我已经查看了Scala question about emulating Java's enum
和case classes vs. Enumeration但是效果太差,似乎付出了太大的努力。
基本上我希望values
方法返回DayOfWeek
的所有单例对象而不重复几次。
这就是我的代码应该是这样的:
object DayOfWeek extends MyEnum {
object MONDAY extends DayOfWeek(1)
object TUESDAY extends DayOfWeek(2)
object WEDNESDAY extends DayOfWeek(3)
object THURSDAY extends DayOfWeek(4)
object FRIDAY extends DayOfWeek(5)
object SATURDAY extends DayOfWeek(6)
object SUNDAY extends DayOfWeek(7)
}
class DayOfWeek(ordinal: Int)
方法values
应该返回类似这样的内容:
val values = Array(MONDAY, TUESDAY, WEDNESDAY, THURSDAY,
FRIDAY, SATURDAY, SUNDAY)
所有内容都应该在MyEnum
特征中发生,所以我只需要扩展它以获得功能。
trait MyEnum {
val values = this.getClass.getField("MODULE$") etc. etc.
}
有什么建议可以做到这一点吗?
这个想法是values
访问类并找到它们正在扩展的类的所有单例对象。
编辑:看起来所有建议都没有考虑到用户也可以创建对象,当然这些对象应该与定义的对象相当。
我会尝试再举一个例子,也许它更清楚:
object MonthDay extends MyEnum {
//Some important holidays
object NewYear extends MonthDay( 1, 1)
object UnityDay extends MonthDay(11, 9)
object SaintNicholas extends MonthDay(12, 6)
object Christmas extends MonthDay(12, 24)
}
class MonthDay(month: Int, day: Int)
//Of course the user can create other MonthDays
val myBirthDay = new MonthDay(month, day)
if(!MonthDay.values.contains(myBirthDay)) "Well, I probably have to work"
else "Great, it is a holiday!"
我希望有一个特性(MyEnum
),我可以将其混合到持有我的“枚举对象”的对象中,并使用方法返回它们的列表(def values: List[MonthDay]
)或迭代它们({ {1}}或def next: MonthDay
)。
PPS:I created a new question for the second part of this question 按照Ken Bloom的要求。
答案 0 :(得分:5)
这个怎么样?它要求您实际为每个新值调用add
方法,但values
会返回正确的类型。
abstract class MyEnum{
type Value //define me to be the value type for this MyEnum
private var _values:List[Value] = Nil
def values = _values
protected def add(newValue:Value) = {
_values = newValue::_values
newValue
}
}
object DayOfWeek extends MyEnum{
class Value(val dayNum:Int)
val SUNDAY = add(new Value(1))
val MONDAY = add(new Value(2))
val TUESDAY = add(new Value(3))
val WEDNESDAY = add(new Value(4))
val THURSDAY = add(new Value(5))
val FRIDAY = add(new Value(6))
val SATURDAY = add(new Value(7))
}
您现在可以致电
println(DayOfWeek.values map (_.dayNum))
如果您需要在不同对象上具有不同方法定义的单例对象,则可以创建如下的匿名类:
add(new Value{
override def dayNum=8
})
答案 1 :(得分:3)
scala.Enumeration完全符合您的要求。
我认为你可能会对Scala 2.7和Scala 2.8感到困惑。您引用emulating Java's enum
的旧问题是在Scala 2.7的时代编写的,尽管我无法测试Scala 2.7的Enumeration
所构成的功能,Scala 2.8的{{1}肯定拥有你正在寻找的一切。
您无法使用Enumeration
定义值,因为object SUNDAY extends Value(1)
已被懒惰地初始化。
答案 2 :(得分:2)
我能提出的最接近的是:
abstract class MyEnum(val displayName:String){
protected object Value{
var _values:List[Value] = Nil
def values = _values
}
protected class Value (val value:Int){
Value._values = this::Value._values
override def toString = "%s(%d)".format(displayName,value)
}
def values = Value.values
}
trait Methods{
def dayName
}
object DayOfWeek extends MyEnum("DayOfWeek"){
val SUNDAY = new Value(1) with Methods{
override def dayName = "Sunday"
}
val MONDAY = new Value(2) with Methods{
override def dayName = "Monday"
}
val TUESDAY = new Value(3) with Methods{
override def dayName = "Tuesday"
}
val WEDNESDAY = new Value(4) with Methods{
override def dayName = "Wednesday"
}
val THURSDAY = new Value(5) with Methods{
override def dayName = "Thursday"
}
val FRIDAY = new Value(6) with Methods{
override def dayName = "Friday"
}
val SATURDAY = new Value(7) with Methods{
override def dayName = "Saturday"
}
}
我还没想出如何更改_values
变量的类型以捕获完整类型Value with Methods
。