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How to split a 32‑bit integer into its four bytes in Kotlin

1 Answer

0 votes
@OptIn(ExperimentalUnsignedTypes::class)

/*
    split_bytes(n)
    --------------
    Splits a 32-bit unsigned integer into its four bytes.

    Layout (little-endian order):
        byte[0] = lowest  8 bits
        byte[1] = next     8 bits
        byte[2] = next     8 bits
        byte[3] = highest  8 bits

    Uses bitwise AND and shifts:
        n & 0xFF        → extract lowest byte
        (n >> 8) & 0xFF → extract next byte
        ...
*/
fun split_bytes(n: UInt): UIntArray {
    return uintArrayOf(
        n and 0xFFu,                 // lowest byte
        (n shr 8) and 0xFFu,
        (n shr 16) and 0xFFu,
        (n shr 24) and 0xFFu         // highest byte
    )
}

/*
    print_bits(label, value)
    ------------------------
    Prints an 8-bit or 32-bit value in binary.

    Uses value.toString(radix = 2) and manual zero-padding.
*/
fun print_bits(label: String, value: UInt, bits: Int) {
    var binary = value.toString(2)

    // Pad with leading zeros
    if (binary.length < bits) {
        binary = "0".repeat(bits - binary.length) + binary
    }

    println("$label ($bits bits): $binary")
}

@OptIn(ExperimentalUnsignedTypes::class)
fun main() {
    val value: UInt = 3298312u

    val bytes: UIntArray = split_bytes(value)

    println("Bytes (little-endian order):")
    for (i in bytes.indices) {
        println("byte[$i]: ${bytes[i]}")
    }

    println("\nBit representation:")

    // Print full 32-bit value
    print_bits("Full value", value, 32)

    // Print each byte in binary
    for (i in bytes.indices) {
        print_bits("byte[$i]", bytes[i], 8)
    }
}


/*
run:

Bytes (little-endian order):
byte[0]: 8
byte[1]: 84
byte[2]: 50
byte[3]: 0

Bit representation:
Full value (32 bits): 00000000001100100101010000001000
byte[0] (8 bits): 00001000
byte[1] (8 bits): 01010100
byte[2] (8 bits): 00110010
byte[3] (8 bits): 00000000

*/

 



answered Jul 20 by avibootz
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