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:
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 Object
so you have to cast it back with asInstanceOf[T]
The closest I could get to your instantiate
function is this:
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).
Edit: 20 September 2012
Three years on, the instantiate
method can be improved to return a properly typed object.
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
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