The stack of a table driven predictive parser initially contains
Answer: A
Pushing the marker first and the start symbol above it lets the parser recognise successful completion when the marker surfaces at the same moment the input is exhausted. Using an explicit marker removes the need for a special case in the driver loop.
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Arrange the actions of a table driven predictive parser when a nonterminal is on top of the stack.
1. pop the nonterminal from the stack
2. read the current lookahead token
3. push the right side of the chosen production in reverse
4. consult the table cell for that nonterminal and token
Answer: B
Correct answer: 2 - 4 - 1 - 3
The lookahead is read first because the table cannot be consulted without it. The cell for the nonterminal and that token gives the production, the nonterminal is popped, and the right side is pushed in reverse so that its leftmost symbol ends up on top - which is what makes the parser expand leftmost.
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The right side of a chosen production is pushed onto the stack in reverse order because
Answer: C
A stack yields symbols in the reverse of the order they were pushed. Pushing the right side backwards therefore presents its symbols left to right, matching the input order. Getting the direction wrong is a classic bug when the driver is written by hand.
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When a terminal sits on top of the parser stack, the parser
Answer: D
Matching consumes one input token and one stack symbol. A mismatch is a syntax error, and the recovery routine decides whether to skip input or to pop the stack symbol. Choosing well between those two options is what makes recovery feel intelligent.
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A predictive parser accepts the input when
Answer: A
Both conditions must hold together. Reaching the marker with input remaining, or exhausting the input with symbols still on the stack, are both errors rather than acceptance. Checking both conditions costs nothing and catches a whole class of mistakes.
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An epsilon production is applied by a predictive parser by
Answer: B
Correct answer: popping the nonterminal and pushing nothing at all
An epsilon production derives the empty string, so applying it removes the nonterminal from the stack and contributes nothing in its place. No input is consumed, and there is no epsilon symbol to push - epsilon is the absence of symbols rather than a symbol of its own.
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Assertion (A): FOLLOW sets are needed to build a predictive parsing table even though FIRST sets alone identify the productions.
Reason (R): A production deriving epsilon matches no input, so the decision must rest on what may come after the nonterminal.
Answer: C
Correct answer: Both A and R are true, and R is the correct explanation of A
Both are true and the reason gives the mechanism. A production deriving epsilon has an empty FIRST set, so nothing in the input can indicate that it is the right choice. The parser must instead ask whether the lookahead is something that could legitimately follow the nonterminal, and FOLLOW is exactly that set.
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Left recursion is fatal to a recursive descent parser because
Answer: D
Correct answer: the procedure for the nonterminal calls itself before consuming input
A production such as \( E \rightarrow E + T \) makes the procedure for E call itself as its first action, with the input position unchanged. The recursion therefore never terminates and the stack overflows. Removing the left recursion by rewriting it as right recursion is the standard preparation.
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After left recursion has been removed, the auxiliary nonterminal introduced by the transformation always
Answer: A
Without the epsilon alternative the new nonterminal could never be discharged and the repetition would be obliged to continue for ever. The epsilon alternative is precisely what converts unbounded left recursion into a loop that the parser can leave, and it is also why FOLLOW must be consulted when the table is filled
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Indirect left recursion, such as A deriving B and B deriving A alpha, is removed by
Answer: B
The standard algorithm orders the nonterminals and substitutes earlier ones into later ones, exposing the recursion as immediate. Only then is the usual transformation applied. The algorithm terminates because each substitution reduces the remaining cycles.
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