How to convert a array of bits to a byte, (not to byte string)?

Question

I have the follow array:

val cells = arrayOf(
  0b1, 0b1, 0b1,
  0b0, 0b0, 0b0,
  0b0, 0b0, 0b0,
)

And I want to convert it to number e.g: 0b111000000, to be able to compare with another byte after:

val cells = arrayOf(
  0b1, 0b1, 0b1,
  0b0, 0b0, 0b0,
  0b0, 0b0, 0b0,
)

print(arrayOfBitsToByte(cells) == 0b111000000)

I could make this:

val cells = arrayOf(
  0b1, 0b1, 0b1,
  0b0, 0b0, 0b0,
  0b1, 0b1, 0b1,
)

print(cells.joinToString("") == 0b111000000.toString(2))

But in the above case, the 2 values were cast to String

And I would like to know if it is possible to compare the numbers without cast to string

You can use bitwise operators to create an Int with the bits defined in the array. See stackoverflow.com/questions/48474520/… — Slaw
– Slaw, Commented Jan 8, 2023 at 0:45

adnan_e · Accepted Answer · 2023-01-08 00:48:32Z

2

Convert the array of bits into a single Int using bitwise operations. This assumes that each element is either 0 or 1, and will break otherwise (due to only shifting by 1 bit).

val cells = arrayOf(
        0b1, 0b1, 0b1,
        0b0, 0b0, 0b0,
        0b0, 0b0, 0b0,
)

var joined = cells.reduce{ acc, curr -> ((acc shl 1) or curr) }
print(joined == 0b111000000)

answered Jan 8, 2023 at 0:48

adnan_e

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4 Comments

lukas.j Over a year ago

Try fold(0) { acc, int -> (acc shl 1) or int }. It seems to be drastically faster. Why I do not know.

adnan_e Over a year ago

@lukas.j Cannot reproduce, it seems to be slightly faster with fold (19 nanoseconds) - pl.kotl.in/3NbDzG0VO

lukas.j Over a year ago

It seems to be rather quite a lot of faster when running that in the playground too. reduce takes at leats 50% longer than fold.

lukas.j Over a year ago

Looking at the generated Kotlin bytecode it seems that reduce creates an IntIterator while fold uses a 'goto-loop'. That might explain why fold is faster.

lukas.j · Accepted Answer · 2023-01-09 12:57:35Z

1

Simple conversion of the array to an int (avoiding bitwise operations):

import kotlin.math.pow

val cells = arrayOf(
    0b1, 0b1, 0b1,
    0b0, 0b0, 0b0,
    0b1, 0b1, 0b1,
)

fun Array<Int>.toInt() = this
    .foldIndexed(0) { index, acc, int ->
      acc + if (int == 0b1) 2.0.pow(index).toInt() else 0
    }

println(cells.toInt() == 0b111000000)   // false
println(cells.toInt() == 0b111000111)   // true

Note that @adnan_e's solutions is faster than this one.

edited Jan 9, 2023 at 12:57

answered Jan 8, 2023 at 12:39

lukas.j

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