[英]Best way to instantiate a type parameter class in Scala
I wanted to have a Queue class that could be used the same way as a list. 我想要一个可以与列表相同使用的Queue类。
For instance, 例如,
val q = Queue()
would instantiate an empty queue. 将实例化一个空队列。
For that purpose I tried using a companion class : 为此,我尝试使用一个伴随类:
object Queue {
def apply() = new Queue[Any]
}
Is that the right way to do it ? 那是正确的方法吗?
Using the apply
method of the companion object is the right way to do it, but you could also add a type parameter on apply
itself: 使用伴随对象的
apply
方法是正确的方法,但是您也可以在apply
自身上添加类型参数:
object Queue {
def apply[T]() = new Queue[T]
}
So that you can create a Queue
of the right type: 这样您就可以创建正确类型的
Queue
:
val q = Queue[Int]()
Usually you also allow populating the sequence on creation, so that the element type can be inferred, as in: 通常,您还允许在创建时填充序列,以便可以推断元素类型,如下所示:
def apply[T](elms: T*) = ???
So that you can do: 这样您就可以:
val q = Queue(1,2,3) // q is a Queue[Int]
Yes. 是。
If you want to initialise an object without using new
, then using apply()
as a factory method in the companion is absolutely the right way to go about it. 如果要初始化对象而不使用
new
,则在伴侣中使用apply()
作为工厂方法绝对是正确的方法。
You might also want to consider a more specific factory (or factories) to help make your code more self-documenting. 您可能还需要考虑一个或更具体的工厂(或多个工厂),以帮助使您的代码更具自记录性。
object Queue {
def apply[T](xs: T*) = new Queue(xs: _*)
def empty[T] = new Queue[T]()
}
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