Maximum Length for scala queue

An often preferable alternative to subclassing is the (unfortunately named) "pimp my library" pattern. You can use it to add an enqueueFinite method to Queue, like so:

import scala.collection.immutable.Queue
class FiniteQueue[A](q: Queue[A]) {
  def enqueueFinite[B >: A](elem: B, maxSize: Int): Queue[B] = {
    var ret = q.enqueue(elem)
    while (ret.size > maxSize) { ret = ret.dequeue._2 }
    ret
  }
}
implicit def queue2finitequeue[A](q: Queue[A]) = new FiniteQueue[A](q)

Whenever queue2finitequeue is in scope, you can treat Queue objects as though they have the enqueueFinite method:

val maxSize = 3
val q1 = Queue(1, 2, 3)
val q2 = q1.enqueueFinite(5, maxSize)
val q3 = q2.map(_+1)
val q4 = q3.enqueueFinite(7, maxSize)

The advantage of this approach over subclassing is that enqueueFinite is available to all Queues, including those that are constructed via operations like enqueue, map, ++, etc.

Update: As Dylan says in the comments, enqueueFinite needs also to take a parameter for the maximum queue size, and drop elements as necessary. I updated the code.

Why don't you just subclass a FIFO queue? Something like this should work: (pseudocode follows...)

class Limited(limit:Int) extends FIFO {
  override def enqueue() = {
    if (size >= limit) {
      //remove oldest element
    }
    super.enqueue()
  }
}