[英]Single Abstract Method Equivalent to a Trait?
考慮以下特征:
trait Foo {
def m1(id : Int) : Try[String]
}
並舉例說明:
val g : Foo = new Foo {
override def m1(id: Int): Try[String] = Success("Good job")
}
Intellij建議將其轉換為單一抽象方法:
val g : Foo = (id: Int) => Success("Good job")
這兩個等效嗎?
讓我們檢查一下scalac
作用:
import scala.util.{Success, Try}
trait Foo {
def m1(id : Int) : Try[String]
}
object MainClass {
val g : Foo = new Foo {
override def m1(id: Int): Try[String] = Success("Good job")
}
}
跑:
$ scalac -print src/main/scala/MainClass.scala
並輸出:
[[syntax trees at end of cleanup]] // MainClass.scala
package <empty> {
abstract trait Foo extends Object {
def m1(id: Int): scala.util.Try
};
object MainClass extends Object {
private[this] val g: Foo = _;
<stable> <accessor> def g(): Foo = MainClass.this.g;
def <init>(): MainClass.type = {
MainClass.super.<init>();
MainClass.this.g = {
new <$anon: Foo>()
};
()
}
};
final class anon$1 extends Object with Foo {
override def m1(id: Int): scala.util.Try = new scala.util.Success("Good job");
def <init>(): <$anon: Foo> = {
anon$1.super.<init>();
()
}
}
}
並對以下步驟執行相同的步驟:
import scala.util.{Success, Try}
trait Foo {
def m1(id : Int) : Try[String]
}
object MainClass {
def main(args: Array[String]): Unit = {
val g : Foo = (id: Int) => Success("Good job")
}
}
跑:
$ scalac -print src/main/scala/MainClass.scala
輸出:
[[syntax trees at end of cleanup]] // MainClass.scala
package <empty> {
abstract trait Foo extends Object {
def m1(id: Int): scala.util.Try
};
object MainClass extends Object {
private[this] val g: Foo = _;
<stable> <accessor> def g(): Foo = MainClass.this.g;
final <artifact> private[this] def $anonfun$g$1(id: Int): scala.util.Try = new scala.util.Success("Good job");
def <init>(): MainClass.type = {
MainClass.super.<init>();
MainClass.this.g = {
((id: Int) => MainClass.this.$anonfun$g$1(id))
};
()
}
}
}
如您所見,它們對於編譯器而言是不同的。
在第一種情況下,它是一個匿名對象,其次是一個匿名函數。
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.