CODEWITHSUNDEEP
scalaioarrays
fgysin
fgysin

Reputation: 11943

How to read a file as a byte array in Scala

I can find tons of examples but they seem to either rely mostly on Java libraries or just read characters/lines/etc.

I just want to read in some file and get a byte array with scala libraries - can someone help me with that?

Upvotes: 86

Views: 76189

Answers (8)

Antonio Maria Sanchez Berrocal
Antonio Maria Sanchez Berrocal

Reputation: 1769

Asynchronous File reading using Scala Future and Java NIO2

  def readFile(path: Path)(implicit ec: ExecutionContext): Future[Array[Byte]] = {
    val p = Promise[Array[Byte]]()
    try {
      val channel = AsynchronousFileChannel.open(path, StandardOpenOption.READ)
      val buffer = ByteBuffer.allocate(channel.size().toInt);
      channel.read(buffer, 0L, buffer, onComplete(channel, p))
    }
    catch {
      case t: Exception => p.failure(t)
    }
    p.future
  }

  private def onComplete(channel: AsynchronousFileChannel, p: Promise[Array[Byte]]) = {
    new CompletionHandler[Integer, ByteBuffer]() {
      def completed(res: Integer, buffer: ByteBuffer): Unit = {
        p.complete(Try {
          buffer.array()
        })
      }

      def failed(t: Throwable, buffer: ByteBuffer): Unit = {
        p.failure(t)
      }
    }
  }

Upvotes: -1

Kareek
Kareek

Reputation: 1

I have used below code to read a CSV file.

import scala.io.StdIn.readLine
import scala.io.Source.fromFile

readFile("C:/users/xxxx/Downloads/", "39025968_ccccc_1009.csv")

def readFile(loc :String,filenm :String): Unit ={

  var flnm = fromFile(s"$loc$filenm") // Imported fromFile package

  println("Files testing")
  /*for (line <- flnm.getLines()) {
    printf("%4d %s\n", line.length, line)
  }*/
  flnm.getLines().foreach(println) // getLines() is imported from readLines.
  flnm.close() 
}

Upvotes: -2

Sagi
Sagi

Reputation: 9294

You can use the Apache Commons Compress IOUtils

import org.apache.commons.compress.utils.IOUtils

val file = new File("data.bin")
IOUtils.toByteArray(new FileInputStream(file))

Upvotes: 2

Vladimir Matveev
Vladimir Matveev

Reputation: 128111

Java 7:

import java.nio.file.{Files, Paths}

val byteArray = Files.readAllBytes(Paths.get("/path/to/file"))

I believe this is the simplest way possible. Just leveraging existing tools here. NIO.2 is wonderful.

Upvotes: 147

fengyun liu
fengyun liu

Reputation: 101

The library scala.io.Source is problematic, DON'T USE IT in reading binary files.

The error can be reproduced as instructed here: https://github.com/liufengyun/scala-bug

In the file data.bin, it contains the hexidecimal 0xea, which is 11101010 in binary and should be converted to 234 in decimal.

The main.scala file contain two ways to read the file:

import scala.io._
import java.io._

object Main {
  def main(args: Array[String]) {
    val ss = Source.fromFile("data.bin")
    println("Scala:" + ss.next.toInt)
    ss.close

    val bis = new BufferedInputStream(new FileInputStream("data.bin"))
    println("Java:" + bis.read)
    bis.close
  }
}

When I run scala main.scala, the program outputs follows:

Scala:205
Java:234

The Java library generates correct output, while the Scala library not.

Upvotes: 6

OlivierBlanvillain
OlivierBlanvillain

Reputation: 7768

You might also consider using scalax.io:

scalax.io.Resource.fromFile(fileName).byteArray

Upvotes: 4

reivzy
reivzy

Reputation: 87

val is = new FileInputStream(fileName)
val cnt = is.available
val bytes = Array.ofDim[Byte](cnt)
is.read(bytes)
is.close()

Upvotes: 5

Jus12
Jus12

Reputation: 18034

This should work (Scala 2.8): 

val bis = new BufferedInputStream(new FileInputStream(fileName))
val bArray = Stream.continually(bis.read).takeWhile(-1 !=).map(_.toByte).toArray

Upvotes: 47

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