scala - 如何将可选的闭包参数传​​递给函数？

Question

我想将闭包作为选项传递给方法，我正在执行如下所示的操作。我收到如下所示的编译错误。是否可以将可选的闭包参数传​​递给函数？

def sampleMethod(a: String, b: String, optionalMethod: Option[(String, Int) => Unit]) {
    // do some processing with a and b
    optionalMethod match {
      case Some(optionalMethod) => {
        optionalMethod("a",3)
      }
      case _
      log("no optional method passed")
    }
}

// definition of optMethod in some other place
val optMethod = (c: String, d: Int) => {
  // some processing with c, d and external values 
}

// invoke
sampleMethod("hi", "bye", optMethod) => FAILS TO COMPILE

ERROR = type mismatch. expecting Option[(String, Int) => Unit] found (String, Int) => Unit

Answer 1

错误消息非常明确:sampleMethod期待 Option ，但您传递的是直接函数值(未包含在 Some 中)。

解决此问题的最简单方法是包装 optMethod成Some :

sampleMethod("hi", "bye", Some(optMethod))

但如果你想能够简单地做 sampleMethod("hi", "bye", optMethod) ，您可以添加 sampleMethod 的重载定义:

object Test {
  def sampleMethod(a: String, b: String, optionalMethod: Option[(String, Int) => Unit]) {
    // do some processing with a and b
    optionalMethod match {
      case Some(optionalMethod) => {
        optionalMethod("a",3)
      }
      case _ => log("no optional method passed")
    }
  }
  def sampleMethod(a: String, b: String) { sampleMethod(a, b, None) }
  def sampleMethod(a: String, b: String, optionalMethod: (String, Int) => Unit) {
    sampleMethod(a, b, Some(optionalMethod)) 
  }
}

val optMethod = (c: String, d: Int) => {
  // some processing with c, d and external values 
}

// invoke
Test.sampleMethod("hi", "bye", optMethod) // Now Compiles fine
Test.sampleMethod("hi", "bye") // This too