继承：在类构造函数中覆盖

Question

我正在阅读此链接中的示例http://www.tutorialspoint.com/scala/scala_classes_objects.htm 示例：

class Point(val xc: Int, val yc: Int) {
   var x: Int = xc
   var y: Int = yc
   def move(dx: Int, dy: Int) {
      x = x + dx
      y = y + dy
      println ("Point x location : " + x);
      println ("Point y location : " + y);
   }
}

class Location(override val xc: Int, override val yc: Int,
   val zc :Int) extends Point(xc, yc){
   var z: Int = zc

   def move(dx: Int, dy: Int, dz: Int) {
      x = x + dx
      y = y + dy
      z = z + dz
      println ("Point x location : " + x);
      println ("Point y location : " + y);
      println ("Point z location : " + z);
   }
}

object Test {
   def main(args: Array[String]) {
      val loc = new Location(10, 20, 15);

      // Move to a new location
      loc.move(10, 10, 5);
   }
}

我不明白类位置构造函数中 override 关键字的作用！为什么要提到它，因为我们在这里有一个扩展？谢谢 ！

Answer 1

给定的

scala> class A(val a: Int)
defined class A

如果您尝试这样定义B：

scala> class B(val a: Int) extends A(a)
<console>:11: error: overriding value a in class A of type Int;
 value a needs `override' modifier
       class B(val a: Int) extends A(a)

编译器抱怨，因为看起来您正在尝试a在类中定义一个成员，B并且它已经通过继承（来自A）存在于那里。override在这种情况下，您需要添加以明确您的意图：

scala> class B(override val a: Int) extends A(a)
defined class B

更具体地说，override如果您要覆盖抽象成员，则不必提供：

scala> trait A { def a: Int }
defined trait A

scala> class B(override val a: Int) extends A
defined class B

scala> class B(val a: Int) extends A
defined class B

但是，为了避免在混合特征时意外覆盖，Scala 通过要求明确的override.

考虑这个例子：

这里没有问题：

scala> trait A { def a: Int = 1 }
defined trait A

scala> class B
defined class B

scala> new B with A
res0: B with A = $anon$1@3e29739a

您已免于歧义：

scala> trait A { def a: Int = 1 }
defined trait A

scala> class B { def a: Int = 2 }
defined class B

scala> new B with A
<console>:13: error: <$anon: B with A> inherits conflicting members:
  method a in class B of type => Int  and
  method a in trait A of type => Int
(Note: this can be resolved by declaring an override in <$anon: B with A>.)
       new B with A
           ^

手动解决冲突：

scala> trait A { def a: Int = 1 }
defined trait A

scala> class B { def a: Int = 2 }
defined class B

scala> new B with A { override val a = super[B].a }
res6: B with A{val a: Int} = $anon$1@76f6896b