在 Scala 中应用和取消应用构造函数 [英] Is apply and unapply a constructor in scala

问题描述

在 Java 中，我们可以使用类名创建构造函数.但是在 Scala 中，我看到了 apply 和 unapply 的不同用法.这和Java中构造函数的用法一样，可以使用apply和unapply作为构造函数吗?

In Java we can make a constructor using the class name. But in scala I have seen different uses of apply and unapply. Is this the same like the usage of a constructor in Java, in which we can use apply and unapply as a constructor?

推荐答案

apply() 可以用作工厂方法(而不是构造函数)，它通常是，但不受限制或必须如此.apply() 的特殊之处在于它可以被悄悄地"调用.

apply() can be used as factory method (not a constructor), and it often is, but it is not restricted or required to be so. What makes apply() special is just that it can be "silently" invoked.

class MyClass(arg: String) {
  def apply(n: Int):String = arg*n
}

val mc = new MyClass("blah")
mc.apply(2)  // res0: String = blahblah
mc(3)        // res1: String = blahblahblah

您经常会看到 apply() 在伴随对象中用作工厂方法.

You'll often see apply() used as a factory method in a companion object.

object MyClass {
  def apply(s: String): MyClass = new MyClass(s)
}
val mc = MyClass("bliss")  // call the object's apply() method

但是，同样，这只是一个方便的约定，而不是要求.

But, again, that's just a convenient convention, not a requirement.

unapply()(和 unapplySeq())的不同之处在于，每当您尝试在类实例上进行模式匹配时，它都会静默"调用.

unapply() (and unapplySeq()) is different in that it is "silently" called whenever you attempt to pattern match on a class instance.

class MyClass(val arg: String) {}
object MyClass {
  def unapply(x: MyClass): Option[String] = Some(x.arg)
}
val mc = new MyClass("bland")
mc match {
  case MyClass(s) => println(s"got $s")  // output: "got bland"
  case _          => println("not")
}

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