Read entire file in Scala?

Question

What's a simple and canonical way to read an entire file into memory in Scala? (Ideally, with control over character encoding.)

The best I can come up with is:

scala.io.Source.fromPath("file.txt").getLines.reduceLeft(_+_)

or am I supposed to use one of Java's god-awful idioms, the best of which (without using an external library) seems to be:

import java.util.Scanner
import java.io.File
new Scanner(new File("file.txt")).useDelimiter("\\Z").next()

From reading mailing list discussions, it's not clear to me that scala.io.Source is even supposed to be the canonical I/O library. I don't understand what its intended purpose is, exactly.

... I'd like something dead-simple and easy to remember. For example, in these languages it's very hard to forget the idiom ...

Ruby    open("file.txt").read
Ruby    File.read("file.txt")
Python  open("file.txt").read()

Answer 1

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

By the way, "scala." isn't really necessary, as it's always in scope anyway, and you can, of course, import io's contents, fully or partially, and avoid having to prepend "io." too.

Answer 2

// for file with utf-8 encoding
val lines = scala.io.Source.fromFile("file.txt", "utf-8").getLines.mkString

Answer 3

The obvious question being "why do you want to read in the entire file?" This is obviously not a scalable solution if your files get very large. The scala.io.Source gives you back an Iterator[String] from the getLines method, which is very useful and concise.

It's not much of a job to come up with an implicit conversion using the underlying java IO utilities to convert a File, a Reader or an InputStream to a String. I think that the lack of scalability means that they are correct not to add this to the standard API.

Answer 4

Just to expand on Daniel's solution, you can shorten things up tremendously by inserting the following import into any file which requires file manipulation:

import scala.io.Source._

With this, you can now do:

val lines = fromFile("file.txt").getLines

I would be wary of reading an entire file into a single String. It's a very bad habit, one which will bite you sooner and harder than you think. The getLines method returns a value of type Iterator[String]. It's effectively a lazy cursor into the file, allowing you to examine just the data you need without risking memory glut.

Oh, and to answer your implied question about Source: yes, it is the canonical I/O library. Most code ends up using java.io due to its lower-level interface and better compatibility with existing frameworks, but any code which has a choice should be using Source, particularly for simple file manipulation.

Answer 5

I've been told that Source.fromFile is problematic. Personally, I have had problems opening large files with Source.fromFile and have had to resort to Java InputStreams.

Another interesting solution is using scalax. Here's an example of some well commented code that opens a log file using ManagedResource to open a file with scalax helpers: http://pastie.org/pastes/420714

Answer 6

Use trunk:

scala> io.File("/etc/passwd").slurp
res0: String = 
##
# User Database
# 
... etc

Answer 7

as a few people mentioned scala.io.Source is best to be avoided due to connection leaks.

Probably scalax and pure java libs like commons-io are the best options until the new incubator project (ie scala-io) gets merged.