从Scala中的类型参数创建对象 [英] Creating an object from a type parameter in Scala

问题描述

我想创建一个接受类型参数的方法，该方法的构造函数上显然没有任何参数，然后返回用该构造函数构造的虚拟对象.基本上是某种工厂模式.

I want to create a method that takes a type parameter, obviously with no parameters on its constructor, and returns a dummy object constructed with that constructor. Basically some kind of factory pattern.

• 在Scala中甚至有可能吗?
• 这是个好主意吗?有没有更好的模式?
• 有没有办法仅在编译时(即不进行反射)实现这一目标?

代码示例:

trait Model
class A extends Model
class B extends Model

def dummy[T <: Model] = new T   // Fails compilation with "class type required but T found"

dummy[A]  // Returns an instance of A
dummy[B]  // Returns an instance of B

推荐答案

这可以使用 ClassManifest s完成，该类旨在克服擦除的作用:

This can be done using ClassManifests, which are designed to overcome erasure:

def dummy[T <: Model : ClassManifest] = classManifest[T].erasure.newInstance

至于在编译时不进行反射，我想它可以使用scala 2.10宏来完成.

As for doing it at compile time without reflection, I guess that it could be done using scala 2.10 macros.

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