Test by Bikram | Compiler Design | Test 1 | Question: 1

Question

Match the following:

$$\begin{array}{|l|l|l|l|} \hline a. & \text{Canonical Parser} & 1. & \text{No adjacent nonterminals} \\ \hline b. & \text{SLR(1) Parser} & 2. & \text{Follow sets must be disjoint} \\ \hline c. & \text{LL(1) Parser} & 3. & \text{Most powerful parser} \\ \hline d. & \text{Operator Precedence Grammar} & 4. & \text{Top-down parser} \\ \hline \end{array}$$

• A. $a-2; b-3; c-1; d-4$
• B. $a-3; b-4; c-2; d-1$
• C. $a-2; b-1; c-4; d-3$
• D. $a-3; b-2; c-4; d-1$

Answer 1

1. Operator precedence grammar have no two non-terminals are adjacent and no epsilon move.
2. If in case one final item and one reduce move  $A\rightarrow p.ad , B\rightarrow b.\:FOLLOW(A) \cap FOLLOW (B)=\phi$ ,That means follow(A) and follow(B) set are disjoint(means no common terminals ) set.
3. Canonical parser means CLR(1)=LR(1)= most powerful parser.
4. LL(1) parser is Top-down parser.

So, the correct answer is $(D)$.

References:

• http://www.cs.ecu.edu/karl/5220/spr16/Notes/Top-down/LL1.html
• http://www.cs.ecu.edu/karl/5220/spr16/Notes/Bottom-up/lr1.html
• http://www.cs.ecu.edu/karl/5220/spr16/Notes/Bottom-up/slr1table.html
• https://cse.iitkgp.ac.in/~bivasm/notes/scribe/11CS10042.pdf
• https://www.cs.clemson.edu/course/cpsc827/material/Operator%20Precedence/Parsing.pdf