Question

Kotlin为什么抱怨这个：

class MyActivity : Activity {
  private var handler:Handler = Handler()

  private var runnable: Runnable = Runnable {
    /* Do something very important */
    handler.postDelayed(this@MyActivity.runnable, 5000)
  }
}

编译器抱怨说，线路中的Variable 'runnable' must be initialized是由处理程序再次发布的。这在纯Java中起作用：

private Handler handler = new Handler();

private Runnable runnable = new Runnable() {
    @Override
    public void run() {
        handler.postDelayed(runnable, 5000);
    }
};

Answer 1

Kotlin认为属性未初始化直到其初始化程序结束，因此它不能在自己的初始化程序中使用，即使在lambdas中也是如此。这种语义类似于local variable usage inside its initializer的限制。

有几种解决方法：

使用object expression可以引用声明对象的this：

private var runnable: Runnable = object : Runnable {
    override fun run() {
        /* Do something very important */
        handler.postDelayed(this, 5000)
    }
}

这只适用于作为lambdas的替代品的接口，并且不是很完美。

lateinit var使用delegated property或Delegates.notNull()：

private lateinit var runnable: Runnable
init {
    runnable = Runnable { 
        /* Do something very important */
        handler.postDelayed(runnable, 5000)
    }
}

相同的初始化程序将使用此声明：

private var runnable: Runnable by Delegates.notNull()

自行实施和使用self-reference for initializers：

class SelfReference<T>(val initializer: SelfReference<T>.() -> T) {
    val self: T by lazy {
        inner ?: throw IllegalStateException("Do not use `self` until initialized.")
    }

    private val inner = initializer()
}

fun <T> selfReference(initializer: SelfReference<T>.() -> T): T {
    return SelfReference(initializer).self
}

然后你可以写点像

private var runnable: Runnable = selfReference { 
    Runnable {
        /* Do something very important */
        handler.postDelayed(self, 5000)
    } 
}

Answer 2

您也可以使用

private var runnable: Runnable = Runnable {
    /* Do something very important */
    handler.postDelayed(runnable(), 5000)
}

private fun runnable() = runnable

变量＆＃39; runnable＆＃39;必须初始化

2 个答案: