Question

使用Scala REPL，我可以得到

的类型

scala> def tryit(thing: Null): Unit = { println("That worked!"); }
tryit: (thing: Null)Unit

然后，如何在REPL之外的Scala源中获取相同的信息？我试过了getClass，但是我收到了一个错误。

<console>:12: error: missing arguments for method tryit;
follow this method with `_' if you want to treat it as a partially applied function
       tryit.getClass
       ^

修改

从另一个堆栈溢出问题，我可以得到这个代码。

// http://stackoverflow.com/questions/11486298/how-can-i-get-the-functionality-of-the-scala-repl-type-command-in-my-scala-prog

import reflect.runtime.universe._

implicit class ColonTypeExtender [T : TypeTag] (x : T) {
    def colonType = typeOf[T].toString
}

object Main extends App {
    // Null is a type, and null is a value
    def tryit(thing: Null): Unit = { println("That worked!"); }
    println((tryit _).colonType)
    tryit(null)  
}

运行代码将显示我期望的结果。

Null => Unit

如何在Scala源中获取函数的类型

修改

0 个答案: