Compiler Design LR(0) Checking

Question

Is E --> E+T / T    ,   T --> i   is LR(0) grammar ?

Answer 1

1.   E'->.E

      E->.E+T / .T

      T-> .i

2. on shift E :

    E'->E. , E->E.+T  -  There is no conflict because E'->E. is like as E'->E.$, it is not recuding, it shows it is the end.

And there is no other conflict

So, It is LR(0)

Comments

This is correct answ

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u r correct

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correct

Answer 2

It is SLR grammar. Shift reduce conflict there but reduce reduce conflict will not be there if we derive it.

 S'-->E.  ,  E--> E. + T contains SR conflict.So, it is SLR.

Comments

it is both

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in LR(0) we are writing a reduced  term in all non-terminal while in a case of SLR(1) we are writing in the follow of nonterminal .is it a reason why you are telling not LR(0) but SLR(1).but as per my information due to augmented grammar we will write "accept" in just $ column so, in that case, it is both LR(0) and SLR(1) as well.

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srestha It's both LR(0) and SLR(1).

Answer 3

I think it is LR(0).