Best Way To Represent A Readline Loop In Scala?

Answer :

How about this?

val lines = Iterator.continually(reader.readLine()).takeWhile(_ != null).mkString 

Well, in Scala you can actually say:

val lines = scala.io.Source.fromFile("file.txt").mkString 

But this is just a library sugar. See Read entire file in Scala? for other possiblities. What you are actually asking is how to apply functional paradigm to this problem. Here is a hint:

Source.fromFile("file.txt").getLines().foreach {println} 

Do you get the idea behind this? foreach line in the file execute println function. BTW don't worry, getLines() returns an iterator, not the whole file. Now something more serious:

lines filter {_.startsWith("ab")} map {_.toUpperCase} foreach {println} 

See the idea? Take lines (it can be an array, list, set, iterator, whatever that can be filtered and which contains an items having startsWith method) and filter taking only the items starting with "ab". Now take every item and map it by applying toUpperCase method. Finally foreach resulting item print it.

The last thought: you are not limited to a single type. For instance say you have a file containing integer number, one per line. If you want to read that file, parse the number and sum them up, simply say:

lines.map(_.toInt).sum 

To add how the same can be achieved using the formerly new nio files which I vote to use because it has several advantages:

val path: Path = Paths.get("foo.txt") val lines = Source.fromInputStream(Files.newInputStream(path)).getLines()  // Now we can iterate the file or do anything we want, // e.g. using functional operations such as map. Or simply concatenate. val result = lines.mkString 

Don't forget to close the stream afterwards.