SLR(1) Parsing

Slr Parsing Table Program In C Programming
Slr Parsing Table Program In C++
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SLR Parser Generator. In other words, they use different algorithms to derive the parsing tables from the grammar. A SLR parser is quite restrictive in practical terms and it is not very used. A LALR parser instead works for most practical grammars and it is widely employed. In fact the popular tools yacc and bison works with LALR parser tables. Slr Parsing Table Program In C Ti 84 Plus Ce-t Games Wonderware Alarming Acknowledgeg Updates Alarm Comments 3ds Inquiry Number Navionics Web App Can You Get High From Just A Few Crumbs Of Weed Little Weed. And SLR parse table construction Wim Bohm and Michelle Strout CS, CSU CS453 Lecture Building LR Parse Tables 1. Input File For SLR Parser: S S+T S T T T.F T F F (S) F t. Posted by Karthik Alapati. Email This BlogThis! Share to Twitter Share to Facebook Share to Pinterest. CONSTRUCTION OF PREDICTIVE PARSER TABLE USING C PROGRAM: #include #include #include void. SLR parsing SLR parsing is LR(0) parsing, but with a different reduce rule: For each edge (X: (I, J)) if X is terminal, put shift J at (I, X) if I contains A!α. Has rule number n for each terminal x in Follow(A), put reduce reduce n at (I, x) Build an SLR parser for our expression grammar 0:S!E$1:E!E+T!!!2:E!T3:T!

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

The step involves in SLR(1) parsing is given below:

Write a CFG for the given input string
Check if the grammar is ambiguous or not.
Add an augmented grammar.
Create a canonical LR(0) item.
Draw DFA
Construct an SLR(1) parsing table.

Example

E => BB

B => cB / d

Add augment production and insert ‘.’ symbol in the first position for every production in G.

E’ => .E

E => .BB

B => .cB

B => .d

I0 state

Add starting production to the I0 State and Compute the Closure.

I0 = closure (E’ => .E)

Since “.” is on the left side of the symbol. It means we have not seen anything on the right-hand side. Therefore, all productions beginning with E will be added in the I0 State because “.” is followed by the non-terminal. So, the modified I0 State will be:

I0 = E’ => .E

E => .BB

Here we can see that “.” is on the left of the variable B. Therefore, all productions beginning with B will be added in the I0 State because “•” is followed by the non-terminal. So, the modified I0 State becomes:

E’ => .E

E => .BB

B => .cB

B => .d

I1 = Here we have applied Goto on (I0 E) = closure (E’ => E.)

Here, the “.” is on the right side of the symbol, so production is reduced so close the state.

I1 becomes E’ => E.

I2 = we have applied Goto on (I0 B) = Closure (E => B.B)

We apply closure and will get all the A production with “.” in the beginning. So I2 becomes:

E => B.B

B => .cB

B => .d

Goto (I2 c) = Closure (B => c.B) (Same as I3)

Goto on (I2 d) = Closure (B => .d) (Same as I4)

I3 = Goto on (I0 c) = Closure (B => c.B)

Add productions starting with B in I3.

B => c.B

B => .cB

B => .d

Go to on (I3 c) = Closure (B => c. B) (Same as I3)

Go to on (I3 d) = Closure (B => d.) (Same as I4)

I4 = Go to on (I0 d) = Closure (B => d.) = B => d.

I5 = Go to on (I2 B) = Closure (E => BB.)

I6 = Go to on (I3 B) = Closure (B =>cB.)

Productions are numbered as follows:

E =>BB (1)

B => cB (2)

B => d (3)

Drawing DFA

SLR(1) Table

Explanations:

Slr Parsing Table Program In C Programming

Follow of B = First of B = c and d and Follow of E = $

I1 state have the final item which drives E => E and follow(E) = ($), so action (I1 $) = Accept
I4 state have the final item which drives B => d and follow(B) = c, d and follow(E) = $ , So action c, d, and $ will be r3
I5 state have the final item which drives E => BB and follow(E) = $, So action $ in I5 will be r1
I6 state have the final item which drives B => cB and follow(B) = c, d and follow(E) =$, So action c, d, and $ in I6 will be r2.

NOTE:

When a state moves to some other state on terminals, it will correspond to a shift move in the action part.

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When a state moves to another state on a variable, it will correspond to the goto move in the go-to part.

Slr Parsing Table Program In Ct

When a state has a final item with no transition to the next state, the production is known as reduce production.