Lexical Analysis Questions and Answers

71.

Handling identifiers containing characters outside ASCII requires

A.rejecting such identifiers as lexical errors in every language
B.extending the character classes and the alphabet of the automaton
C.abandoning the finite automaton approach for token recognition
D.delegating identifier recognition to the semantic analyser

Answer: B

Correct answer: extending the character classes and the alphabet of the automaton

Nothing about the method breaks: the automaton simply works over a larger alphabet, and the character classes that define what may start or continue an identifier are widened to include the additional characters. The transition table grows, which is one more reason for grouping characters into classes rather than tabulating each one.

Discuss in Forum

72.

Character classes are used in scanner specifications chiefly to

A.let the parser rather than the scanner handle all the punctuation
B.avoid the need for a longest match rule during scanning
C.keep the specification short and the transition table compact
D.allow the scanner to recognise context free constructs

Answer: C

Correct answer: keep the specification short and the transition table compact

Grouping the twenty-six letters into a single class means the specification says letter once instead of listing them, and the transition table needs one column for the class rather than twenty-six. Both effects matter: the first for writing and maintaining the rules, the second for the size of the generated scanner.

Discuss in Forum

73.

A token for a numeric literal typically carries as its attribute

A.a pointer to a symbol table entry created for the literal
B.the line number on which the literal appeared in the source
C.the name of the type that the literal will eventually take
D.the numeric value obtained by converting the matched text

Answer: D

Converting the text to a value in the scanner saves later phases from repeating the work. Some compilers instead intern literals in a table so that identical constants share one entry. Interning saves space when the same constant appears many times in one program.

Discuss in Forum

74.

Distinguishing the unary minus from the binary minus is

A.beyond the scanner, since it depends on the surrounding context
B.done by the scanner using one character of lookahead
C.impossible in any language whatever, so both are treated alike
D.the responsibility of the code generator in the back end

Answer: A

Correct answer: beyond the scanner, since it depends on the surrounding context

The two uses of the minus sign are lexically identical, so no pattern can tell them apart; what distinguishes them is what came before - an operand implies binary, an operator or an opening bracket implies unary. That is contextual information the scanner does not have, so it emits one token and the parser decides which production applies.

Discuss in Forum

75.

State whether the following is true or false.
A scanner can determine whether an identifier has been declared.

A.True
B.False
C.True only at large scales
D.False only at small scales

Answer: B

The scanner may create or find a symbol table entry, but declaration status depends on scope and on statements the scanner has no view of. Semantic analysis makes that determination once the tree and table are complete.

Discuss in Forum

76.

The token returned for a keyword such as while is normally

A.no token at all
B.the same token as any other identifier in the program
C.a distinct token name for that keyword with no attribute needed
D.an identifier token whose attribute names the keyword

Answer: C

Giving each keyword its own token name lets the grammar mention it directly in productions. No attribute is required because the token name already determines the lexeme completely. Identifiers by contrast need an attribute, since the name itself must be carried along.

Discuss in Forum

77.

Scanning of a language with significant indentation, such as Python, requires the scanner to

A.ignore all leading white space exactly as other languages do
B.hand the indentation to the parser as ordinary space characters
C.reject any line at all whose indentation differs from the previous one
D.emit synthetic indent and dedent tokens by tracking a stack of levels

Answer: D

The scanner keeps a stack of indentation widths and emits tokens as levels open and close. The grammar then treats block structure with ordinary productions over those tokens. The stack of open levels is the extra state that a pure finite automaton could not hold.

Discuss in Forum

78.

Assertion (A): Indentation sensitive languages need a scanner that maintains state between lines.
Reason (R): The meaning of a line's leading white space depends on the indentation of the lines before it.

A.Both A and R are true, and R is the correct explanation of A
B.Both A and R are true, but R is not the correct explanation of A
C.A is true but R is false
D.A is false but R is true

Answer: A

Both are true and the reason gives the mechanism, and this pushes the scanner slightly beyond a pure finite automaton. The stack of levels is extra state carried in the semantic actions rather than in the automaton itself.

Discuss in Forum

79.

String literals containing escape sequences are usually handled by

A.rejecting any literal that contains a backslash character
B.recognising the whole literal and translating the escapes in the action
C.emitting a separate token for each escape sequence encountered
D.leaving the escapes untranslated for the code generator to deal with

Answer: B

The pattern matches the literal including its escapes, and the semantic action builds the actual string value. Doing the translation in the scanner means later phases see a plain value. The original spelling is usually retained as well so diagnostics can quote the source.

Discuss in Forum

80.

A scanner that must handle both /* */ and // comments will

A.require the parser to decide where each comment ends
B.need a pushdown automaton rather than a finite one
C.use separate patterns or start conditions for the two forms
D.be unable to recognise them and must reject one of the forms

Answer: C

Neither form nests, so both are regular and can be matched by patterns. Start conditions are a convenient way to switch the scanner into a mode in which only the closing delimiter matters. Without modes the patterns would have to describe every possible comment body explicitly.

Discuss in Forum

Loading…

Topics

Other Subjects

Tags: Lexical Analysis Questions and Answers || Lexical Analysis MCQ Questions and Answers || Lexical Analysis GK Questions and Answers || Lexical Analysis GK MCQ Questions || Lexical Analysis Multiple Choice Questions and Answers || Lexical Analysis GK || GK on Lexical Analysis || Compiler Design Questions and Answers || Compiler Design MCQ Questions and Answers || Compiler Design GK Questions and Answers || GK on Compiler Design

Lexical Analysis Questions and Answers

Topics

Other Subjects

About Us

Stay informed about all important Dates & Notifications