Scala 库将数字（Int、Long、Double）转换为/从 Array[Byte]
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/9810010/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Scala library to convert numbers (Int, Long, Double) to/from Array[Byte]
提问by em70
As the title says, is there any Scala library that exports functions to convert, preferably fluently, a byte array to an Int, to a Long or to a Double?
正如标题所说，是否有任何 Scala 库可以导出函数以将字节数组（最好是流畅地）转换为 Int、Long 或 Double？
I need something compatible with 2.9.1 and FOSS.
我需要与 2.9.1 和 FOSS 兼容的东西。
If you happen to know exactly what I need and where to find it, a line for SBT and a line for an example will be enough! :)
如果您恰好知道我需要什么以及在哪里可以找到它，那么一行 SBT 和一行示例就足够了！:)
If there's no such thing as what I'm looking for, the closest thing in Java will also work...
如果没有我正在寻找的东西，那么 Java 中最接近的东西也将起作用......
回答by Travis Brown
You can use Java NIO's ByteBuffer:
您可以使用Java NIO 的ByteBuffer：
import java.nio.ByteBuffer
ByteBuffer.wrap(Array[Byte](1, 2, 3, 4)).getInt
ByteBuffer.wrap(Array[Byte](1, 2, 3, 4, 5, 6, 7, 8)).getDouble
ByteBuffer.wrap(Array[Byte](1, 2, 3, 4, 5, 6, 7, 8)).getLong
No extra dependencies required.
不需要额外的依赖。
回答by Themerius
You can also use BigIntfrom the scala standard library.
您也可以BigInt从 Scala 标准库中使用。
import scala.math.BigInt
val bytearray = BigInt(1337).toByteArray
val int = BigInt(bytearray)
回答by Rex Kerr
Java's nio.ByteBufferis the way to go for now:
Javanio.ByteBuffer是目前要走的路：
val bb = java.nio.ByteBuffer.allocate(4)
val i = 5
bb.putInt(i)
bb.flip // now can read instead of writing
val j = bb.getInt
bb.clear // ready to go again
You can also putarrays of bytes, etc.
您还可以put使用字节数组等。
Keep in mind the little/big-endian thing. bb.order(java.nio.ByteOrder.nativeOrder)is probably what you want.
请记住小/大端的事情。 bb.order(java.nio.ByteOrder.nativeOrder)可能是你想要的。
回答by prosseek
For Double <-> ByteArray, you can use java.lang.Double.doubleToLongBits and java.lang.Double.longBitsToDouble.
对于 Double <-> ByteArray，您可以使用 java.lang.Double.doubleToLongBits 和 java.lang.Double.longBitsToDouble。
import java.lang.Double
def doubleToByteArray(x: Double) = {
val l = java.lang.Double.doubleToLongBits(x)
val a = Array.fill(8)(0.toByte)
for (i <- 0 to 7) a(i) = ((l >> ((7 - i) * 8)) & 0xff).toByte
a
}
def byteArrayToDouble(x: Array[scala.Byte]) = {
var i = 0
var res = 0.toLong
for (i <- 0 to 7) {
res += ((x(i) & 0xff).toLong << ((7 - i) * 8))
}
java.lang.Double.longBitsToDouble(res)
}
scala> val x = doubleToByteArray(12.34)
x: Array[Byte] = Array(64, 40, -82, 20, 122, -31, 71, -82)
scala> val y = byteArrayToDouble(x)
y: Double = 12.34
Or ByteBuffer can be used:
或者可以使用 ByteBuffer：
import java.nio.ByteBuffer
def doubleToByteArray(x: Double) = {
val l = java.lang.Double.doubleToLongBits(x)
ByteBuffer.allocate(8).putLong(l).array()
}
def byteArrayToDouble(x:Array[Byte]) = ByteBuffer.wrap(x).getDouble
回答by Jkulkarni
The following worked for me using Scala:
以下使用 Scala 对我有用：
import org.apache.kudu.client.Bytes
Bytes.getFloat(valueToConvert)
回答by user2012643
You can also use: Bytes.toInt(byteArray) Worked like a charm!
您还可以使用： Bytes.toInt(byteArray) 像魅力一样工作！
