[英]How to create object/singleton of generic type in Scala?
在下面顯示的代碼中,如何將EmptyTree
轉換為對象(Singleton)?
trait Tree[T] {
def contains(num: T): Boolean
def inc( num: T ): Tree[T]
}
class EmptyTree[T <% Ordered[T] ] extends Tree[T] {
def contains(num:T):Boolean = false
def inc(num:T):Tree[T] = {
new DataTree(num, new EmptyTree, new EmptyTree)
}
override def toString = "."
}
class DataTree[T <% Ordered[T] ](val x:T, val left:Tree[T], val right:Tree[T]) extends Tree[T] {
def contains(num:T):Boolean = {
if( num < x ) left.contains(x)
else if ( num > x ) right.contains(x)
else true
}
def inc(num:T):Tree[T] = {
if(num < x ) new DataTree(x, left.inc(num), right)
else if ( num > x ) new DataTree(x, left, right.inc(num))
else this
}
override def toString = "{" + left + x + right + "}"
}
val t = new DataTree(20, new EmptyTree[Int], new EmptyTree[Int])
//> t : greeting.Test.DataTree[Int] = {.20.}
val p = t.inc(10) //> p : greeting.Test.Tree[Int] = {{.10.}20.}
val a = p.inc(30) //> a : greeting.Test.Tree[Int] = {{.10.}20{.30.}}
val s = a.inc(5) //> s : greeting.Test.Tree[Int] = {{{.5.}10.}20{.30.}}
val m = s.inc(11) //> m : greeting.Test.Tree[Int] = {{{.5.}10{.11.}}20{.30.}}
讓我告訴阿列克謝的回答。 以下是完整的實現,包括一些代碼樣式改進:
首先用協方差的知識來定義你的特征:
trait Tree[+T] {
def contains[U >: T : Ordering](num: U): Boolean
def inc[U >: T : Ordering](num: U): Tree[U]
}
接下來定義您的所有樹類型對象
case object EmptyTree extends Tree[Nothing] {
def contains[U >: Nothing : Ordering](num: U): Boolean = false
def inc[U >: Nothing : Ordering](num: U): Tree[U] =
DataTree(num, EmptyTree, EmptyTree)
override def toString = "."
}
現在改變你的一般案例實現:
case class DataTree[T: Ordering](x: T, left: Tree[T], right: Tree[T]) extends Tree[T] {
import Ordering.Implicits._
def contains[U >: T : Ordering](num: U): Boolean =
if (num < x) left.contains(x)
else if (num > x) right.contains(x)
else true
def inc[U >: T : Ordering](num: U): Tree[U] =
if (num < x) DataTree(x, left.inc(num), right)
else if (num > x) DataTree(x, left, right.inc(num))
else this
override def toString = "{" + left + x + right + "}"
}
因為我替換了Ordered
with Ordering
,你可能會有點沮喪,但你應該知道視圖邊界已被棄用
您必須修復泛型參數,因為這是您唯一可以提供的參數:
scala> trait A[T]
defined trait A
scala> object B extends A[Int]
defined object B
顯然你想為所有類型的T
重用EmptyTree
,所以不要為每種類型定義A[SOMETYPE]
,只需使用底部類型Nothing
:
scala> object B extends A[Nothing]
defined object B
此對象可以與任何樹一起使用。
這正是Scala中實現Option[T]
方式。 以下是None
定義方式:
case object None extends Option[Nothing]
如果保留泛型,也可以選擇添加空工廠 - 就像為Map和Vector完成的那樣。 當然,對於這樣的實現,它不會是每個創建的唯一實例對象,但是當使用inc
方法時,它不會生成新對象,它只會引用自身。
object DataTree {
def empty[T <% Ordered[T]] = new Tree[T] {
def contains(num: T):Boolean = false
def inc(num: T): Tree[T] = {
new DataTree(num, this, this)
}
override def toString = "."
}
}
所以你可以將它實例化如下:
val t = new DataTree(20, DataTree.empty[Int], DataTree.empty[Int])
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