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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2021/imperative-programming-resit/6/e/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2021 / Imperative programming (resit) / 6 / e / Solution by

Codex  0 2026-10-25

Use short-circuit evaluation so that even an availability test of the second iterator is delayed until the first is exhausted:

class ConcatIterator[T](first: Iterator[T], second: Iterator[T])
    extends Iterator[T] {
  def hasNext: Boolean = first.hasNext || second.hasNext

  def next(): T = {
    if (first.hasNext) first.next()
    else if (second.hasNext) second.next()
    else throw new java.util.NoSuchElementException("empty concatenation")
  }
}

Each successful call returns from the first producer while it has entries, then from the second. There is no constructor-time request or element prefetch. This implements concatenation of the producers' remaining sequences, subject to their ordinary iterator contracts.

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