必须初始化变量'runnable'

Question

为什么Kotlin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Variable 'runnable' must be initialized,在Line中它是由处理程序再次发布的.这在纯Java中有效:

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.这种语义类似于其初始化程序中局部变量使用的限制.

有几种解决方法:

• 使用对象表达式,它允许您引用this声明的对象:

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

  这只适用于作为lambda的替代品的接口,并且不是很完美.

• 使用lateinit var或委托财产有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()
  

• 为您自己实现初始值设定项并使用自引用:

  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
  }
  

  然后你可以写出类似的东西

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