在抽象构造函数中访问overriden val时出现NullPointerException

Question

考虑以下(简化)示例:

abstract class Bar[T] {
    val f: PartialFunction[T, T]
    val default: PartialFunction[T, T] = { case x => x }
    val chained = f orElse default
}

class Foo extends Bar[Int] {
    val f: PartialFunction[Int, Int] = { case 1 => 2 }
}

看着它崩溃:

scala> val foo = new Foo
java.lang.NullPointerException
        at Bar.<init>(<console>:8)
        at Foo.<init>(<console>:6)
        at .<init>(<console>:7)
        at .<clinit>(<console>)
        at RequestResult$.<init>(<console>:9)
        at RequestResult$.<clinit>(<console>)
        at RequestResult$scala_repl_result(<console>)
        ....

但是,如果我们chained输入具体的类:

abstract class Bar[T] {
    val f: PartialFunction[T, T]
    val default: PartialFunction[T, T] = { case x => x }
}

class Foo extends Bar[Int] {
    val f: PartialFunction[Int, Int] = { case 1 => 2 }
    val chained = f orElse default
}

它按预期工作:

scala> val foo = new Foo
foo: Foo = Foo@16132c4

我不得不承认我完全不知道这里发生了什么.错误？(这是在Scala 2.8.1上.)

Answer 1

"为什么我的抽象或覆盖val为空？" https://github.com/paulp/scala-faq/wiki/Initialization-Order