Question

我想初始化我的班级属性。因为我大量使用Kotlin的功能元素，所以我想将这些初始化放在命名良好的函数中，以提高代码的可读性。问题是我不能分配val属性，如果代码不在init块中，而是在从init块调用的函数中。

如果属性是val，是否可以将类的初始化拆分为不同的函数？

以下是代码：

    val socket: DatagramSocket = DatagramSocket()
    val data: ByteArray = "Cassiopeiae server discovery packet".toByteArray()
    val broadcastAddresses: List<InetAddress>

    init {
        socket.broadcast = true
        val interfaceAddresses = ArrayList<InterfaceAddress>()
        collectValidNetworkInterfaces(interfaceAddresses)
        collectBroadcastAddresses(interfaceAddresses)
    }

    private fun collectValidNetworkInterfaces(interfaceAddresses: ArrayList<InterfaceAddress>) {
        NetworkInterface.getNetworkInterfaces().toList()
                .filter { validInterface(it) }
                .forEach { nInterface -> nInterface.interfaceAddresses.toCollection(interfaceAddresses) }
    }

    private fun collectBroadcastAddresses(interfaceAddresses: ArrayList<InterfaceAddress>) {
        broadcastAddresses = interfaceAddresses
                .filter { address -> address.broadcast != null }
                .map { it.broadcast }
    }

当然它没有编译，因为 collectBroadcastAddresses 函数试图重新分配 broadcastAddresses val。虽然我不想把这个函数的代码放到init块中，因为代码在做什么并不明显，而且函数名称非常好。

在这种情况下我该怎么办？我想保持我的代码干净，这是最重要的一点！

Answer 1

解决问题的一种方法是使用pure functions初始化字段：

class Operation {
    val socket = DatagramSocket().apply { broadcast = true }
    val data: ByteArray = "Cassiopeiae server discovery packet".toByteArray()
    val broadcastAddresses = collectBroadcastAddresses(collectValidNetworkInterfaces()) 

    private fun collectValidNetworkInterfaces() =
        NetworkInterface.getNetworkInterfaces().toList()
            .filter { validInterface(it) }
            .flatMap { nInterface -> nInterface.interfaceAddresses }

    private fun validInterface(it: NetworkInterface?) = true

    private fun collectBroadcastAddresses(interfaceAddresses: List<InterfaceAddress>) {
        interfaceAddresses
            .filter { address -> address.broadcast != null }
            .map { it.broadcast }
    }
}

注意socket字段初始化如何使用apply扩展名。

我经常发现将集合操作例程提取到扩展方法中很有用：

class Operation {
    val socket = DatagramSocket().apply { broadcast = true }
    val data: ByteArray = "Cassiopeiae server discovery packet".toByteArray()
    val broadcastAddresses = NetworkInterface.getNetworkInterfaces()
        .collectValidNetworkInterfaces { validInterface(it) }
        .collectBroadcastAddresses()

    private fun validInterface(it: NetworkInterface?) = true
}

fun Iterable<InterfaceAddress>.collectBroadcastAddresses(): List<InetAddress> =
    filter { address -> address.broadcast != null }.map { it.broadcast }

fun Enumeration<NetworkInterface>.collectValidNetworkInterfaces(isValid: (NetworkInterface) -> Boolean = { true }) =
    toList()
        .filter { isValid(it) }
        .flatMap { nInterface -> nInterface.interfaceAddresses }

由于val属性，无法从init块调用函数

1 个答案: