I would like to make a case class Bla that takes a type parameter A and it knows the type of A at runtime (it stores it in its info field).
My attempt is shown in the example below. The problem is that this example does not compile.
case class Bla[A] (){
val info=Run.paramInfo(this) // this does not compile
}
import scala.reflect.runtime.universe._
object Run extends App{
val x=Bla[Int]
def paramInfo[T](x:T)(implicit tag: TypeTag[T]): String = {
val targs = tag.tpe match { case TypeRef(_, _, args) => args }
val tinfo=s"type of $x has type arguments $targs"
println(tinfo)
tinfo
}
paramInfo(x)
}
However when I comment val info=Run.paramInfo(this) then the program runs fine and prints:
type of Bla() has type arguments List(Int)
Is there a way to make this example below compile ? (or in some other way achieve the same goal, i.e. that a case class is self aware of the type of it's type parameter?)
There's little point in using reflection based APIs for this, shapeless has a typeclass that exposes compile time information to runtime using an implicit macro.
import shapeless.Typeable
class Test[T : Typeable] {
def info: String = implicitly[Typeable[T]].describe
}
It's also relatively easy to roll your own thing here, with the added inconvenience of having to compile the implicit macro in a different compilation unit than whatever is using it.
You just need to pass the implicit type tag parameter to the case class constructor (otherwise the type information is lost before calling paraInfo which requires it):
case class Bla[A : TypeTag]() { ... }
Which is shorthand for:
case class Bla[A](implicit tag: TypeTag[A]) { ... }
I'm new to both Scala and Akka and have been following this tutorial. I came across the following and wondering what exactly this syntax does/mean?
import akka.actor.Props
val props1 = Props[MyActor] //Not sure what this means???
val props2 = Props(new ActorWithArgs("arg")) // careful, see below
val props3 = Props(classOf[ActorWithArgs], "arg")
I'm not sure what the line commented with //Not sure what this means does? It seems like a generic trait that gives a parameterised type. If I look at the source code, akka.actor.Props is defined as an Object that extends the trait AbstractProps. However, AbstractProps is not defined with a type parameter i.e. AbstractProps[T]. Can someone explain how that above line works and what it does?
In Scala, any object which implements an apply method can be called without the new keyword, simply by calling MyObject(), which will automatically lookup for it's apply.
If you look at the companion object for Props, you'll see the following method defined:
/**
* Scala API: Returns a Props that has default values except for "creator"
* which will be a function that creates an instance
* of the supplied type using the default constructor.
*/
def apply[T <: Actor: ClassTag](): Props =
apply(defaultDeploy, implicitly[ClassTag[T]].runtimeClass, List.empty)
This apply takes one type parameter and no arguments. T <: Actor means that T, the type you're passing, must extend Actor. That's how Scala knows how to create the object.
Additionally, any method with arity-0 in Scala may drop it's parenthesis. That's how you're seeing Props[MyActor] actually compile, as it is equivalent of Props[MyActor](), which is equivalent to Props.apply[MyActor]().
akka.actor.Props is defined as an Object that extends the trait
AbstractProps
Its also defined as a case class:
final case class Props(deploy: Deploy, clazz: Class[_], args: immutable.Seq[Any])
This is a common pattern in Scala, a class with a companion object. The companion object frequently houses factory methods, and thats what your actually calling in your example.
val props1 = Props[MyActor]
This simply calls apply() of the companion object. You can omit the parentheses in Scala if no arguments are neccessary and apply is a special method that can be invoked directly on the object/instance. Say you have a sequence and want the element at index 1:
val s = Seq("one", "two", "three")
// These two are equivalent
s(1) // -> "two"
s.apply(1) // -> "two"
Ultimately your code can be rewritten as
val props1 = Props.apply[MyActor]()
The following code does not compile:
import scala.language.implicitConversions
trait Base {
class Wrp[+T](val v: T) // wrapper / internal representation
}
trait BooleanOps extends Base {
// implicit conversion
implicit def lift2BooleanOpsCls(x: Boolean): BooleanOpsCls =
new BooleanOpsCls(new Wrp[Boolean](x))
class BooleanOpsCls(wx: Wrp[Boolean]) {
def ||(wy: =>Wrp[Boolean]): Wrp[Boolean] = new Wrp[Boolean](wx.v || wy.v)
}
}
trait MyExample extends BooleanOps {
// test method
def foo(): Wrp[Boolean] = {
val ret: Wrp[Boolean] = false || new Wrp[Boolean](true)
ret
}
}
Output:
MyExample.scala:18: error: type mismatch;
found : MyExample.this.Wrp[Boolean]
required: Boolean
val ret: Wrp[Boolean] = false || new Wrp[Boolean](true)
^
But if I:
1) put the class Wrp outside of Base
or
2) move the body of BooleanOps to MyExample
everything compiles.
Why does not the original example work? If you have some insight in this behavior, help would be appreciated. Thank you.
One issue is the call-by-name nature of the argument in the def ||(wy: =>Wrp[Boolean])
if you rewite it to def ||(wy: Wrp[Boolean]) it works
but I agree that it is weird that it works if you move around Wrp or BooleanOpsCls! Intended or bug of implicit resolution??
The original example will work if you rename the || method. The compiler finds the false.||() method and doesn't bother to look for an implicit that might also work there.
The problem is that there is no single class called Wrp (ignore the T for a moment) -- Base does not define Wrp but rather defines a named subclass of each and every concrete class that extends Base. The implicits are a red herring, too. The error that's the giveaway is the mention of MyExample.this.Wrp -- remember that there is no such class even as MyExample -- val x = new MyExample would have type Object with MyExample.
Why wouldn't the scala compiler dig this:
class Clazz
class Foo[C <: Clazz] {
val foo = new C
}
class type required but C found
[error] val a = new C
[error] ^
Related question - How to get rid of : class type required but T found
This is a classic generic problem that also happens in Java - you cannot create an instance of a generic type variable. What you can do in Scala to fix this, however, is to introduce a type evidence to your type parameter that captures the runtime type:
class Foo[C <: Clazz](implicit ct: ClassTag[C]) {
val foo = ct.runtimeClass.newInstance
}
Note that this only works if the class has a constructor without any arguments. Since the parameter is implicit, you don't need to pass it when calling the Foo constructor:
Foo[Clazz]()
I came up with this scheme, couldn't simplify it through a companion object thought.
class Clazz
class ClazzFactory {
def apply = new Clazz
}
class Foo(factory: ClazzFactory) {
val foo: Clazz = factory.apply
}
It's very annoying that ClazzFactory can't be an object rather than a class though. A simplified version:
class Clazz {
def apply() = new Clazz
}
class Foo(factory: Clazz) {
val foo: Clazz = factory.apply
}
This requires the caller to use the new keyword in order to provide the factory argument, which is already a minor enough annoyance relative to the initial problem. But, scala could have made this scenario all more elegant; I had to fallback here to passing a parameter of the type I wish to instantiate, plus the new keyword. Maybe there's a better way.
(motivation was to instantiate that type many times within the real Foo, that's why this is at all a solution; otherwise my pattern above is just redundantly meaningless).
I want to do something like this:
sealed abstract class Base(val myparam:String)
case class Foo(override val myparam:String) extends Base(myparam)
case class Bar(override val myparam:String) extends Base(myparam)
def getIt( a:Base ) = a.copy(myparam="changed")
I can't, because in the context of getIt, I haven't told the compiler that every Base has a 'copy' method, but copy isn't really a method either so I don't think there's a trait or abstract method I can put in Base to make this work properly. Or, is there?
If I try to define Base as abstract class Base{ def copy(myparam:String):Base }, then case class Foo(myparam:String) extends Base results in class Foo needs to be abstract, since method copy in class Base of type (myparam: String)Base is not defined
Is there some other way to tell the compiler that all Base classes will be case classes in their implementation? Some trait that means "has the properties of a case class"?
I could make Base be a case class, but then I get compiler warnings saying that inheritance from case classes is deprecated?
I know I can also:
def getIt(f:Base)={
(f.getClass.getConstructors.head).newInstance("yeah").asInstanceOf[Base]
}
but... that seems very ugly.
Thoughts? Is my whole approach just "wrong" ?
UPDATE I changed the base class to contain the attribute, and made the case classes use the "override" keyword. This better reflects the actual problem and makes the problem more realistic in consideration of Edmondo1984's response.
This is old answer, before the question was changed.
Strongly typed programming languages prevent what you are trying to do. Let's see why.
The idea of a method with the following signature:
def getIt( a:Base ) : Unit
Is that the body of the method will be able to access a properties visible through Base class or interface, i.e. the properties and methods defined only on the Base class/interface or its parents. During code execution, each specific instance passed to the getIt method might have a different subclass but the compile type of a will always be Base
One can reason in this way:
Ok I have a class Base, I inherit it in two case classes and I add a
property with the same name, and then I try to access the property on
the instance of Base.
A simple example shows why this is unsafe:
sealed abstract class Base
case class Foo(myparam:String) extends Base
case class Bar(myparam:String) extends Base
case class Evil(myEvilParam:String) extends Base
def getIt( a:Base ) = a.copy(myparam="changed")
In the following case, if the compiler didn't throw an error at compile time, it means the code would try to access a property that does not exist at runtime. This is not possible in strictly typed programming languages: you have traded restrictions on the code you can write for a much stronger verification of your code by the compiler, knowing that this reduces dramatically the number of bugs your code can contain
This is the new answer. It is a little long because few points are needed before getting to the conclusion
Unluckily, you can't rely on the mechanism of case classes copy to implement what you propose. The way the copy method works is simply a copy constructor which you can implement yourself in a non-case class. Let's create a case class and disassemble it in the REPL:
scala> case class MyClass(name:String, surname:String, myJob:String)
defined class MyClass
scala> :javap MyClass
Compiled from "<console>"
public class MyClass extends java.lang.Object implements scala.ScalaObject,scala.Product,scala.Serializable{
public scala.collection.Iterator productIterator();
public scala.collection.Iterator productElements();
public java.lang.String name();
public java.lang.String surname();
public java.lang.String myJob();
public MyClass copy(java.lang.String, java.lang.String, java.lang.String);
public java.lang.String copy$default$3();
public java.lang.String copy$default$2();
public java.lang.String copy$default$1();
public int hashCode();
public java.lang.String toString();
public boolean equals(java.lang.Object);
public java.lang.String productPrefix();
public int productArity();
public java.lang.Object productElement(int);
public boolean canEqual(java.lang.Object);
public MyClass(java.lang.String, java.lang.String, java.lang.String);
}
In Scala, the copy method takes three parameter and can eventually use the one from the current instance for the one you haven't specified ( the Scala language provides among its features default values for parameters in method calls)
Let's go down in our analysis and take again the code as updated:
sealed abstract class Base(val myparam:String)
case class Foo(override val myparam:String) extends Base(myparam)
case class Bar(override val myparam:String) extends Base(myparam)
def getIt( a:Base ) = a.copy(myparam="changed")
Now in order to make this compile, we would need to use in the signature of getIt(a:MyType) a MyType that respect the following contract:
Anything that has a parameter myparam and maybe other parameters which
have default value
All these methods would be suitable:
def copy(myParam:String) = null
def copy(myParam:String, myParam2:String="hello") = null
def copy(myParam:String,myParam2:Option[Option[Option[Double]]]=None) = null
There is no way to express this contract in Scala, however there are advanced techniques that can be helpful.
The first observation that we can do is that there is a strict relation between case classes and tuples in Scala. In fact case classes are somehow tuples with additional behaviour and named properties.
The second observation is that, since the number of properties of your classes hierarchy is not guaranteed to be the same, the copy method signature is not guaranteed to be the same.
In practice, supposing AnyTuple[Int] describes any Tuple of any size where the first value is of type Int, we are looking to do something like that:
def copyTupleChangingFirstElement(myParam:AnyTuple[Int], newValue:Int) = myParam.copy(_1=newValue)
This would not be to difficult if all the elements were Int. A tuple with all element of the same type is a List, and we know how to replace the first element of a List. We would need to convert any TupleX to List, replace the first element, and convert the List back to TupleX. Yes we will need to write all the converters for all the values that X might assume. Annoying but not difficult.
In our case though, not all the elements are Int. We want to treat Tuple where the elements are of different type as if they were all the same if the first element is an Int. This is called
"Abstracting over arity"
i.e. treating tuples of different size in a generic way, independently of their size. To do it, we need to convert them into a special list which supports heterogenous types, named HList
Conclusion
Case classes inheritance is deprecated for very good reason, as you can find out from multiple posts in the mailing list: http://www.scala-lang.org/node/3289
You have two strategies to deal with your problem:
If you have a limited number of fields you require to change, use an approach such as the one suggested by #Ron, which is having a copy method. If you want to do it without losing type information, I would go for generifying the base class
sealed abstract class Base[T](val param:String){
def copy(param:String):T
}
class Foo(param:String) extends Base[Foo](param){
def copy(param: String) = new Foo(param)
}
def getIt[T](a:Base[T]) : T = a.copy("hello")
scala> new Foo("Pippo")
res0: Foo = Foo#4ab8fba5
scala> getIt(res0)
res1: Foo = Foo#5b927504
scala> res1.param
res2: String = hello
If you really want to abstract over arity, a solution is to use a library developed by Miles Sabin called Shapeless. There is a question here which has been asked after a discussion : Are HLists nothing more than a convoluted way of writing tuples? but I tell you this is going to give you some headache
If the two case classes would diverge over time so that they have different fields, then the shared copy approach would cease to work.
It is better to define an abstract def withMyParam(newParam: X): Base. Even better, you can introduce an abstract type to retain the case class type upon return:
scala> trait T {
| type Sub <: T
| def myParam: String
| def withMyParam(newParam: String): Sub
| }
defined trait T
scala> case class Foo(myParam: String) extends T {
| type Sub = Foo
| override def withMyParam(newParam: String) = this.copy(myParam = newParam)
| }
defined class Foo
scala>
scala> case class Bar(myParam: String) extends T {
| type Sub = Bar
| override def withMyParam(newParam: String) = this.copy(myParam = newParam)
| }
defined class Bar
scala> Bar("hello").withMyParam("dolly")
res0: Bar = Bar(dolly)
TL;DR: I managed to declare the copy method on Base while still letting the compiler auto generate its implementations in the derived case classes. This involves a little trick (and actually I'd myself just redesign the type hierarchy) but at least it goes to show that you can indeed make it work without writing boiler plate code in any of the derived case classes.
First, and as already mentioned by ron and Edmondo1984, you'll get into troubles if your case classes have different fields.
I'll strictly stick to your example though, and assume that all your case classes have the same fields (looking at your github link, this seems to be the case of your actual code too).
Given that all your case classes have the same fields, the auto-generated copy methods will have the same signature which is a good start. It seems reasonable then to just add the common definition in Base, as you did:
abstract class Base{ def copy(myparam: String):Base }
The problem is now that scala won't generate the copy methods, because there is already one in the base class.
It turns out that there is another way to statically ensure that Base has the right copy method, and it is through structural typing and self-type annotation:
type Copyable = { def copy(myParam: String): Base }
sealed abstract class Base(val myParam: String) { this : Copyable => }
And unlike in our earlier attempt, this will not prevent scala to auto-generate the copy methods.
There is one last problem: the self-type annotation makes sure that sub-classes of Base have a copy method, but it does not make it publicly availabe on Base:
val foo: Base = Foo("hello")
foo.copy()
scala> error: value copy is not a member of Base
To work around this we can add an implicit conversion from Base to Copyable. A simple cast will do, as a Base is guaranteed to be a Copyable:
implicit def toCopyable( base: Base ): Base with Copyable = base.asInstanceOf[Base with Copyable]
Wrapping up, this gives us:
object Base {
type Copyable = { def copy(myParam: String): Base }
implicit def toCopyable( base: Base ): Base with Copyable = base.asInstanceOf[Base with Copyable]
}
sealed abstract class Base(val myParam: String) { this : Base. Copyable => }
case class Foo(override val myParam: String) extends Base( myParam )
case class Bar(override val myParam: String) extends Base( myParam )
def getIt( a:Base ) = a.copy(myParam="changed")
Bonus effect: if we try to define a case class with a different signature, we get a compile error:
case class Baz(override val myParam: String, truc: Int) extends Base( myParam )
scala> error: illegal inheritance; self-type Baz does not conform to Base's selftype Base with Base.Copyable
To finish, one warning: you should probably just revise your design to avoid having to resort to the above trick.
In your case, ron's suggestion to use a single case class with an additional etype field seems more than reasonable.
I think this is what extension methods are for. Take your pick of implementation strategies for the copy method itself.
I like here that the problem is solved in one place.
It's interesting to ask why there is no trait for caseness: it wouldn't say much about how to invoke copy, except that it can always be invoked without args, copy().
sealed trait Base { def p1: String }
case class Foo(val p1: String) extends Base
case class Bar(val p1: String, p2: String) extends Base
case class Rab(val p2: String, p1: String) extends Base
case class Baz(val p1: String)(val p3: String = p1.reverse) extends Base
object CopyCase extends App {
implicit class Copy(val b: Base) extends AnyVal {
def copy(p1: String): Base = b match {
case foo: Foo => foo.copy(p1 = p1)
case bar: Bar => bar.copy(p1 = p1)
case rab: Rab => rab.copy(p1 = p1)
case baz: Baz => baz.copy(p1 = p1)(p1.reverse)
}
//def copy(p1: String): Base = reflect invoke
//def copy(p1: String): Base = macro xcopy
}
val f = Foo("param1")
val g = f.copy(p1="param2") // normal
val h: Base = Bar("A", "B")
val j = h.copy("basic") // enhanced
println(List(f,g,h,j) mkString ", ")
val bs = List(Foo("param1"), Bar("A","B"), Rab("A","B"), Baz("param3")())
val vs = bs map (b => b copy (p1 = b.p1 * 2))
println(vs)
}
Just for fun, reflective copy:
// finger exercise in the api
def copy(p1: String): Base = {
import scala.reflect.runtime.{ currentMirror => cm }
import scala.reflect.runtime.universe._
val im = cm.reflect(b)
val ts = im.symbol.typeSignature
val copySym = ts.member(newTermName("copy")).asMethod
def element(p: Symbol): Any = (im reflectMethod ts.member(p.name).asMethod)()
val args = for (ps <- copySym.params; p <- ps) yield {
if (p.name.toString == "p1") p1 else element(p)
}
(im reflectMethod copySym)(args: _*).asInstanceOf[Base]
}
This works fine for me:
sealed abstract class Base { def copy(myparam: String): Base }
case class Foo(myparam:String) extends Base {
override def copy(x: String = myparam) = Foo(x)
}
def copyBase(x: Base) = x.copy("changed")
copyBase(Foo("abc")) //Foo(changed)
There is a very comprehensive explanation of how to do this using shapeless at http://www.cakesolutions.net/teamblogs/copying-sealed-trait-instances-a-journey-through-generic-programming-and-shapeless ; in case the link breaks, the approach uses the copySyntax utilities from shapeless, which should be sufficient to find more details.
Its an old problem, with an old solution,
https://code.google.com/p/scala-scales/wiki/VirtualConstructorPreSIP
made before the case class copy method existed.
So in reference to this problem each case class MUST be a leaf node anyway, so define the copy and a MyType / thisType plus the newThis function and you are set, each case class fixes the type. If you want to widen the tree/newThis function and use default parameters you'll have to change the name.
as an aside - I've been waiting for compiler plugin magic to improve before implementing this but type macros may be the magic juice. Search in the lists for Kevin's AutoProxy for a more detailed explanation of why my code never went anywhere