如何将可选的闭包参数传​​递给函数？

Question

我想将一个闭包传递给一个方法作为Option，我正在做如下所示的事情。我收到编译错误，如下所示。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需要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将optMethod包裹到Some中：

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

但是，如果您希望能够简单地执行sampleMethod("hi", "bye", optMethod)，则可以添加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

Answer 2

如前所述，您的方法需要包含Option的{​​{1}}值。因此，您必须将optionalMethod值传递给它：

Option

如果您想避开// invoke with method sampleMethod("hi", "bye", Some(optMethod)) // invoke without method sampleMethod("hi", "bye", None) 值（尤其是避免Option），您可以尝试以下操作：

None

Answer 3

怎么样

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

Answer 4

更清楚：

scala> def sampleMethod(a: String, b: String, optionalMethod: Option[(String, Int) => Unit]) {
     | optionalMethod.map(f => f("a", 3))
     | }
sampleMethod: (a: String, b: String, optionalMethod: Option[(String, Int) => Unit])Unit


scala> sampleMethod("A", "A", Some((c:String, d:Int) => println(s"Hello wolrd $c...$d")))
Hello wolrd a...3

您必须在可选功能

周围添加“Some（）”