The Source Class in Scala IO

Things on this page are fragmentary and immature notes/thoughts of the author. Please read with your own judgement!

import scala.collection.JavaConverters._

import scala.collection.JavaConverters._

BufferedSource¶

This object provides convenience methods to create an iterable representation of a source file. A BufferedSource object is iterable! Many of the collection methods apply to it.

val bs = scala.io.Source.fromFile("test.yaml")

non-empty iterator

bs.mkString

aa:
  - x
  - y
bb: z

InputStream to BufferedSource¶

scala.io.Source.fromInputStream(is)

References¶

https://www.scala-lang.org/api/current/scala/io/Source$.html

https://www.scala-lang.org/api/current/scala/io/BufferedSource.html

https://stackoverflow.com/questions/5221524/idiomatic-way-to-convert-an-inputstream-to-a-string-in-scala/5221595