Test: Lexical Analysis- 2 - Computer Science Engineering (CSE) MCQ

In 2-pass assembler during pass-1 pseudo code EQU is to be evaluated.

Operation performed during pass-1 in 2-pass compiler:
1. Processing the DB pseudo-op.
2. Updating the location counter.
3. Processing of EQU pseudo-op.
4. Processing of DS pseudo-op.

The context free grammar is represented as
S → X

where S∈ V(Single symbol non-terminal)

D → (V ∪ T ) * (Any number of terminal and nonterminals)

In incomplete automata some states have no transition on some tokens.

Lexical analysis consists of the phases that
(i) Simplify the subsequent phases by the help of tokens instead of arbitrary elements.
(ii) Since interrelated elements are converted to stream of tokens hence compiler efficiency is improved as no arbitrary storage or value is to be checked independently.
(iii) Computer portability is enhanced.

Grammar Production: S → SaS | aSb | bSa | SS | ∈

Option 1: String: abab

S → aSb

S → abSab 

S → ab∈ab

S → abab

So, abab can be generated by the given grammar.

Option 2: String: aaab

S → SS

S → SaSS 

S → SaSaSS 

S → aaaSbS 

S → aaa∈bS 

S → aaa∈b∈ 

S → aaab

Option 3: String:  abbaa

S → SS

S → aSbS 

S → a∈bS 

S → abbSa  

S → abbSaSa 

S → abb∈aSa

S → abba∈a

S → abbaa  

All three abab, aaab and abbaa are generated by the given grammar.

Only babba can’t be generated.

Given two DFA’s M₁ and M₂. They are equivalent if the language generated by them is same i,e, L(M₁) = L(M₂)

S → aAS, S→ a, A → SbA, A→  SS, A → ba}, S

Input: aabbaa

Modules may be compiled separately and all linking between references is made at link time.

Error is underlined function fro()
Which is a syntactical error rather than lexical.

Here grammar, G₂ is not ambiguous 

Because there exist only one parse tree.

Hash table can contribute to an efficient average case solution for symbol table look up problem

S — ABc — abc

Mixup notation of left linear and right linear grammar is not allowed in regular grammar.