Register

Conflict

• retagged by
814 views

1 Answer

Best answer
3 votes
3 votes

It is true.

Because once we construct CLR table and it has no conflict then we go for merging the

states(LALR) and still there are chances for Reduce-Reduce Conflict in LALR table.

- But question says LALR has no conflict, then surely CLR has no conflict.

• selected by
User Avatar
Voting & Accepting an answer helps improve the community
← PreviousNext →
← Previous in categoryNext in category →

Related questions

2 votes
2 votes
1 answer
1
arch asked Nov 19, 2017
308 views
parser
every LL(k) grammer is also SLR(k) grammer. (true / false)
every LL(k) grammer is also SLR(k) grammer. (true / false)
User Avatar
2 votes
2 votes
2 answers
2
kallu singh asked Sep 5, 2017
692 views
Please solve my doubt
Q if any Grammar is LL(1) definitely LALR(1) ? It is true or false Please ans explain in detail.
Q if any Grammar is LL(1) definitely LALR(1) ?It is true or falsePlease ans explain in detail.
User Avatar
1 votes
1 votes
2 answers
3
gate_forum asked Dec 31, 2016
612 views
parsing
State True/False : "LR(0)⊂SLR(1)⊂LALR(1)⊂LR(1)".
State True/False : "LR(0)⊂SLR(1)⊂LALR(1)⊂LR(1)".
User Avatar
0 votes
0 votes
2 answers
4
sripo asked Nov 1, 2018
2,397 views
Is the given grammar CLR(1) or not please explain me if there is a shift reduce conflict
S→(X S→E] S→E) X→E) X→E] E→ϵ Is this grammar CLR(1)? The answer says it is but I find a shift reduce conflict for E-> epsilon with lookup symbols ),]
S→(XS→E]S→E)X→E)X→E]E→ϵIs this grammar CLR(1)? The answer says it is but I find a shift reduce conflict for E- epsilon with lookup symbols ),]
User Avatar
Link Copied!