I think I had a good understanding of Scala till I find myself in this simple scenario
sealed abstract case class Name private(name: String)
object Name {
def make(name: String): Option[Name] =
if (name.nonEmpty) Some(new Name(name) {}) else None
}
my question is about the private modifier for the class.
If I use it like this, everything works, but if I move the private keyword at the start, something like
private sealed abstract case class Name(name: String) it doesn't compile becuase gives me the following error
private class Name escapes its defining scope as part of type
Option[example.package.Name]
where example.package is the package object I'm working in.
I think I need some clarification because I'm not sure what's happening here
In
sealed abstract case class Name(name: String)
Name effectively denotes both
the name of the class
the default constructor for that class
sealed abstract case class Name private(name: String)
declares that the class is public, but the constructor is private (i.e. can only be called from within that class or its companion object).
private sealed abstract case class Name(name: String)
declares that the class is private (which implies that the constructor is also private).
This code compiles with no error:
private sealed abstract case class Name(name: String)
The problem is that you are then returning a value that contains a public value of this class, so the definition is leaking out and it isn't really private. You need to make sure that all references to Name are also private.
The first syntax marks the constructor as private;
sealed abstract case class Name private(name: String)
This means that the constructor can only be called from inside the Name companion object. So this is not allowed
val name = new Name("name") // constructor Name in class Name cannot be accessed
I am getting a compile error of:
value txn is not a member of Charge
new Charge(this.txn + that.txn)
^
with the following Scala class definition:
class Charge(txn: Double = 0){
def combine(that:Charge): Charge =
new Charge(this.txn + that.txn)
}
Explicitly declaring txn as a val allows it to work:
class Charge(val txn: Double = 0){
def combine(that:Charge): Charge =
new Charge(this.txn + that.txn)
}
I thought val was assumed? Can somebody explain this? Is it a problem with my understanding of the default constructor or the scope of the method?
In scala, you can define classes in two forms, for ex.
class Charge(txn: Double) -> In this case scala compiler compiles it to java like below
public class Charge {
....
public Charge combine(Charge);
....
public Charge(double);
....
}
As you can notice in compiled java code, there is no public accessor for txn
Let's look at another variation of Charge class,
If you define like this class Charge(val txn: String), it gets compiled to below
public class Charge {
...
public double txn();
...
public Charge combine(Charge);
...
public Charge(double);
...
}
As you can see, in this case compiler generates public accessor for txn hence you are able to access that.txn when you mark it as val
case class Charge(txn: Double): This is data class for which scala generates getters, equals and toString for you.
You can compile this class
scalac Charge.scala
javap -c Charge.class
And then see what it generates
What you pass to constructor is essentially constructor parameters whose scope is limited to the constructor. If you want to make them visible from the outside, you have to declare them as vals or reassign to some other vals in the constructor body.
This topic has been discussed before but not answered satisfactorily (in my opinion).
Consider the following scala code:
class A(a:Int) { val _a=a }
class A1(val a:Int) { val _a=a }
class B(a:Int) extends A(a) // OK
class C(val a:Int) extends A(a) // OK
class B1(a:Int) extends A1(a) // OK
class C1(val a:Int) extends A1(a) // fails
class D1(override val a:Int) extends A1(a) // OK
I believe that declaring the class parameter as val only has an effect on the constructor call:
the parameter is copied instead of passing a reference. However in each case the class field is allocated as a val. is this correct?
Now what I do not understand is why we need the override keyword in the last line.
Note that we do not declare the classes as case classes so no automatic allocation of fields is going on.
Finally is there a good reason why one even would want to define a class like A1 with a val
class parameter?
Thanks in advance for all replies.
I believe that declaring the class parameter as val only has an
effect on the constructor call: the parameter is copied instead of
passing a reference. However in each case the class field is allocated
as a val. is this correct?
Not at all.
If there is a val or var in the constructor, that causes a val or var of the same name to be declared in the class, and assign the constructor parameter to it at construction. Otherwise, a (private) val may still be created in the class, if the constructor parameter is used outside initialization, i.e in a method.
In class A1, the member _a is really useless, because if you write
class A1(val a: Int) {}
it is equivalent to
class A1(someFreshName: Int) {val a = someFreshName}
So in C1, you are trying to declare a new member a, while there is already one. Hence the override.
In your particular instance, both member would have the same value, but you could well do (maybe not a good idea)
class C1(override val a: Int) extends A1(12)
Then, the new member a would have the constructor parameter as value, and the previous a would have value 12 and be hidden (but still, code written in A1 would access it).
In Scala, if we have
class Foo(bar:String)
We can create a new object but cannot access bar
val foo = new Foo("Hello")
foo.bar // gives error
However, if we declare Foo to be a case class this works:
case class Foo(bar:String)
val foo = Foo("hello")
foo.bar // works
I am forced to make many of my classes as case classes because of this. Otherwise, I have to write boilerplate code for accessing bar:
class Foo(bar:String) {
val getbar = bar
}
So my questions are:
Is there any way to "fix" this without using case classes or boilerplate code?
Is using case classes in this context a good idea? (or what are the disadvantages of case classes?)
I guess the second one deserves a separate question.
Just use val keyword in constructor to make it publicly accessible:
class Foo(val bar:String) {
}
As for your question: if this is the only feature you need, don't use case classes, just write with val.
However, would be great to know why case classes are not good.
In case classes all arguments by default are public, whereas in plain class they're all private. But you may tune this behaviour:
scala> class Foo(val bar:String, baz: String)
defined class Foo
scala> new Foo("public","private")
res0: Foo = Foo#31d5e2
scala> res0.bar
res1: String = public
scala> res0.baz
<console>:10: error: value baz is not a member of Foo
res0.baz
And even like that:
class Foo(private[mypackage] val bar:String) {
// will be visible only to things in `mypackage`
}
For case classes (thanks to #JamesIry):
case class Bar(`public`: String, private val `private`: String)
You can use the BeanProperty annotation to automatically generate java-like getters
import scala.reflect.BeanProperty
case class Foo(#BeanProperty bar:String)
Now Foo has a getBar method that returns the bar value.
Note though that this is only useful if you have a good reason to use java-like getters (typical reasons being that you need your class to be a proper java bean, so as to work with java libraries that expect java beans and use reflection to access the bean's properties).
Otherwise, just access the bar value directly, this is "the scala way".
See http://www.scala-lang.org/api/current/index.html#scala.reflect.BeanProperty
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