scala 使用构造函数参数通过反射实例化对象

instantiate object with reflection using constructor arguments

提问by justin

I'm trying to figure out how to instantiate a case class object with reflection. Is there any support for this? The closest I've come is looking at scala.reflect.Invocation, but this seems more for executing methods that are a part of an object.

我想弄清楚如何用反射实例化一个案例类对象。是否有任何支持？我最接近的是查看 scala.reflect.Invocation，但这似乎更适合执行作为对象一部分的方法。

case class MyClass(id:Long, name:String)

def instantiate[T](className:String)(args:Any*) : T = { //your code here }

Is close to the API I'm looking for.

接近我正在寻找的 API。

Any help would be appreciated.

任何帮助，将不胜感激。

回答by michael.kebe

scala> case class Foo(id:Long, name:String)
defined class Foo

scala> val constructor = classOf[Foo].getConstructors()(0)
constructor: java.lang.reflect.Constructor[_] = public Foo(long,java.lang.String)

scala> val args = Array[AnyRef](new java.lang.Integer(1), "Foobar")
args: Array[AnyRef] = Array(1, Foobar)

scala> val instance = constructor.newInstance(args:_*).asInstanceOf[Foo]
instance: Foo = Foo(1,Foobar)

scala> instance.id
res12: Long = 1

scala> instance.name
res13: String = Foobar

scala> instance.getClass
res14: java.lang.Class[_] = class Foo

Currently there is not much reflection support in Scala. But you can fall back to th Java Reflection API. But there are some obstacles:

目前 Scala 中没有太多反射支持。但是您可以退回到 Java 反射 API。但是有一些障碍：

• You have to create a Array[AnyRef]and box your "primitive types" in the wrapper classes (java.lang.Integer, java.lang.Character, java.lang.Double, ...)

• newInstance(Object ... args)gets an varargs array of Object, so you should give the type inferer a hint with :_*

• newInstance(...)returns an Objectso you have to cast it back with asInstanceOf[T]

• 您必须Array[AnyRef]在包装类 ( java.lang.Integer, java.lang.Character, java.lang.Double, ...) 中创建一个并装箱您的“原始类型”

• newInstance(Object ... args)得到一个可变参数数组Object，所以你应该给类型推断器一个提示:_*

• newInstance(...)返回一个Object所以你必须把它扔回去asInstanceOf[T]

The closest I could get to your instantiatefunction is this:

我能得到的最接近你的instantiate功能是这样的：

def instantiate(clazz: java.lang.Class[_])(args:AnyRef*): AnyRef = {
  val constructor = clazz.getConstructors()(0)
  return constructor.newInstance(args:_*).asInstanceOf[AnyRef]
}

val instance = instantiate(classOf[MyClass])(new java.lang.Integer(42), "foo")
println(instance)           // prints: MyClass(42,foo)
println(instance.getClass)  // prints: class MyClass

You cannot get the get class from a generic type. Java erases it (type erasure).

您无法从泛型类型获取 get 类。Java 擦除它（类型擦除）。

Edit: 20 September 2012

编辑：2012 年 9 月 20 日

Three years on, the instantiatemethod can be improved to return a properly typed object.

三年过去了，instantiate可以改进该方法以返回正确类型的对象。

def instantiate[T](clazz: java.lang.Class[T])(args:AnyRef*): T = {
  val constructor = clazz.getConstructors()(0)
  return constructor.newInstance(args:_*).asInstanceOf[T]
}

See http://www.nabble.com/How-do-I-get-the-class-of-a-Generic--td20873455.html

见http://www.nabble.com/How-do-I-get-the-class-of-a-Generic--td20873455.html

回答by Eugene Yokota

See answers to Scala: How do I dynamically instantiate an object and invoke a method using reflection?as well, especially regarding type erasure.

请参阅Scala 的答案：如何使用反射动态实例化对象并调用方法？同样，尤其是在类型擦除方面。

回答by justin

This is what I've ended up with so far, I'd like to not have to deal directly with AnyRef if possible. So if anyone knows a way to get around that I'd appreciate the help.

到目前为止，这就是我最终得到的结果，如果可能的话，我不想直接与 AnyRef 打交道。因此，如果有人知道解决方法，我将不胜感激。

case class MyClass(id:Long,name:String)

def instantiate[T](classArgs: List[AnyRef])(implicit m : Manifest[T]) : T ={
      val constructor = m.erasure.getConstructors()(0)
      constructor.newInstance(classArgs:_*).asInstanceOf[T]
    }

val list = List[AnyRef](new java.lang.Long(1),"a name")
val result = instantiate[MyClass](list)
println(result.id)