I have a byte array and want to convert it into short array, say input: [1, 2, 3, 4] output: [0x102, 0x304]
Is there any method to call?
2 Answers
The grouped(2).map approach of Peter Neyens is what I would do, although just using simply bitshift instead of all the effort to go through a ByteBuffer:
def convert(in: Array[Byte]): Array[Short] =
in.grouped(2).map { case Array(hi, lo) => (hi << 8 | lo).toShort } .toArray
val in = Array(1.toByte, 2.toByte, 3.toByte, 4.toByte)
val out = convert(in)
out.map(x => s"0x${x.toHexString}") // 0x102, 0x304
If your input might have an odd size, use an extra case in the pattern match as in Peter's answer.
1 Comment
niczky12
If this is a signed short then I believe a slight change is needed:
(hi << 8 | (0x00ff & lo)You can use the toShort method of Byte:
val bytes = Array[Byte](1, 2, 3, 4, 192.toByte)
bytes: Array[Byte] = Array(1, 2, 3, 4, -64)
bytes.map(_.toShort)
res1: Array[Short] = Array(1, 2, 3, 4, -64)
I see you want to combine two Bytes into one Short :
import java.nio.{ByteBuffer, ByteOrder}
def bytesToShort(byte1: Byte, byte2: Byte) : Short = {
ByteBuffer
.allocate(2)
.order(ByteOrder.LITTLE_ENDIAN)
.put(byte1)
.put(byte2)
.getShort(0)
}
val bytes = Array[Byte](1, 2, 3, 4)
bytes.grouped(2).map {
case Array(one) => bytesToShort(one, 0.toByte)
case Array(one, two) => bytesToShort(one, two)
}.toArray
