scala 如何在Scala中将文件作为字节数组读取
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How to read a file as a byte array in Scala
提问by fgysin reinstate Monica
I can find tons of examples but they seem to either rely mostly on Java libraries or just read characters/lines/etc.
我可以找到大量示例，但它们似乎要么主要依赖于 Java 库，要么只是读取字符/行/等。
I just want to read in some file and get a byte array with scala libraries - can someone help me with that?
我只想读入一些文件并获得一个带有 Scala 库的字节数组 - 有人可以帮我吗？
回答by Vladimir Matveev
Java 7:
爪哇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.
我相信这是最简单的方法。只是利用这里的现有工具。NIO.2很棒。
回答by Jus12
This should work (Scala 2.8):
这应该有效（Scala 2.8）：
val bis = new BufferedInputStream(new FileInputStream(fileName))
val bArray = Stream.continually(bis.read).takeWhile(-1 !=).map(_.toByte).toArray
回答by reivzy
val is = new FileInputStream(fileName)
val cnt = is.available
val bytes = Array.ofDim[Byte](cnt)
is.read(bytes)
is.close()
回答by OlivierBlanvillain
回答by fengyun liu
The library scala.io.Sourceis problematic, DON'T USE IT in reading binary files.
库scala.io.Source有问题，请勿在读取二进制文件时使用它。
The error can be reproduced as instructed here: https://github.com/liufengyun/scala-bug
错误可以按照这里的说明重现：https: //github.com/liufengyun/scala-bug
In the file data.bin, it contains the hexidecimal 0xea, which is 11101010in binary and should be converted to 234in decimal.
在文件中data.bin，它包含十六进制0xea，它是11101010二进制的，应该转换为234十进制。
The main.scalafile contain two ways to read the file:
该main.scala文件包含两种读取文件的方式：
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:
当我运行时scala main.scala，程序输出如下：
Scala:205
Java:234
The Java library generates correct output, while the Scala library not.
Java 库生成正确的输出，而 Scala 库则没有。
回答by Sagi
You can use the Apache Commons CompressIOUtils
您可以使用Apache Commons CompressIOUtils
import org.apache.commons.compress.utils.IOUtils
val file = new File("data.bin")
IOUtils.toByteArray(new FileInputStream(file))
