Scala 方法类型和方法作为参数 [英] Scala method types and methods as parameters

问题描述

在下面的代码示例中，我不明白为什么函数 fun 可以作为参数传递给方法 addAction.方法 fun 是 Unit 类型，而 addAction 方法需要 () => Unit 类型的函数.

In the following code example, I do not understand why the function fun can be passed as an argument to the method addAction. The method fun is of type Unit, while the method addAction expects a function of type () => Unit.

如果 fun 是 () => Unit 类型，那么为什么编译器会抱怨 fun 是 Unit 类型，当我尝试将 fun 添加到操作列表时: actions = fun :: actions?

If fun is of type () => Unit, then why does the compiler complain that fun is of type Unit, when I try to add fun to the actions list: actions = fun :: actions?

package myscala

object MyScala {

  def fun() { println("fun1 executed.") }

  def addAction(a: () => Unit) {
    actions = a :: actions
  }

  var actions: List[() => Unit] = List()

  def main(args: Array[String]) {
    // the following line would produce a compiler error (found: Unit, required: () => Unit), it's OK
    // actions = fun :: actions
    actions = (() => fun) :: actions // OK
    // I would expect the same compiler error here (found: Unit, required: () => Unit), but it's OK why?
    addAction(fun)
    actions.foreach(_()) // prints twice "fun1 executed"
  }
}

推荐答案

以此为介绍性示例:

def fun() { println("fun1 executed.") }

val a1 = fun
val a2: () => Unit = fun

这两行都可以编译并且(由于类型推断)它们看起来是等效的.但是 a1 是 Unit 类型，而 a2 是 () => 类型.单位...这怎么可能?

Both lines compile and (thanks to type inference) they look equivalent. However a1 is of type Unit while a2 is of type () => Unit... How is this possible?

由于您没有明确提供a1的类型，编译器将fun解释为Unitfun调用code>，因此a1的类型与fun的类型相同.这也意味着这一行将打印 fun1 执行.

Since you are not explicitly providing type of a1, compilers interprets fun as a method fun call of type Unit, hence the type of a1 is the same as type of fun. It also means that this line will print fun1 executed.

然而，a2 已经明确声明了 () => 的类型.单位.编译器在这里帮助你，它理解因为上下文需要一个 () => 类型的函数；Unit 并且你提供了一个匹配这个类型的方法，它不应该调用那个方法，而是把它当作一等函数！

However, a2 has explicitly declared type of () => Unit. The compiler helps you here and it understands that since the context requires a function of type () => Unit and you provided a method matching this type, it shouldn't call that method, but treat it as first class function!

你不是注定要明确指定a1的类型.说:

You are not doomed to specify type of a1 explicitly. Saying:

val a1 = fun _

您现在明白问题出在哪里了吗?

Do you now understand where your problem is?

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