这应该可以做到：

var source: ULong = 0uL            
val bytes = ByteArray(5)
bytes[0x0] = (source shr 32).toByte()
bytes[0x1] = (source shr 24).toByte()
bytes[0x2] = (source shr 16).toByte()
bytes[0x3] = (source shr 8).toByte()
bytes[0x4] = source.toByte()

这取决于您是否希望在索引0处更高阶地填充字节，等等。

fun ULong.toByteArray(): ByteArray {
    val result = ByteArray(ULong.SIZE_BYTES)
    (0 until ULong.SIZE_BYTES).forEach {
        result[it] = this.shr(Byte.SIZE_BITS * it).toByte()
    }
    return result
}

然后用途：someULongValue.toByteArray()