The dangling else ambiguity is normally resolved by
Answer: C
Either the grammar is rewritten to distinguish matched from unmatched statements, or the parser generator is told to prefer shifting the else. Both realise the same nearest if rule. Languages requiring an explicit end keyword avoid the ambiguity altogether.
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State whether the following is true or false.
An ambiguous grammar can sometimes be rewritten as an unambiguous grammar for the same language.
Answer: D
Correct answer: True
The dangling else is the standard demonstration: splitting the statement nonterminal into matched and unmatched forms gives an unambiguous grammar generating exactly the same language. It cannot always be done - some languages are inherently ambiguous - but for programming language constructs it usually can, and precedence and associativity are handled the same way.
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An expression grammar is made to reflect operator precedence by
Answer: A
Correct answer: introducing one nonterminal level per precedence level
Each level of the hierarchy - expression, term, factor - corresponds to one precedence level, and because a lower level can only be reached through a higher one, the tighter-binding operators end up deeper in the tree. Associativity follows from whether the recursion is on the left or the right of the production.
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A production written as E goes to E plus T makes the plus operator
Answer: B
Recursion on the left produces left leaning trees, so a plus b plus c groups as (a plus b) plus c. Recursion on the right would make the operator right associative instead. Associativity is therefore a property of how the production is written.
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A production written as E goes to T equals E makes the equals operator
Answer: C
Right recursion produces right leaning trees, so a equals b equals c groups as a equals (b equals c). Assignment is conventionally right associative for exactly this reason. Comparison operators by contrast are usually left associative or non associative.
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Left recursion in a grammar is a problem for
Answer: D
Correct answer: top-down parsers, which would recurse without consuming input
A top-down parser expanding A by the production A → Aα immediately faces A again with no token consumed, so it recurses forever. A bottom-up parser has no such difficulty and in fact prefers left recursion, since it keeps the stack shallow.
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Eliminating immediate left recursion from A goes to A alpha or beta gives
Answer: A
Correct answer: A goes to beta A prime, with A prime going to alpha A prime or epsilon
The original generates β followed by any number of αs, and so does the replacement: A produces the β once, and A′ produces the repetitions and then stops with ε. The recursion has been turned from left to right, which is what a top-down parser needs, and the language is unchanged.
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Assertion (A): Removing left recursion changes the shape of the parse tree.
Reason (R): The transformed grammar is right recursive, so the tree leans the other way.
Answer: B
Both are true and the reason gives the mechanism, and this is why the semantic actions must be rewritten too. Associativity is recovered by carrying an inherited attribute through the new nonterminal rather than by the tree shape.
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Left factoring is applied when
Answer: C
Correct answer: two productions for a nonterminal share a common prefix
If two alternatives begin with the same symbols, one token of lookahead cannot tell them apart. Left factoring pulls the common prefix out into a single production and defers the choice to a new nonterminal, so the decision is postponed until the input has distinguished the alternatives.
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Left factoring A goes to alpha beta1 or alpha beta2 produces
Answer: D
The common prefix is consumed first and the choice is deferred to a new nonterminal. The language is unchanged, but the parser can now decide after seeing what follows the prefix. Repeated factoring may be needed when several alternatives share prefixes.
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