Scala Iterator ++ blows the stack - scala

Question

I recently notice this bug causing StackOverFlowError in Scala Iterator++ caused by lazy init. Here's the code to make the bug appear.
var lines = Source.fromFile("file").getLines()
var line = lines.next()
lines = Array(line).toIterator ++ lines
lines.foreach { println(_) }
System.exit(0)
What I get is
Exception in thread "main" java.lang.StackOverflowError
at scala.collection.Iterator$JoinIterator.hasNext(Iterator.scala:219)
at scala.collection.Iterator$JoinIterator.hasNext(Iterator.scala:219)
at scala.collection.Iterator$JoinIterator.hasNext(Iterator.scala:219)
at scala.collection.Iterator$JoinIterator.hasNext(Iterator.scala:219)
...
It should be caused by this line in scala source (scala.collection.Iterator.scala:208)
lazy val rhs: Iterator[A] = that.toIterator
As rhs is a lazy init val, when the iterator is used what the name "lines" refers to already changed, and caused a loop reference, which leads to the error.
I noticed this post talks about the problem in 2013. However it seems it has not been fully repaired. I am running Scala 2.11.8 from Maven Repo.
My Question: I can rename the iterator, e.g. "lines2" to avoid this bug, but is this the only way to solve the problem? I feel like using the name "lines" is more natural and don't want to forsake it if possible.

Answer 1

If you want to reload an Iterator using the same var, this appears to work. [Tested on 2.11.7 and 2.12.1]
scala> var lines = io.Source.fromFile("file.txt").getLines()
lines: Iterator[String] = non-empty iterator
scala> var line = lines.next()
line: String = this,that,other,than
scala> lines = Iterator(line +: lines.toSeq:_*)
lines: Iterator[String] = non-empty iterator
scala> lines.foreach(println)
this,that,other,than
here,there,every,where
But it might make more sense to use a BufferedIterator where you can call head on it to peek at the next element without consuming it.
explanation
lines.toSeq <-- turn the Iterator[String] into a Seq[String] (The REPL will show this as a Stream but that's because the REPL has to compile and represent each line of input separately.)
line +: lines.toSeq <-- create a new Seq[String] with line as the first element (i.e. prepended)
(line +: lines.toSeq:_*) <-- turns a single Seq[T] into a parameter list that can be passed to the Iterator.apply() method. #som-snytt has cleverly pointed out that this can be simplified to (line +: lines.toSeq).iterator
BufferedIterator example
scala> var lines = io.Source.fromFile("file.txt").getLines.buffered
lines: scala.collection.BufferedIterator[String] = non-empty iterator
^^^^^^^^^^^^^^^^^^^^^^^^ <-- note the type
scala> lines.head
res5: String = this,that,other,than
scala> lines foreach println
this,that,other,than
here,there,every,where

Answer 2

Simple capture:
scala> var lines = Iterator.continually("x")
lines: Iterator[String] = non-empty iterator
scala> lines = { val z = lines ; Iterator.single("y") ++ z }
lines: Iterator[String] = non-empty iterator
scala> lines.next
res0: String = y
scala> lines.next
res1: String = x