ScalaNumber 的底层实现是如何工作的？
scala.math.ScalaNumber 是一个 Java 文件，如下所示：
public abstract class ScalaNumber extends java.lang.Number {
protected abstract boolean isWhole();
public abstract Object underlying();
}
scala.math.BigDecimal 实现它的方式为：
class BigDecimal(val bigDecimal: BigDec, val mc: MathContext)
extends ScalaNumber with ScalaNumericConversions with Serializable {
...
def underlying = bigDecimal
}
以及 scala.math.BigInt< /code>：
class BigInt(val bigInteger: BigInteger) extends ScalaNumber with ScalaNumericConversions with Serializable {
...
def underlying = bigInteger
}
令人困惑的是，underlying 的类型是 java.math.BigDecimal/java.math.BigInt 而不是对象。
我是否错过了一些非常明显的东西，或者这里有什么特殊情况吗？
编辑：当然我错过了一些明显的事情......你没问题。协变返回类型。谢谢！
scala.math.ScalaNumber is a Java file which looks like this:
public abstract class ScalaNumber extends java.lang.Number {
protected abstract boolean isWhole();
public abstract Object underlying();
}
And scala.math.BigDecimal implements it with:
class BigDecimal(val bigDecimal: BigDec, val mc: MathContext)
extends ScalaNumber with ScalaNumericConversions with Serializable {
...
def underlying = bigDecimal
}
as well as scala.math.BigInt:
class BigInt(val bigInteger: BigInteger) extends ScalaNumber with ScalaNumericConversions with Serializable {
...
def underlying = bigInteger
}
Confusingly, underlying is of type java.math.BigDecimal/java.math.BigInt instead of Object.
Do I miss something pretty obvious or is there something special-cased here?
EDIT: Of course I missed something obvious ... You're all right. Co-variant return types. Thanks!
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它只是一个协变返回类型，在 Scala 和 Java 中都是允许的。
其背后的基本原理是：如果一个类
Base承诺从某个方法返回A，那么子类Derived <: Base就会尊重里氏替换原则如果它返回A或任何子类B <: A< /代码>。当然，如果BigInt#underlying返回一个BigInteger，对于ScalaNumber的客户来说这不是问题，他们可能只希望得到一个简单的对象。It is simply a covariant return type, which is allowed both in Scala and Java.
The rationale behind it is: if a class
Basepromises to returnAform a certain method, then a subclassDerived <: Baserespects the Liskov substitution principle if it returnsAor any subclassB <: A. Certainly, ifBigInt#underlyingreturns aBigInteger, this is no problem for clients ofScalaNumber, who may only hope for a plainObject.在 Java 和 Scala 方法中，重写时返回类型可以是协变的。也就是说，如果您重写一个方法，则可以使其返回类型成为被重写方法的返回类型的子类型。
In both Java and Scala method return types can be covariant when overriding. That is to say if you override a method you can make its return type be a subtype of the overriden method's return type.