I know this question has been asked before here. But the answers there do not satisfy my doubt.
I was told that they prevent mix-up of class types, the code below shows that they're not mixed up at all.
So, it shouldn't matter right?
Classes:
package Practice
abstract class Animal {
def name: String
}
abstract class Pet extends Animal {}
class Cat extends Pet {
override def name: String = "Cat"
}
class Dog extends Pet {
override def name: String = "Dog"
}
Here is the real confusion:
//Class with Upper Bound
class PetContainer[P <: Pet](p: P) {
def pet: P = p
}
//Class with Subtyping(Or Upcasting, I think they're the same)
class SimplePetContainer(p: Pet){
def pet: Pet = p
}
Driver Code:
val CatContainer: PetContainer[Cat] = new PetContainer[Cat](new Cat)
val DogContainer: SimplePetContainer = new SimplePetContainer(new Dog
println(CatContainer.pet.getClass)
println(DogContainer.pet.getClass)
Output:
class Practice.Cat
class Practice.Dog
//Practice was the package
Like I mentioned before, the classes are preserved.
So my question is, What advantage does Upper Bound have on Subtyping?
With your CatContainer, you know that CatContainer.pet is a Cat at compile-time. Meaning that the compiler also knows that. So you can say
CatContainer.pet.meow()
For the SimplePetContainer you do not have static type information about the pet inside anymore.
Like I mentioned before, the classes are preserved.
At runtime, the pet of course still knows its type (well, almost, it knows its class, which in your case would have been enough, any extra type information such as the generic types of that class has been erased).
But the variable DogContainer.pet lacks information about what sort of Pet it contains.
I was told that they prevent mix-up of class types
The compiler won't stop you from writing
val DogContainer = new SimplePetContainer(new Cat())
but it will reject this
val DogContainer = new PetContainer[Dog](new Cat())
Let's say I have the following trait
trait Named {
def name: String
}
and the following Algebraic Data Type
sealed trait Animal extends Named
case object Dog extends Animal {
override val name: String = "dog man"
}
case object Cat extends Animal {
override val name: String = "cat man"
}
case object Owl extends Animal {
override val name: String = "I am an owl left in the dark"
}
Now, I can deserialize an instance of string into my Animal ADT with the following method.
object Animal {
def apply(name: String): Animal = name match {
case Dog.name => Dog
case Cat.name => Cat
}
}
#oxbow_lakes mentions at the end of his answer that:
Can't instantiate easily from persisted value. This is also true but,
except in the case of huge enumerations (for example, all currencies),
this doesn't present a huge overhead.
I find that the fact that when you add a new value it needs to be added to the deserialization code explicitly as error prone (I thought that the compiler would warn me of an in-exhaustive match, but take a look at Owl above and the apply method - there was no warning issued...)
Is there no better way? (If not with the standard scala toolset, a third party one?)
This problem already solved by enumeratum library:
https://github.com/lloydmeta/enumeratum
Your code could be written like this:
import enumeratum._
import enumeratum.EnumEntry.Lowercase
sealed trait Animal extends EnumEntry with Lowercase
object Animal extends Enum[Animal] {
val values = findValues
case object Dog extends Animal
case object Cat extends Animal
case object Owl extends Animal
}
val dogName = Animal.Dog.entryName
val dog = Animal.withNameInsensitive(dogName)
One thing you could try is to use reflection to obtain the set of types that extend Animal, then use that to create a Map[String,Animal] using name to lookup object values, then use the map in your Animal.apply function.
Refer to this question for more information on obtaining the Animal subclasses.
I have a basic example from chapter 20.7 of Programming in Scala (Martin Odersky, Lex Spoon and Bill Venners), on the topic of Abstract Types. The code below is from Listing 20.10, except that I added the two last lines which seem ostensibly implied by the previous example:
class Food
abstract class Animal {
type SuitableFood <: Food
def eat(food: SuitableFood)
}
class Grass extends Food
class Cow extends Animal {
type SuitableFood = Grass
override def eat(food: Grass) {}
}
class Fish extends Food
val bossy: Animal = new Cow // If the compiler were helpful, this would error.
bossy.eat(new Grass) // error!
// type mismatch; found: Grass, required: bossy.SuitableFood
As I stated above, the two lines where bossy is declared as an Animal are not actually in the example, but seem a very reasonable conceptual leap. At the level of the abstract class Animal (the declared type of bossy), the type member SuitableFood is still abstract. So, nothing will satisfy the compiler, even though it looks as if it wants a path-dependent-type at the method call.
If I declare my val to be of type Cow, the method call works, as follows:
val bessy: Cow = new Cow
bessy.eat(new Grass) // happy-happy
Given that there is nothing I could put in the 'eat()' method call for bossy (declared as an Animal) to satisfy the compiler, why does the compiler even allow bossy to be declared as an Animal/instantiated as a Cow? In other words, what possible use allowing the object declaration/instantiation, but not the method call, have?
Are there "best practices" for this feature in Scala, given that abstract member type refining seems to deliberately allow something normally forbidden in OO programming? Perhaps someone has found a killer-use?
I very much desire to see this behavior as something that makes perfect sense. What is the motivating use-case for this feature, i.e., declaring an abstract type, then refining that type in a derived class such that the subtype has a more refined type than the supertype?
Given that there is nothing I could put in the 'eat()' method call for bossy (declared as an Animal) to satisfy the compiler, why does the compiler even allow bossy to be declared as an Animal/instantiated as a Cow?
There is: bossy.eat((new Grass).asInstanceOf[bossy.SuitableFood]). Of course, this doesn't mean you should ever write code like this.
Even if there weren't, there are a lot of things you can do with bossy without calling eat method: put it into a List, get its hash code, etc. etc.
You can still do other useful stuff. You can make an Animal eat Food if you can prove that it is SuitableFood.
When you can make an Animal throw up, you know that everything he throws up is something he can eat, because he's eaten is before. And you know it's Food even if you don't know for sure whether it's Grass or Fish. So you can do operations on it that are possible for every type of Food.
scala> :paste
// Entering paste mode (ctrl-D to finish)
abstract class Food { def name: String }
class Grass extends Food { def name = "Grass" }
abstract class Animal {
type SuitableFood <: Food
def eat(food: SuitableFood): Unit
def throwUp: Option[SuitableFood]
}
class Cow extends Animal {
type SuitableFood = Grass
private[this] var digesting: List[Grass] = Nil
def eat(food: Grass) {
digesting = food :: digesting
}
def throwUp = digesting match {
case Nil => None
case food :: rest =>
digesting = rest
Some(food)
}
}
def dispose(food: Food) = println(s"Disposing of some ${food.name}.")
// Exiting paste mode, now interpreting.
scala> val animal: Animal = { val cow = new Cow; cow.eat(new Grass); cow }
animal: Animal = Cow#506dcf55
scala> animal.throwUp foreach animal.eat // can eat his own vomit :s
scala> animal.throwUp foreach dispose // I can dispose of all food
Disposing of some Grass.
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
I think it would be easier to describe a problem with concrete example. Suppose I have have Fruit class hierarchy and Show type class:
trait Fruit
case class Apple extends Fruit
case class Orange extends Fruit
trait Show[T] {
def show(target: T): String
}
object Show {
implicit object AppleShow extends Show[Apple] {
def show(apple: Apple) = "Standard apple"
}
implicit object OrangeShow extends Show[Orange] {
def show(orange: Orange) = "Standard orange"
}
}
def getAsString[T](target: T)(implicit s: Show[T]) = s show target
I also have list of fruits that I would like to show to the user using Show (this is my main goal in this question):
val basket = List[Fruit](Apple(), Orange())
def printList[T](list: List[T])(implicit s: Show[T]) =
list foreach (f => println(s show f))
printList(basket)
This will not compile because List is parametrized with Fruit and I have not defined any Show[Fruit]. What is the best way to achieve my goal using type classes?
I tried to find solution for this problem, but unfortunately have not found any nice one yet. It's not enough to know s in printList function - somehow it needs to know Show[T] for each element of the list. This means, that in order to be able to make this, we need some run-time mechanism in addition to the compile-time one. This gave me an idea of some kind of run-time dictionary, that knows, how to find correspondent Show[T] at run-time.
Implementation of implicit Show[Fruit]can serve as such dictionary:
implicit object FruitShow extends Show[Fruit] {
def show(f: Fruit) = f match {
case a: Apple => getAsString(a)
case o: Orange => getAsString(o)
}
}
And actually very similar approach can be found in haskell. As an example, we can look at Eq implementation for Maybe:
instance (Eq m) => Eq (Maybe m) where
Just x == Just y = x == y
Nothing == Nothing = True
_ == _ = False
The big problem with this solution, is that if I will add new subclass of Fruit like this:
case class Banana extends Fruit
object Banana {
implicit object BananaShow extends Show[Banana] {
def show(banana: Banana) = "New banana"
}
}
and will try to print my basket:
val basket = List[Fruit](Apple(), Orange(), Banana())
printList(basket)
then scala.MatchError would be thrown because my dictionary does not know anything about bananas yet. Of course, I can provide updated dictionary in some context that knows about bananas:
implicit object NewFruitShow extends Show[Fruit] {
def show(f: Fruit) = f match {
case b: Banana => getAsString(b)
case otherFruit => Show.FruitShow.show(otherFruit)
}
}
But this solution is far from perfect. Just imagine that some other library provides another fruit with it's own version of dictionary. It will just conflict with NewFruitShow if I try to use them together.
Maybe I'm missing something obvious?
Update
As #Eric noticed, there is one more solution described here: forall in Scala . It's really looks very interesting. But I see one problem with this solution.
If I use ShowBox, then it will remember concrete type class during it's creation time. So I generally building list with objects and correspondent type classes (so dictionary in present in the list). From the other hand, scala has very nice feature: I can drop new implicits in the current scope and they will override defaults. So I can define alternative string representation for the classes like:
object CompactShow {
implicit object AppleCompactShow extends Show[Apple] {
def show(apple: Apple) = "SA"
}
implicit object OrangeCompactShow extends Show[Orange] {
def show(orange: Orange) = "SO"
}
}
and then just import it in current scope with import CompactShow._. In this case AppleCompactShow and OrangeCompactShow object would be implicitly used instead of defaults defined in the companion object of Show. And as you can guess, list creation and printing happens in different places. If I will use ShowBox, than most probably I will capture default instances of type class. I would like to capture them at the last possible moment - the moment when I call printList, because I even don't know, whether my List[Fruit] will ever be shown or how it would be shown, in the code that creates it.
The most obvious answer is to use a sealed trait Fruit and a Show[Fruit]. That way your pattern matches will complain at compile time when the match is not exhaustive. Of course, adding a new kind of Fruit in an external library will not be possible, but this is inherent in the nature of things. This is the "expression problem".
You could also stick the Show instance on the Fruit trait:
trait Fruit { self =>
def show: Show[self.type]
}
case class Apple() extends Fruit { self =>
def show: Show[self.type] = showA
}
Or, you know, stop subtyping and use type classes instead.