I1 state have the final item which drives E => E and follow(E) = ($), so action (I1 $) = Accept.

I5 = Go to on (I2 B) = Closure (E => BB.)įollow of B = First of B = c and d and Follow of E = $ I4 = Go to on (I0 d) = Closure (B => d.) = B => d. Go to on (I3 d) = Closure (B => d.) (Same as I4)

Goto (I2 c) = Closure (B => c.B) (Same as I3) We apply closure and will get all the A production with '.' in the beginning. I2 = we have applied Goto on (I0 B) = Closure (E => B.B) Here, the '.' is on the right side of the symbol, so production is reduced so close the state. I1 = Here we have applied Goto on (I0 E) = closure (E’ => E.) Therefore, all productions beginning with B will be added in the I0 State because ' Here we can see that “.” is on the left of the variable B. Therefore, all productions beginning with E will be added in the I0 State because '.' is followed by the non-terminal. It means we have not seen anything on the right-hand side. Since '.' is on the left side of the symbol. Check if the grammar is ambiguous or not.Īdd augment production and insert '.' symbol in the first position for every production in G.Īdd starting production to the I0 State and Compute the Closure.

The step involves in SLR(1) parsing is given below: The reduce move is placed only in the FOLLOW of those variables whose production is reduced. The SLR(1) parsing use canonical LR(0) item. The parsing table for both the parser vary. Most of the function of this parsing is the same as LR(0) parsing.

Slr parsing table program in c