Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2015/imperative-programming/6/b/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2015 / Imperative programming / 6 / b / Solution by Codex 0 2026-10-31
A covariant type parameter may be produced but not consumed directly. Widen the output type of add through a lower type bound. In compose, the first parameter of CoPartialFn is invariant, so replace its occurrence of U by a fresh type parameter W with lower type bound U.
class CoPartialFn[T,+U](val data: List[(T,U)]) {
  require(data.map(_._1).distinct.size == data.size)

  def get(x: T): Option[U] =
    data.find(p => p._1 == x).map(_._2)

  def add[V >: U](x: T, y: V): CoPartialFn[T,V] = {
    val old: List[(T,V)] = data.filterNot(p => p._1 == x)
    new CoPartialFn[T,V]((x,y) :: old)
  }

  def remove(x: T): CoPartialFn[T,U] =
    new CoPartialFn[T,U](data.filterNot(p => p._1 == x))

  def compose[V, W >: U](that: CoPartialFn[W,V]): CoPartialFn[T,V] = {
    val pairs: List[(T,V)] = data.flatMap { case (x,y) =>
      that.get(y).toList.map(z => (x,z))
    }
    new CoPartialFn[T,V](pairs)
  }
}
List, Tuple2, and Option are covariant in their element/output parameters, so data, get, and remove are valid producers. The variance-position rule permits U in the lower type bounds; every value of U can be passed to that.get, since . No cast or mutation through a widened reference is needed.[ref]
(1)
The input-key representation invariant and the partial function semantics are unchanged.
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