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The Source Class in Scala IO

In [11]:
import scala.collection.JavaConverters._
Out[11]:
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.

In [2]:
val bs = scala.io.Source.fromFile("test.yaml")
Out[2]:
non-empty iterator
In [3]:
bs.mkString
Out[3]:
aa:
  - x
  - y
bb: z

InputStream to BufferedSource

In [ ]:
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

In [ ]:

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