Question

我正在引导自己进入Scala，并且正在努力解决为什么编译器在以下来源中禁止classOf[Hello]。

package example

import org.slf4j.LoggerFactory

object Hello extends Greeting with App {
  val logger = LoggerFactory.getLogger(classOf[Hello])
  logger.info(s"greeting is $greeting")
  println(greeting)
}

trait Greeting {
  lazy val greeting: String = "hello"
}

编译器抱怨

[info] Compiling 2 Scala sources to /Users/robert.kuhar/dev/hellosbt/target/scala-2.12/classes ...
[error] /Users/robert.kuhar/dev/hellosbt/src/main/scala/example/Hello.scala:6:48: not found: type Hello
[error]   val logger = LoggerFactory.getLogger(classOf[Hello])
[error]                                                ^
[error] one error found
[error] (compile:compileIncremental) Compilation failed
[error] Total time: 1 s, completed Dec 17, 2017 1:20:28 PM

Hello是类Hello的单例，不是吗？在我的应用程序中利用常见的“Logger Name is Class Name”来实例化记录器需要什么？

Answer 1

因为您无法获得object的类型。

scala> object MySingleton
defined object MySingleton

scala> classOf[MySingleton]
<console>:12: error: not found: type MySingleton
       classOf[MySingleton]
               ^

另外，请注意当您将单例分配给变量时，键入的内容为MySingleton.type

scala> object MySingleton { def foo = "bar" }
defined object MySingleton

scala> val x = MySingleton
x: MySingleton.type = MySingleton$@4ffe3d42

scala> x.foo
res3: String = bar

通常情况下，我在单个对象中创建记录器，但是参考该类。

import org.slf4j.LoggerFactory

class Hello

object Hello extends Greeting with App {

  val logger = LoggerFactory.getLogger(classOf[Hello])

  logger.info(s"greeting is $greeting")
  println(greeting)

}

trait Greeting {
  lazy val greeting: String = "hello"
}

另一种方法是使用方法.getClass

获取课程

scala> object MySingleton
defined object MySingleton

scala> MySingleton.getClass
res1: Class[_ <: MySingleton.type] = class MySingleton$

读数

http://ktoso.github.io/scala-types-of-types/

为什么classOf [T]在我的对象中不起作用？

1 个答案:

读数