Section 4.8 Aggelos Kiayias Computer Science & Engineering Department

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Section 4.8 Aggelos Kiayias Computer Science & Engineering Department The University of Connecticut 371 Fairfield Road, Box U-155 Storrs, CT 06269-1155 aggelos@cse.uconn.edu http://www.cse.uconn.edu/~akiayias

Ambiguous Grammars Ambiguous Grammars produce “conflicts” Shift/reduce, or reduce/reduce. Three “typical examples” The dangling else The “expression” grammar Eqn pre-processor of Troff

Dangling Else Ambiguity Recall Grammar Can be rewritten as: S  iSeS | iS | a We compute LR(0) items + Goto and the SLR Parsing table. stmt  if expr then stmt | if expr then stmt else stmt | other (any other statement)

Canonical sets of LR(0) items I0 I3 S’  .S S  a. S  .iSeS S  .iS I4 S  .a S  iS.eS S  iS. I1 S’ S. I5 S  iSe.S I2 S  .iSeS S  i.SeS S  .iS S  i.S S  .a S  .iS I6 S  .a S  iSeS. Follow(S)= $,e

Parsing Table i e a $ S 0 s2 s3 1 1 acc 2 s2 s3 4 3 r3 r3 4 s5/r2 r2 1. S iSeS 2. S iS 3. S a i e a $ S 0 s2 s3 1 1 acc 2 s2 s3 4 3 r3 r3 4 s5/r2 r2 5 s2 s3 6 6 r1 r1 Resolve: choose s5 e.g. parse iiaea to understand the conflict resolution.

Tracing $0 iiaea$ STACK INPUT Remark

Expressions Recall grammar: Originally written as: E  E + E | E *E | (E) | id In fact we can try to construct the SLR parsing table and see what will happen E’  E E  E + T | T T  T * F | F F  ( E ) | id

Canonical Sets I0 I2 I5 I8 I1 I3 I6 I9 Follow(E’)=$ Follow(E)=+*)$ E’  .E E  .( E ) E  E * . E E  E * E . E  .E + E E  .E + E E  .E + E E  E . + E E  .E * E E  .E * E E  .E * E E  E . * E E  .( E ) E  .( E ) E  .( E ) E  .id E  .id E  .id I1 I3 I6 I9 E’  E. E  id. E  ( E . ) E  ( E ) . E  E . + E E  E . + E E  E . * E I4 E  E . * E E  E + . E E  .E + E I7 E  .E * E E  E + E . E  .( E ) E  E . + E E  .id E  E . * E Follow(E’)=$ Follow(E)=+*)$

The Parsing Table id + * ( ) $ E 0 s3 s2 1 1 s4 s5 acc 2 s3 s2 6 1. E E+E 2. E E*E 3. E (E) 4. E id id + * ( ) $ E 0 s3 s2 1 1 s4 s5 acc 2 s3 s2 6 3 r4 r4 r4 r4 4 s3 s2 7 5 s3 s2 8 6 s4 s5 s9 7 s4/r1 s5/r1 r1 r1 8 s4/r2 s5/r2 r2 r2 9 r3 r3 r3 r3

Resolving Ambiguity If the state contains E  E op E . and we find op on the input we Reduce (left-association) Shift (right-association) If the state contains E  E op E . and we find op’ on the input we Reduce (op’ has lower precedence than op) Shift (op’ has higher precedence than op)

Subscript/Superscript 1. E EsubEsupE 2. E EsubE 3. E EsupE 4. E {E} 5. E c Even if we resolve conflicts using precedence and associations Rule 1 will still produce problems, i.e. when seeing a } or $ Being at the state that contains the item E EsubEsupE. We will have two possibilities for reduce: Either r3 or r1. (reduce/reduce conflict) Resolving for r1 allows a special meaning for it.