I have a Base class have some function and val, I want to inherent them in my inherent case class how to do it ?
This is my base class:
class Base(val name:String, val number:int) extends Sometrait {
def copy(name:String=this.name, number:int=this.number){
new Base(name, number)
}
}
I want to write the:
case class SomeCase(val name:String, val number:int, val id:int)extends Base(String, number){
...
}
But the compiler always told me:
value **** needs `override' modifier social.scala /scalatest/src/scalatest line 35 Scala
But I really want to is just do as inherent not override, how to do it.
(I need to put the child class as case class, as it is easy for me to use in slick. (here is my another question for how to use class as table content class in slick, someone give me really great answer, but I still mass.))
Because name and number fields in both Base and SomeCase are defined with val without any modifiers like private, they are both public members and participate in inheritance. Because these fields have same names in base and child classes, you have to add override modifier before val name and val number in the child class:
case class SomeCase(override val name: String,
override val number: Int,
val id: Int) extends Base(name, number) { ... }
Related
Suppose I have two classes, Person and Business, that are extended by the trait Entity.
trait Entity
case class Person(name: String) extends Entity
case class Business(id: String) extends Entity
Assuming I cannot change Entity, Person and Business (they are in a different file and not to be changed) how can I define a function, say a printEntity, that prints the field name or id, depending on the entity? For example, given instances of Person and Business, how can I do something like this:
object Main extends App {
val person1: Person = Person("Aaaa Bbbb")
val business1: Business = Business("0001")
// How can I do something like this?
person1.printEntity // would call a function that executes println(id)
business1.printEntity // would call a function that executes println(name)
}
Any ideas are appreciated! Sorry for the lack of context, I am still learning!
This is done via so called "extension methods". In scala 2 this is achieved using implicit wrapper class:
trait Entity
case class Person(name: String) extends Entity
case class Business(id: String) extends Entity
implicit class PersonWrapper(val p: Person) extends AnyVal {
def printEntity(): Unit = {
println(p.name)
}
}
implicit class BusinessWrapper(val b: Business) extends AnyVal {
def printEntity(): Unit = {
println(b.id)
}
}
val person1: Person = Person("Aaaa Bbbb")
val business1: Business = Business("0001")
person1.printEntity()
business1.printEntity()
// prints:
//Aaaa Bbbb
//0001
Note, x.printEntity can be called without parentheses, but, by convention, methods with Unit result type and side effects should be called with explicit empty parentheses.
UPD: As #DmytroMitin pointed out, you should extend implicit wrapper classes from AnyVal. This allows the compiler to avoid actually allocating wrapper class instances at runtime, improving performance.
Scala throws "reassignment to val" error for the following code.
abstract case class Gun(var bulletCount:Int)
class Pistol(bulletCount:Int) extends Gun(bulletCount){
def fire() { bulletCount=bulletCount-1 }
}
Anything I missed here?
For starters, you should consider case class as final, and not extend them.
Second, do not use var with case class, you should rather create a copy of a case class to get one of its field changed.
Third, if you want a common type, you can use a base trait.
All in one, here's what it could look like:
sealed trait Gun {
def bulletCount: Int
}
case class Pistol(bulletCount: Int) extends Gun {
def fire(): Pistol = copy(bulletCount=bulletCount)
}
You're referring to bulletCount field generated by Pistol primary constructor parameter. To set base class variable, you need to directly call field using super:
class Pistol(bulletCount: Int) extends Gun(bulletCount) {
def fire(): Unit = {
super.bulletCount = super.bulletCount - 1
}
}
Alternatively, you can label parameter-generated field with override var:
class Pistol(override var bulletCount: Int) extends Gun(bulletCount) {
def fire(): Unit = {
bulletCount = bulletCount - 1
}
}
On a side note, as Frederic A. suggested in his answer, you should avoid inheriting case classes. They are syntactic sugar, and code generation don't work over inheritance - you'll need to implement all the fancy stuff like apply or unapply methods in companion class all by yourself. Scala compiler team tried to support case class to case class inheritance, but discovered that it breaks structural equality and lots of other things.
Say I have 2 case classes:
case class Basic(id: String, name) extends SomeBasicTrait
case class Info (age: Int, country: String, many other fields..) extends SomeInfoTrait
and want to create a case class that has all fields from both of those case classes. This is a possible way:
case class Full(bs: Basic, meta: Info) extends SomeBasicTrait with SomeInfoTrait {
val id = bs.id
val name = bs.name
val age = meta.age
val country = meta.country
// etc
}
But it's a lot of boilerplate code. Is there any way to avoid this?
I couldn't find a way to achieve this in Shapeless but maybe there is..
[Update]
#jamborta 's comment helps and is basically this:
case class FullTwo(id: String, name: String, age:Int, country:String)
val b = Basic("myid", "byname")
val i = Info(12, "PT")
Generic[FullTwo].from(Generic[Basic].to(b) ++ Generic[Info].to(i))
The problem with this solution is that it still requires defining every field in the arguments of the FullTwo class, so that every time a change to Basic or Info is made, we also have to remember to change FullTwo as well.
Is there any way to create dynamically at compile time a case class equal to FullTwo?
I have an application based on Squeryl. I define my models as case classes, mostly since I find convenient to have copy methods.
I have two models that are strictly related. The fields are the same, many operations are in common, and they are to be stored in the same DB table. But there is some behaviour that only makes sense in one of the two cases, or that makes sense in both cases but is different.
Until now I only have used a single case class, with a flag that distinguishes the type of the model, and all methods that differ based on the type of the model start with an if. This is annoying and not quite type safe.
What I would like to do is factor the common behaviour and fields in an ancestor case class and have the two actual models inherit from it. But, as far as I understand, inheriting from case classes is frowned upon in Scala, and is even prohibited if the subclass is itself a case class (not my case).
What are the problems and pitfalls I should be aware in inheriting from a case class? Does it make sense in my case to do so?
My preferred way of avoiding case class inheritance without code duplication is somewhat obvious: create a common (abstract) base class:
abstract class Person {
def name: String
def age: Int
// address and other properties
// methods (ideally only accessors since it is a case class)
}
case class Employer(val name: String, val age: Int, val taxno: Int)
extends Person
case class Employee(val name: String, val age: Int, val salary: Int)
extends Person
If you want to be more fine-grained, group the properties into individual traits:
trait Identifiable { def name: String }
trait Locatable { def address: String }
// trait Ages { def age: Int }
case class Employer(val name: String, val address: String, val taxno: Int)
extends Identifiable
with Locatable
case class Employee(val name: String, val address: String, val salary: Int)
extends Identifiable
with Locatable
Since this is an interesting topic to many, let me shed some light here.
You could go with the following approach:
// You can mark it as 'sealed'. Explained later.
sealed trait Person {
def name: String
}
case class Employee(
override val name: String,
salary: Int
) extends Person
case class Tourist(
override val name: String,
bored: Boolean
) extends Person
Yes, you have to duplicate the fields. If you don't, it simply would not be possible to implement correct equality among other problems.
However, you don't need to duplicate methods/functions.
If the duplication of a few properties is that much of an importance to you, then use regular classes, but remember that they don't fit FP well.
Alternatively, you could use composition instead of inheritance:
case class Employee(
person: Person,
salary: Int
)
// In code:
val employee = ...
println(employee.person.name)
Composition is a valid and a sound strategy that you should consider as well.
And in case you wonder what a sealed trait means — it is something that can be extended only in the same file. That is, the two case classes above have to be in the same file. This allows for exhaustive compiler checks:
val x = Employee(name = "Jack", salary = 50000)
x match {
case Employee(name) => println(s"I'm $name!")
}
Gives an error:
warning: match is not exhaustive!
missing combination Tourist
Which is really useful. Now you won't forget to deal with the other types of Persons (people). This is essentially what the Option class in Scala does.
If that does not matter to you, then you could make it non-sealed and throw the case classes into their own files. And perhaps go with composition.
case classes are perfect for value objects, i.e. objects that don't change any properties and can be compared with equals.
But implementing equals in the presence of inheritance is rather complicated. Consider a two classes:
class Point(x : Int, y : Int)
and
class ColoredPoint( x : Int, y : Int, c : Color) extends Point
So according to the definition the ColorPoint(1,4,red) should be equal to the Point(1,4) they are the same Point after all. So ColorPoint(1,4,blue) should also be equal to Point(1,4), right? But of course ColorPoint(1,4,red) should not equal ColorPoint(1,4,blue), because they have different colors. There you go, one basic property of the equality relation is broken.
update
You can use inheritance from traits solving lots of problems as described in another answer. An even more flexible alternative is often to use type classes. See What are type classes in Scala useful for? or http://www.youtube.com/watch?v=sVMES4RZF-8
In these situations I tend to use composition instead of inheritance i.e.
sealed trait IVehicle // tagging trait
case class Vehicle(color: String) extends IVehicle
case class Car(vehicle: Vehicle, doors: Int) extends IVehicle
val vehicle: IVehicle = ...
vehicle match {
case Car(Vehicle(color), doors) => println(s"$color car with $doors doors")
case Vehicle(color) => println(s"$color vehicle")
}
Obviously you can use a more sophisticated hierarchy and matches but hopefully this gives you an idea. The key is to take advantage of the nested extractors that case classes provide
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