2 Goto (I7,( ) S-> (.L) L->.S L>.L,S S->.a S->.(L) I3 label Production Ending iteration R0 S’-> S . I1 R1 S->a . I2 R2 S->(L) . I6 R3 L->S . I5 R4 L->L,S . I8
- 5. Step :5 construct parsing table Label the rows of table with no. of iterations Divide the Column into two parts • Action part: terminal symbols • Goto part: Nonterminal symbols Terminals Non terminals a ( ) , $ S L 0 1 2 3 4 5 6 7 8
- 6. Step :5 construct parsing table from step 4: Label the rows of table with no. of iterations Divide the Column into two parts • Action part: terminal symbols • Goto part: Nonterminal symbols ACTION GOTO a ( ) , $ S L 0 S2 S3 1 1 2 3 S2 S3 5 4 4 S6 S7 5 6 7 S2 S3 8 8 Goto (I0, S) 1 Goto (I3, ( ) 3 Goto (I0, a) 2 Goto (I4, ) ) 6 Goto (I0, ( ) 3 Goto (I4, ,) 7 Goto (I3, L) 4 Goto (I7, S) 8 Goto (I3, S) 5 Goto (I7, a) 2 Goto (I3, a) 2 Goto (I7, ( ) 3
- 7. Step :5 construct parsing table Label the rows of table with no. of iterations Divide the Column into two parts • Action part: terminal symbols • Goto part: Nonterminal symbols ACTION GOTO a ( ) , $ S L 0 S2 S3 1 1 accept 2 R1 R1 R1 3 S2 S3 5 4 4 S6 S7 5 R3 R3 6 R2 R2 R2 7 S2 S3 8 8 R4 R4 label Production Ending iteration Follow(LHS) R0 S’-> S . I1 Follow(S’)={ $ } R1 S-> a . I2 Follow(S)={ $ ) , } R2 S-> ( L ) . I6 Follow(S)={ $ ) , } R3 L-> S . I5 Follow(L)={ ) , } R4 L-> L , S . I8 Follow(L)={ ) , }
- 8. Step :5 construct parsing table Label the rows of table with no. of iterations Divide the Column into two parts • Action part: terminal symbols • Goto part: Nonterminal symbols ACTION GOTO a ( ) , $ S L 0 S2 S3 1 1 accept 2 R1 R1 R1 3 S2 S3 5 4 4 S6 S7 5 R3 R3 6 R2 R2 R2 7 S2 S3 8 8 R4 R4 label Production Ending iteration Follow(LHS) R0 S’-> S . I1 Follow(S’)={ $ } R1 S-> a . I2 Follow(S)={ $ ) , } R2 S-> ( L ) . I6 Follow(S)={ $ ) , } R3 L-> S . I5 Follow(L)={ ) , } R4 L-> L , S . I8 Follow(L)={ ) , } Goto (I0, S) 1 Goto (I3, ( ) 3 Goto (I0, a) 2 Goto (I4, ) ) 6 Goto (I0, ( ) 3 Goto (I4, ,) 7 Goto (I3, L) 4 Goto (I7, S) 8 Goto (I3, S) 5 Goto (I7, a) 2 Goto (I3, a) 2 Goto (I7, ( ) 3
- 9. Step :6 String parsing Let string w=(a,a) derived from the given grammar S – Shift action * shift input symbol to the stack * shift number to the stack R – Reduce action * pop 2 times of RHS symbols * shift LHS to the stack * find the number in the parsing table for the last two elements in the stack ACTION GOTO a ( ) , $ S L 0 S2 S3 1 1 accept 2 R1 R1 R1 3 S2 S3 5 4 4 S6 S7 5 R3 R3 6 R2 R2 R2 7 S2 S3 8 8 R4 R4 Stack Input Action $0 (a,a)$ S3 label Production R0 S’-> S . R1 S-> a . R2 S-> ( L ) . R3 L-> S . R4 L-> L , S .
- 10. Step :6 String parsing Let string w=(a,a) derived from the given grammar S – Shift action * shift input symbol to the stack * shift number to the stack R – Reduce action * pop 2 times of RHS symbols * shift LHS to the stack * find the number in the parsing table for the last two elements in the stack ACTION GOTO a ( ) , $ S L 0 S2 S3 1 1 accept 2 R1 R1 R1 3 S2 S3 5 4 4 S6 S7 5 R3 R3 6 R2 R2 R2 7 S2 S3 8 8 R4 R4 Stack Input Action $0 (a,a)$ S3 $0(3 a,a)$ S2 label Production R0 S’-> S . R1 S-> a . R2 S-> ( L ) . R3 L-> S . R4 L-> L , S .
- 11. Step :6 String parsing Let string w=(a,a) derived from the given grammar S – Shift action * shift input symbol to the stack * shift number to the stack R – Reduce action * pop 2 times of RHS symbols * shift LHS to the stack * find the number in the parsing table for the last two elements in the stack ACTION GOTO a ( ) , $ S L 0 S2 S3 1 1 accept 2 R1 R1 R1 3 S2 S3 5 4 4 S6 S7 5 R3 R3 6 R2 R2 R2 7 S2 S3 8 8 R4 R4 Stack Input Action $0 (a,a)$ S3 $0(3 a,a)$ S2 $0(3a2 ,a)$ R1 label Production R0 S’-> S . R1 S-> a . R2 S-> ( L ) . R3 L-> S . R4 L-> L , S .
- 12. Step :6 String parsing Let string w=(a,a) derived from the given grammar S – Shift action * shift input symbol to the stack * shift number to the stack R – Reduce action * pop 2 times of RHS symbols * shift LHS to the stack * find the number in the parsing table for the last two elements in the stack ACTION GOTO a ( ) , $ S L 0 S2 S3 1 1 accept 2 R1 R1 R1 3 S2 S3 5 4 4 S6 S7 5 R3 R3 6 R2 R2 R2 7 S2 S3 8 8 R4 R4 Stack Input Action $0 (a,a)$ S3 $0(3 a,a)$ S2 $0(3a2 ,a)$ R1 $0(3S5 ,a)$ R3 label Production R0 S’-> S . R1 S-> a . R2 S-> ( L ) . R3 L-> S . R4 L-> L , S .
- 13. Step :6 String parsing Let string w=(a,a) derived from the given grammar S – Shift action * shift input symbol to the stack * shift number to the stack R – Reduce action * pop 2 times of RHS symbols * shift LHS to the stack * find the number in the parsing table for the last two elements in the stack ACTION GOTO a ( ) , $ S L 0 S2 S3 1 1 accept 2 R1 R1 R1 3 S2 S3 5 4 4 S6 S7 5 R3 R3 6 R2 R2 R2 7 S2 S3 8 8 R4 R4 Stack Input Action $0 (a,a)$ S3 $0(3 a,a)$ S2 $0(3a2 ,a)$ R1 $0(3S5 ,a)$ R3 $0(3L4 ,a)$ S7 label Production R0 S’-> S . R1 S-> a . R2 S-> ( L ) . R3 L-> S . R4 L-> L , S .
- 14. Step :6 String parsing Let string w=(a,a) derived from the given grammar S – Shift action * shift input symbol to the stack * shift number to the stack R – Reduce action * pop 2 times of RHS symbols * shift LHS to the stack * find the number in the parsing table for the last two elements in the stack ACTION GOTO a ( ) , $ S L 0 S2 S3 1 1 accept 2 R1 R1 R1 3 S2 S3 5 4 4 S6 S7 5 R3 R3 6 R2 R2 R2 7 S2 S3 8 8 R4 R4 Stack Input Action $0 (a,a)$ S3 $0(3 a,a)$ S2 $0(3a2 ,a)$ R1 $0(3S5 ,a)$ R3 $0(3L4 ,a)$ S7 label Production R0 S’-> S . R1 S-> a . R2 S-> ( L ) . R3 L-> S . R4 L-> L , S .
- 15. Step :6 String parsing Let string w=(a,a) derived from the given grammar S – Shift action * shift input symbol to the stack * shift number to the stack R – Reduce action * pop 2 times of RHS symbols * shift LHS to the stack * find the number in the parsing table for the last two elements in the stack ACTION GOTO a ( ) , $ S L 0 S2 S3 1 1 accept 2 R1 R1 R1 3 S2 S3 5 4 4 S6 S7 5 R3 R3 6 R2 R2 R2 7 S2 S3 8 8 R4 R4 Stack Input Action $0 (a,a)$ S3 $0(3 a,a)$ S2 $0(3a2 ,a)$ R1 $0(3S5 ,a)$ R3 $0(3L4 ,a)$ S7 $0(3L4 , 7 a)$ S2 label Production R0 S’-> S . R1 S-> a . R2 S-> ( L ) . R3 L-> S . R4 L-> L , S .
- 16. Step :6 String parsing Let string w=(a,a) derived from the given grammar S – Shift action * shift input symbol to the stack * shift number to the stack R – Reduce action * pop 2 times of RHS symbols * shift LHS to the stack * find the number in the parsing table for the last two elements in the stack ACTION GOTO a ( ) , $ S L 0 S2 S3 1 1 accept 2 R1 R1 R1 3 S2 S3 5 4 4 S6 S7 5 R3 R3 6 R2 R2 R2 7 S2 S3 8 8 R4 R4 Stack Input Action $0 (a,a)$ S3 $0(3 a,a)$ S2 $0(3a2 ,a)$ R1 $0(3S5 ,a)$ R3 $0(3L4 ,a)$ S7 $0(3L4 , 7 a)$ S2 $0(3L4 , 7a2 )$ R1 label Production R0 S’-> S . R1 S-> a . R2 S-> ( L ) . R3 L-> S . R4 L-> L , S .
- 17. Step :6 String parsing Let string w=(a,a) derived from the given grammar S – Shift action * shift input symbol to the stack * shift number to the stack R – Reduce action * pop 2 times of RHS symbols * shift LHS to the stack * find the number in the parsing table for the last two elements in the stack ACTION GOTO a ( ) , $ S L 0 S2 S3 1 1 accept 2 R1 R1 R1 3 S2 S3 5 4 4 S6 S7 5 R3 R3 6 R2 R2 R2 7 S2 S3 8 8 R4 R4 Stack Input Action $0 (a,a)$ S3 $0(3 a,a)$ S2 $0(3a2 ,a)$ R1 $0(3S5 ,a)$ R3 $0(3L4 ,a)$ S7 $0(3L4 , 7 a)$ S2 $0(3L4 , 7a2 )$ R1 $0(3L4 , 7S8 )$ R4 label Production R0 S’-> S . R1 S-> a . R2 S-> ( L ) . R3 L-> S . R4 L-> L , S .
- 18. Step :6 String parsing Let string w=(a,a) derived from the given grammar S – Shift action * shift input symbol to the stack * shift number to the stack R – Reduce action * pop 2 times of RHS symbols * shift LHS to the stack * find the number in the parsing table for the last two elements in the stack ACTION GOTO a ( ) , $ S L 0 S2 S3 1 1 accept 2 R1 R1 R1 3 S2 S3 5 4 4 S6 S7 5 R3 R3 6 R2 R2 R2 7 S2 S3 8 8 R4 R4 Stack Input Action $0 (a,a)$ S3 $0(3 a,a)$ S2 $0(3a2 ,a)$ R1 $0(3S5 ,a)$ R3 $0(3L4 ,a)$ S7 $0(3L4 , 7 a)$ S2 $0(3L4 , 7a2 )$ R1 $0(3L4 , 7S8 )$ R4 $0(3L4 ) $ S6 label Production R0 S’-> S . R1 S-> a . R2 S-> ( L ) . R3 L-> S . R4 L-> L , S .
- 19. Step :6 String parsing Let string w=(a,a) derived from the given grammar S – Shift action * shift input symbol to the stack * shift number to the stack R – Reduce action * pop 2 times of RHS symbols * shift LHS to the stack * find the number in the parsing table for the last two elements in the stack ACTION GOTO a ( ) , $ S L 0 S2 S3 1 1 accept 2 R1 R1 R1 3 S2 S3 5 4 4 S6 S7 5 R3 R3 6 R2 R2 R2 7 S2 S3 8 8 R4 R4 Stack Input Action $0 (a,a)$ S3 $0(3 a,a)$ S2 $0(3a2 ,a)$ R1 $0(3S5 ,a)$ R3 $0(3L4 ,a)$ S7 $0(3L4 , 7 a)$ S2 $0(3L4 , 7a2 )$ R1 $0(3L4 , 7S8 )$ R4 $0(3L4 ) $ S6 $0(3L4)6 $ R2 label Production R0 S’-> S . R1 S-> a . R2 S-> ( L ) . R3 L-> S . R4 L-> L , S .
- 20. Step :6 String parsing Let string w=(a,a) derived from the given grammar S – Shift action * shift input symbol to the stack * shift number to the stack R – Reduce action * pop 2 times of RHS symbols * shift LHS to the stack * find the number in the parsing table for the last two elements in the stack ACTION GOTO a ( ) , $ S L 0 S2 S3 1 1 accept 2 R1 R1 R1 3 S2 S3 5 4 4 S6 S7 5 R3 R3 6 R2 R2 R2 7 S2 S3 8 8 R4 R4 Stack Input Action $0 (a,a)$ S3 $0(3 a,a)$ S2 $0(3a2 ,a)$ R1 $0(3S5 ,a)$ R3 $0(3L4 ,a)$ S7 $0(3L4 , 7 a)$ S2 $0(3L4 , 7a2 )$ R1 $0(3L4 , 7S8 )$ R4 $0(3L4 ) $ S6 $0(3L4)6 $ R2 $0S1 $ accept Hence the string parsed successfully !!!!!! label Production R0 S’-> S . R1 S-> a . R2 S-> ( L ) . R3 L-> S . R4 L-> L , S .