Scala：声明 Any => 类型的延续时出现编译错误没有什么
此代码给出编译错误：
import scala.util.continuations._
object CTest {
def loop: Nothing = reset {
shift {c: (Unit => Nothing) => c()}
loop
}
def main(argv: Array[String]) {loop}
}
错误消息：
error: type mismatch;
found : ((Unit) => Nothing) => (Unit) => Nothing
required: ((Unit) => B) => (Unit) => Nothing
但此代码按预期工作：
import scala.util.continuations._
object CTest {
def loop: Nothing = reset {
shift {c: (Unit => Any) => c.asInstanceOf[Unit => Nothing]()}
loop
}
def main(argv: Array[String]) {loop}
}
问题是：为什么 Scala 编译器讨厌 me 类型 Any => 的延续；没有什么？
This code gives compilation error:
import scala.util.continuations._
object CTest {
def loop: Nothing = reset {
shift {c: (Unit => Nothing) => c()}
loop
}
def main(argv: Array[String]) {loop}
}
Error message:
error: type mismatch;
found : ((Unit) => Nothing) => (Unit) => Nothing
required: ((Unit) => B) => (Unit) => Nothing
But this code works as expected:
import scala.util.continuations._
object CTest {
def loop: Nothing = reset {
shift {c: (Unit => Any) => c.asInstanceOf[Unit => Nothing]()}
loop
}
def main(argv: Array[String]) {loop}
}
The question is: why Scala compiler hates me continuations of type Any => Nothing?
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如果我指定类型参数，它就会编译：
在我看来，编译器应该推断出
B是Nothing，但事实并非如此。It compiles if I specify the type arguments:
It looks to me like the compiler should infer that
BisNothing, but it doesn't.您无法返回
Nothing类型，因为它没有实例。任何预期返回Nothing的代码绝不能返回。例如，总是抛出异常的方法可以被声明为不返回任何内容。Java 中调用
void的返回值在 Scala 中是Unit。欲了解更多信息，为什么不看看 James Iry 对 一无所有。
You can't return the type
Nothing, because it has no instances. Any code that is expected to returnNothingmust never return. For example, a method which always throws exceptions may be declared as returning nothing.A return of what Java calls
voidisUnitin Scala.For more information, why don't you see what James Iry had to say about Getting to the Bottom of Nothing at All.