[英]Overriding a sealed trait in Scala
I am using a library that has a sealed trait. 我正在使用一个具有密封特性的库。 I really need to extend this trait.
我真的需要扩展这个特性。
Is there a way (even a dirty one) to bypass that? 有没有办法(甚至是肮脏的)绕过那个?
For a bit of background, this is what I am trying to work around : https://github.com/ReactiveMongo/ReactiveMongo/issues/247 对于一些背景知识,我正在努力解决这个问题: https : //github.com/ReactiveMongo/ReactiveMongo/issues/247
Don't extend, include, If you need to consider another option. 不要扩展,包括, 如果您需要考虑另一种选择。
// Foo.scala
sealed trait Foo
case class Foo1(n: Int) extends Foo
case class Foo2(s: String) extends Foo
// Bar.scala
sealed trait Bar
case class BarFoo(foo: Foo) extends Bar
case class Bar3(b: Boolean) extends Bar
And the boilerplate can be reduced by implicit conversion 并且可以通过隐式转换来减少样板
implicit def fooToBar(foo: Foo): Bar = BarFoo(foo)
Or if you want to add new methods to an existing class , use implicit classes: 或者, 如果要向现有类添加新方法 ,请使用隐式类:
// Foo.scala
sealed trait Foo
case class Foo1(n: Int) extends Foo
case class Foo2(s: String) extends Foo
// Quux.scala
// This class be called FooOps or FooExtra, if you like, name is irrelevant
implicit class Quux(foo: Foo) {
def quux: Boolean = foo match {
case Foo1(n) => n == 0
case Foo2(s) => s.isEmpty
}
}
Then you can use .quux
on Foo
values: 然后你可以在
Foo
值上使用.quux
:
scala> Foo1(2).quux
res1: Boolean = false
scala> Foo2("").quux
res2: Boolean = true
This is about the same as extending, but with even less boilerplate, compare to: 这与扩展大致相同,但更少的样板,与以下内容相比:
case class Quux2(foo: Foo) {
def quux2: Boolean = foo match {
case Foo1(n) => n == 0
case Foo2(s) => s.isEmpty
}
}
implicit def fooToQuux2(foo: Foo): Quux2 = Quux2(foo)
scala> Foo1(2).quux
res0: Boolean = false
scala> Foo2("").quux
res1: Boolean = true
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