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
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错误消息非常明确:sampleMethod期望a Option,但是您传递的是直接函数值(未包含在内Some).
解决这个问题的最简单方法是包装optMethod成Some:
sampleMethod("hi", "bye", Some(optMethod))
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但是如果你想简单地做sampleMethod("hi", "bye", optMethod),你可以添加重载的定义sampleMethod:
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
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如前所述,您的方法需要一个Option包含的值optionalMethod.因此,您必须将Option值传递给它:
// invoke with method
sampleMethod("hi", "bye", Some(optMethod))
// invoke without method
sampleMethod("hi", "bye", None)
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如果您想避免该Option值(特别是避免None),您可以尝试以下方法:
def sampleMethod(a: String, b: String, optionalMethod: (String, Int) => Unit = (_, _) => log("no optional method passed")) {
optionalMethod("a", 3)
}
// invoke with method
sampleMethod("hi", "bye", optMethod)
// invoke without method
sampleMethod("hi", "bye")
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