scala 从抽象类继承的案例类
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Case classes inheriting from abstract class
提问by Sven Jacobs
I'm currently learning Scala and I have some problems designing my case classes. I need two case classes that have the same properties. So I thought I would be a good idea to inherit from an abstract base class that defines these properties. However this code does not compile
我目前正在学习 Scala，但在设计案例类时遇到了一些问题。我需要两个具有相同属性的案例类。所以我认为从定义这些属性的抽象基类继承是个好主意。但是这段代码不能编译
abstract class Resource(val uri : String)
case class File(uri : String) extends Resource(uri)
case class Folder(uri : String) extends Resource(uri)
because uriin the case class constructors would overwrite the uriproperty of the base class.
因为uri在这种情况下，类构造函数会覆盖uri基类的属性。
What would be the correct way to design this?
设计这个的正确方法是什么？
I want to be able to do something like this
我希望能够做这样的事情
val arr = Array[Resource](File("test"), Folder("test2"))
arr.foreach { r : Resource => r match {
case f : File => println("It's a file")
case f : Folder => println("It's a folder")
} }
The "equivalent" Java code should be something like
“等效的”Java 代码应该类似于
abstract class Resource {
private String uri;
public Resource(String uri) {
this.uri = uri
}
public String getUri() {
return uri;
}
}
// same for Folder
class File extends Resource {
public File(String uri) {
super(uri);
}
}
回答by onof
The correct syntax should be:
正确的语法应该是：
abstract class Resource {
val uri: String
}
case class File(uri : String) extends Resource
case class Folder(uri : String) extends Resource
Stream[Resource](File("test"), Folder("test2")) foreach {
r : Resource => r match {
case f : File => println("It's a file")
case f : Folder => println("It's a folder")
} }
EDIT
编辑
Without case classes:
没有案例类：
abstract class Resource(val uri : String)
class File(uri : String) extends Resource(uri) {
override def toString = "..."
}
object File {
def apply(uri: String) = new File(uri)
}
class Folder(uri : String) extends Resource(uri) {
override def toString = "..."
}
object Folder {
def apply(uri: String) = new Folder(uri)
}
回答by Brian Hsu
Make these two case class extends a common trait which define it interface and it should work.
使这两个 case 类扩展一个共同的特征，定义它的接口并且它应该可以工作。
BTW, you need an identifier before the type clause in casestatement.
顺便说一句，在case语句中的类型子句之前需要一个标识符。
trait Resource {
val uri: String
}
case class File(uri : String) extends Resource
case class Folder(uri : String) extends Resource
val arr = Array[Resource](File("test"), Folder("test2"))
arr.foreach { r : Resource => r match {
case s: File => println("It's a file")
case s: Folder => println("It's a folder")
}}
