How to convert a Seq[Byte] into an Array[Boolean] representing each bit in Scala

Question

Is there a better way to convert a sequence of Bytes into an Seq[Boolean] where each element represents a bit from the Byte sequence?

I'm currently doing this, but byte2Bools seems a little too heavy...

object Main extends App {

  private def byte2Bools(b: Byte) =
    (0 to 7).foldLeft(ArrayBuffer[Boolean]())((bs, i) => bs += isBitSet(b, i))

  private def isBitSet(byte: Byte, bit: Int) =
    ((byte >> bit) & 1) == 1

  val bytes = List[Byte](1, 2, 3)
  val bools = bytes.flatMap(b => byte2Bools(b))

  println(bools)

}

Perhaps the real question is: what's a better implementation of byte2Bools?

4e6 · Accepted Answer · 2013-04-28 21:30:44Z

4

First, accumulator in foldLeft is not necessary need to be a mutable collection.

def byte2Bools(b: Byte): Seq[Boolean] = 
  (0 to 7).foldLeft(Vector[Boolean]()) { (bs, i) => bs :+ isBitSet(b)(i) }

Second, you can just map initial sequence with isBitSet.

def byte2Bools(b: Byte): Seq[Boolean] =
  0 to 7 map isBitSet(b)

def isBitSet(byte: Byte)(bit: Int): Boolean =
  ((byte >> bit) & 1) == 1

answered Apr 28, 2013 at 21:30

4e6

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2 Comments

Kevin Herron Over a year ago

I made it mutable to cut out the overhead of creating a new collection every step... as for your second point: DOH! Yes, that seems like the obvious approach doesn't it. I'm new to Scala and functional programming in general. I often find myself trying to get too tricky for my own good.

4e6 Over a year ago

There is great book about FP in scala. You should definitely check it out.

Keith Pinson · Accepted Answer · 2013-04-29 01:19:11Z

1

For whatever it's worth, you can convert a Byte to a BinaryString and then to sequence of Booleans with:

  val b1 : Byte = 7
  (0x100 + b1).toBinaryString.tail.map{ case '1' => true; case _ => false }

Results in: Vector(false, false, false, false, false, true, true, true)

And, you would go back (Booleans to Byte) with:

  val s1 = Vector(false, false, false, false, false, true, true, true)
  Integer.parseInt( s1.map{ case true => '1'; case false => '0' }.mkString, 2 ).toByte

edited Apr 29, 2013 at 1:19

answered Apr 29, 2013 at 1:05

Keith Pinson

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