似乎可以更改具有如下特征的类上方法的实现:
trait Abstract { self: Result =>
override def userRepr = "abstract"
}
abstract class Result {
def userRepr: String = "wtv"
}
case class ValDefResult(name: String) extends Result {
override def userRepr = name
}
val a = new ValDefResult("asd") with Abstract
a.userRepr
这里提供了实时代码:http://www.scalakata.com/52534e2fe4b0b1a1c4daa436
但现在我想调用函数的前一个或超级实现,如下所示:
trait Abstract { self: Result =>
override def userRepr = "abstract" + self.userRepr
}
或
trait Abstract { self: Result =>
override def userRepr = "abstract" + super.userRepr
}
但是,这些替代方案都不能编译。你知道怎么做到这一点吗?
发布于 2013-10-08 10:32:13
这就是我要找的答案。感谢您使用Scala的abstract override
特性为我指明了正确的方向。
trait Abstract extends Result {
abstract override def userRepr = "abstract " + super.userRepr
}
abstract class Result {
def userRepr: String = "wtv"
}
case class ValDefResult(name: String) extends Result {
override def userRepr = name
}
val a = new ValDefResult("asd") with Abstract
a.userRepr
这里提供了实时代码:http://www.scalakata.com/52536cc2e4b0b1a1c4daa4a4
对于令人困惑的示例代码,很抱歉,我正在编写一个处理Scala AST的库,并且没有足够的灵感来更改名称。
发布于 2013-10-08 09:02:41
我不知道您是否可以进行以下更改,但您想要的效果可以通过引入一个额外的特征(我称之为Repr
)并在Abstract
特征中使用abstract override
来实现:
trait Repr {
def userRepr: String
}
abstract class Result extends Repr {
def userRepr: String = "wtv"
}
case class ValDefResult(name: String) extends Result {
override def userRepr = name
}
trait Abstract extends Repr { self: Result =>
abstract override def userRepr = "abstract-" + super.userRepr // 'super.' works now
}
您的示例用法现在提供:
scala> val a = new ValDefResult("asd") with Abstract
a: ValDefResult with Abstract = ValDefResult(asd)
scala> a.userRepr
res3: String = abstract-asd
发布于 2013-10-08 18:00:58
abstract override
是一种机制,也就是可堆叠的特征。值得一提的是,线性化很重要,因为线性化决定了super
的含义。
这个问题是the canonical Q&A on self-type vs extension的一个很好的补充。
继承对于self类型是不明确的:
scala> trait Bar { def f: String = "bar" }
defined trait Bar
scala> trait Foo { _: Bar => override def f = "foo" }
defined trait Foo
scala> new Foo with Bar { }
<console>:44: error: <$anon: Foo with Bar> inherits conflicting members:
method f in trait Foo of type => String and
method f in trait Bar of type => String
(Note: this can be resolved by declaring an override in <$anon: Foo with Bar>.)
new Foo with Bar { }
^
很明显,你可以选择:
scala> new Foo with Bar { override def f = super.f }
res5: Foo with Bar = $anon$1@57a68215
scala> .f
res6: String = bar
scala> new Foo with Bar { override def f = super[Foo].f }
res7: Foo with Bar = $anon$1@17c40621
scala> .f
res8: String = foo
或
scala> new Bar with Foo {}
res9: Bar with Foo = $anon$1@374d9299
scala> .f
res10: String = foo
https://stackoverflow.com/questions/19237049
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