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).
Related
I have following two classes.
class A (name: String) {
}
object A {
}
According to definition of Singleton, we can have only one object of that type. However I am able to create two different objects of type A using following piece of code.
object B {
def main(args: Array[String]): Unit = {
val a = new A("Vinod")
println(a)
val b = new A("XYZ")
println(b)
}
}
can someone please explain me, where my understanding is not correct?
An object by itself is a singleton. It has its own class and no other instance of the same class exist at runtime.
However, the pattern you describe here is different: object A is not an instance of class A unless you make it so using object A extends A. You could make it the only instance of class A by making class A a sealed class, but this is unnecessary in almost all cases.
If you really want the singleton pattern, drop the class and use only object A, all of its members will be "static" in the sense of Java.
Note that the actual type of object A can be referred to as A.type, which by default is completely unrelated to type A if class A exists. Again, A.type could be a subtype of A if you explicitly make it so.
The companion object is not an instance of the companion class. They're not even the same type.
class A
object A {
var state = 0
def update() :Unit = state = state + 1
}
val abc :A = new A //instance of class A
val xyz :A.type = A //2nd reference to object A
// both reference the same singleton object
xyz.update() //res0: Unit = ()
A.state //res1: Int = 1
abc.state //Error: value state is not a member of A$A2521.this.A
the companion object can be thought of as the static space of a class. if you want to make A a singleton you can make it an object rather than a class
new A refers to class A (which is not a singleton), not to object A. You can easily check it: if you remove class A, the new A lines will no longer compile.
Also note that objects aren't necessarily singletons: they can be nested inside classes or traits, in this case there is one for each instance of the outer type.
Assume the following pair of classes:
class A(arg:String)
class B(argList:Vector[String]) extends A(argList.first)
I want to be able to check for argList being empty before providing the base class constructor with its first element. Unfortunately, placing that check in the default constructor for B (e.g through require, as shown here) is way too late, since the base class' constructor will need to be called first.
This is probably a more general OOP question, but the solution is likely to be Scala-specific.
What do you expect to pass if argList is empty? In any case, you could just use the following:
class B(argList:Vector[String]) extends A(argList.headOption.getOrElse("your default string here")
One way to deal with this is via a companion object. You can mark the constructor for B as private to ensure no-one can by-pass the check, then add a suitable apply method to the companion object that pre-checks the input value(s):
class A(arg:String)
class B private(argList:Vector[String]) extends A(argList.head)
object B {
def apply(argList:Vector[String]): B = argList.headOption.map(_ => new B(argList)).getOrElse(throw new RuntimeException("Oops"))
}
Usage examples:
scala> B(Vector("foo", "bar"))
res2: B = B#328e9109
scala> B(Vector())
java.lang.RuntimeException: Oops
at B$$anonfun$apply$2.apply(<console>:24)
...
Note that for simplicity's sake, I simply throw an exception when handling bad data, but would probably try some other way of handling this situation (a default value per #Zoltan's answer is one such way).
That's why in most places constructors are replaced by factory objects. In Scala it's idiomatic to use companion object as such factories.
class A(arg: String)
abstract class B(arg: String) extends A(arg) {
def argList: IndexedSeq[String]
}
object B {
case object Empty extends B("") {
def argList = IndexedSeq.empty
}
case class NonEmpty private[B](argList: Vector[String]) extends B(argList.head)
def apply(argList: Vector[String]) =
if (argList.isEmpty) Empty else NonEmpty(argList)
def unapplySeq(b:B): Option[IndexedSeq[String]] = b match {
case Empty ⇒ Some(IndexedSeq.empty)
case NonEmpty(args) ⇒ Some(args)
}
}
you could verify that
B(Vector()) == B.Empty
B(Vector("x", "y")).isInstanceOf[B.NonEmpty]
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 have something like this in scala:
abstract class Point[Type](n: String){
val name = n
var value: Type = _
}
So far so good. The problem comes in a class that extends Point.
case class Input[Type](n:String) extends Point(n){
def setValue(va: Type) = value = va
}
On the setValue line I have this problem:
[error] type mismatch;
[error] found : va.type (with underlying type Type)
[error] required: Nothing
[error] def setValue(va: Type) = value = va
I have tried to initialize with null and null.asInstanceOf[Type] but the result is the same.
How can I initialize value so it can be used in setValue?
You should specify that Input implements Point with the generic type Type because for now, as it is not specified, it is considered as Nothing (I guess the compiler can't infer it from the setValue method). So you have to do the following:
case class Input[Type](n:String) extends Point[Type](n){
def setValue(va: Type) = value = va
}
More information
I answered this question for the compilation error (it does compile on scala 2.9.0.1). Moreover I saw this case class as the implementation for an existing type, like 'Int'. The usage of _ is of course a bad idea in the abstract class, however it is not prohibited, but the _ is not always a null, it is the default value, for exemple: var x:Int = _ will assign the value 0 to x.
Try the following:
package inputabstraction
abstract class Point[T](n:String){
def value: T
val name = n
}
case class Input[T](n:String, value:T) extends Point[T](n)
object testTypedCaseClass{
def test(){
val foo = Input("foo", "bar")
println(foo)
}
}
A simple Application to check that it works:
import inputabstraction._
object TestApp extends Application{
testTypedCaseClass.test()
}
Explanation
The first mistake you are making is case class Input[Type](n:String) extends Point(n){. Point is a typed class, and so when you are calling the superclass constructor with extends Point(n) you need to specify the type of Point. This is done like this: extends Point[T](n), where T is the Type you are planning to use.
The second error is that you are both defining and declaring value:T here: var value: Type = _. In this statement, _ is a value. Its value is Nothing. The scala compiler infers from this that Point[T] is Point[Nothing]. Thus when you attempt to set it to a type in the body of your setValue method, you must set it to Nothing, which is probably not what you want. If you attempt to set it to anything besides Nothing, you will get the type mismatch from above, because value is typed as Nothing due to your use of _.
The third mistake is using var instead of val or def. val and def can be overridden interchangeably, which means that subtypes can override with either val or def, and the scala compiler will figure it out for you. It is best practice to define vals as functions using def in abstract classes and traits, because the initialization order of subtype constructors is a very difficult thing to get right (there is an algorithm for how the compiler decides how to construct a class from its supertypes). TL#DR === use def in supertypes. Case class parameters are automatically generate val fields, which, since you are extending a Point, will create a val value field that overrides the def value field in Point[T].
You can get away with all this Type||T abstraction in Scala because of type inference and the fact that Point is abstract, therefore making value extendable via val.
The preferred way of doing dependency injection like this is the cake pattern, but this example I have provided works for your use-case.
I need access to the Class of the object being constructed in its own constructor (for various detailed reasons I don't think are relevant to my question).
I want something like this
class Foo(val i:Int)
class Bar extends Foo(this.getClass.getName.length)
val b = new Bar
println(b.i)
to print 3 ("Bar".length). But it doesn't. If the code above is inside some other object, "this" refers to that object. If the code above is not inside some other object (just in some package), the compiler complains with
error: this can be used only in a class, object, or template
class Bar extends Foo(this.getClass.getName)
^
CLARIFICATION: I can't change Foo to use a val in its body instead of its constructor because Foo's API already exists and is fixed (so, yes, i must be a constructor parameter). It needs an integer argument at constructor time, but that integer can only be calculated with access to the Class.
(I know the example above is still silly and degenerate. If people care, I can explain in detail why I need the class in my real project, http://code.google.com/p/factorie)
Surely, the class of the object being constructed is known to the compiler and runtime during construction. With what syntax can I get it? (Is there no such syntax? If not, I wonder why. I'm surprised it there doesn't seem to be a simple, standard way to get this.)
A lazy val solves this problem:
object Early
{
abstract class Foo { val name: String }
class Bar extends Foo { lazy val name = getClass.getName }
def
main(args: Array[String]): Unit = {
val b = new Bar
println(b.name)
}
}
Yields:
% scala Early
Early$Bar
Not sure whether this is possible in a clean way. If you like hacks you could do
class Bar extends Foo((new Exception).getStackTrace.apply(0).getClassName)
However I strongly advertise against it!
This appears to satisfy your requirement without using a lazy val and without altering the base class:
scala> class Base(val name: String)
defined class Base
scala> class Derived extends Base(classOf[Derived].getName)
defined class Derived
scala> new Derived name
res0: String = Derived
You are going to have to explain your motivation for wanting to do this. Does name have to be a constructor parameter of Foo or can it be an abstract member? Does it have to be a val or will a def do.
You can do this
class Foo(val name: String)
class Bar extends Foo("") {
override val name = getClass.getName
}
and new Bar().name will give Bar
But I suspect that if your true motivation is known then there is a better way to do what you really want.
How about
class Bar extends Foo(classOf[Bar].getName.length)
What about this:
class Foo(otherName: Option[String] = None) {
val name = otherName.getOrElse(this.getClass.getName)
}
class Bar extends Foo()
val b = new Bar
println(b.name)