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Que. Match all items in Group 1 with correct options from those given in Group 2.

Group 1 Group 2

P. Regular expression 1. Syntax analysis
Q. Pushdown automata 2. Code generation
R. Dataflow analysis 3. Lexical analysis
S. Register allocation 4. Code optimization
a. P-4. Q-1, R-2, S-3
b. P-3, Q-1, R-4, S-2
c. P-3, Q-4, R-1, S-2
d. P-2, Q-1, R-4, S-3
Answer:P-3, Q-1, R-4, S-2
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Similar Questions:

Consider the following statements:

(I) The output of a lexical analyzer is groups of characters.
(II) Total number of tokens in printf("i=%d, &i=%x", i, &i); are 11.
(III) Symbol table can be implementation by using array and hash table but not tree.

Which of the following statement(s) is/are correct?
Consider the following grammar G.

S → F ⎪ H
F → p ⎪ c
H → d ⎪ c

Where S, F and H are non-terminal symbols, p, d and c are terminal symbols. Which of the following statement(s) is/are correct?

S1: LL(1) can parse all strings that are generated using grammar G.
S2: LR(1) can parse all strings that are generated using grammar G.
Which one of the following is True at any valid state in shift-reduce parsing?
The number of tokens in the following C statement is

printf("i = %d, &i = %x", i, &i);
Consider the grammar with the following translation rules and E as the start symbol.

E → E1 # T { E.value = E1.value * T.value }
| T{ E.value = T.value }
T → T1 & F { T.value = T1.value + F.value }
| F{ T.value = F.value }
F → num { F.value = num.value }

Compute E.value for the root of the parse tree for the expression: 2 # 3 & 5 # 6 & 4.
Match all items in Group 1 with correct options from those given in Group 2.

Group 1 Group 2

P. Regular expression 1. Syntax analysis
Q. Pushdown automata 2. Code generation
R. Dataflow analysis 3. Lexical analysis
S. Register allocation 4. Code optimization
Consider the grammar defined by the following production rules, with two operators ∗ and +

S --> T * P
T --> U | T * U
P --> Q + P | Q
Q --> Id
U --> Id

Which one of the following is TRUE?
Consider the grammar with non-terminals N = {S,C,S1 },terminals T={a,b,i,t,e}, with S as the start symbol, and the following set of rules:

S --> iCtSS1|a
S1 --> eS|ϵ
C --> b

The grammar is NOT LL(1) because:
Consider the grammar

S → (S) | a

Let the number of states in SLR(1), LR(1) and LALR(1) parsers for the grammar be n1, n2 and n3 respectively. The following relationship holds good
Consider the grammar shown below.

S → C C
C → c C | d

The grammar is

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