scala - val 和单例对象之间的细微差别是什么？

Question

本质上类似于val 和 def 之间细微差异的问题。我想知道拥有成员单例对象之间的语义区别是什么：

class Text {
    ...
    object Whitespace { def unapply(s :String) = 
        if (s.forall(_.isWhitespace)) Some(s) else None
    }
}

和

class Text {
    ...
    val Whitespace = new { def unapply(s :String) = 
        if (s.forall(_.isWhitespace)) Some(s) else None
    }
}

我知道两者是如何转换为字节码的，但是我可以用代码中的一个来做什么，而我不能用另一个呢？

Answer 1

您需要更加努力地覆盖成员对象。

apm@mara:~/tmp$ skala
Welcome to Scala version 2.11.0-20130622-103744-990c2b024a (OpenJDK 64-Bit Server VM, Java 1.7.0_21).
Type in expressions to have them evaluated.
Type :help for more information.

scala> class Foo1 { val foo = () => 1 }
defined class Foo1

scala> class Bar1 extends Foo1 { override val foo = () => 2 }
defined class Bar1

scala> class Foo2 { object foo { def apply() = 3 } }
defined class Foo2

scala> class Bar2 extends Foo2 { override object foo { def apply() = 4 }}
<console>:8: error: overriding object foo in class Foo2;
 object foo cannot override final member
       class Bar2 extends Foo2 { override object foo { def apply() = 4 }}
                                                 ^

scala> :q
apm@mara:~/tmp$ skala -Yoverride-objects
Welcome to Scala version 2.11.0-20130622-103744-990c2b024a (OpenJDK 64-Bit Server VM, Java 1.7.0_21).
Type in expressions to have them evaluated.
Type :help for more information.

scala> class Foo2 { object foo { def apply() = 3 } }
defined class Foo2

scala> class Bar2 extends Foo2 { override object foo { def apply() = 4 }}
defined class Bar2

scala> new Bar2
res0: Bar2 = Bar2@508c825

scala> .foo()
res1: Int = 4

scala> :q