scala – 如何将可选的闭包参数传​​递给函数？

我想将一个闭包传递给一个方法作为一个选项,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

解决方法

错误消息非常明确：sampleMethod需要一个Option,但是您传递的是一个直接函数值(未包含在Some中).

解决这个问题的最简单方法是将optMethod包装成Some：

sampleMethod("hi",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,3)
      }
      case _ => log("no optional method passed")
    }
  }
  def sampleMethod(a: String,b: String) { sampleMethod(a,b,None) }
  def sampleMethod(a: String,optionalMethod: (String,Int) => Unit) {
    sampleMethod(a,Some(optionalMethod)) 
  }
}

val optMethod = (c: String,d and external values 
}

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