This question is somewhat hard to summarize. Following code block shows what I want to do.
I have a base class like this:
`class Base {
def methA:String="ook"
def methB:Int=1
}
Also I have a derived class, where I want each subclass method to call the super class method twice, compare the results and throw an exception on mismatch (this is for a test scenario).
But if I write
class Derived extends Base {
private def callDoublyAndCompare[T](fun:()=>T) : T = {
val fst=fun()
val snd=fun()
if(fst!=snd) throw new RuntimeException(s"Mismatch fst=$fst != snd=$snd")
snd
}
override def methB:Int={
callDoublyAndCompare(() => super[Derived].methB)
}
}
Then this will not compile. The only way out of this problem sofar has been to extract a method in class Derived which only calls the superclass' methB and to call this from the lambda call.
Is there a better way?
I understood you want to call super call method. Hope below is what you want.
You can call that as below with the key word super only
(new Derived).methB . This will call super call method in callDoublyAndCompare twice as per your code .
class Derived extends Base {
private def callDoublyAndCompare[T](fun:()=>T) : T = {
val fst=fun()
val snd=fun()
if(fst!=snd) throw new RuntimeException(s"Mismatch fst=$fst != snd=$snd")
snd
}
override def methB:Int={
callDoublyAndCompare(() => super.methB) //kept only super
}
}
The original example was not fully complete insofar as the Derived class was defined as inner class of another scala class.
After I moved out this inner class to the top level, the example from Praveen above suddenly worked.
Related
I have an abstract superclass with a variety of stateless implementations. It seems like the thing to do is make the abstract superclass a class, and make each implementation an object because I only ever need one of each.
This is a toy example, but it shows the compile error I’m getting:
// Tramsformer.scala
class Transformer {
def transform(value : String) : String
}
object Transformer {
getTransformer(String : name) : Transformer = {
name match {
case "upper" => UpperTransformer
// I'm getting "Cannot Resolve Symbol" on UpperTransformer,
// even though they're in the same package.
case _ => throw new IllegalArgumentException("...")
}
}
}
// ---
// UpperTransformer.scala is in the same package
object UpperTransformer extends Transformer {
override def transform(vlaue : String) = foo.toUpperCase()
}
I’m really shooting for some sort of dynamic dispatch on (dataProvider, desiredAction) here.
Then some class can call Transformer.getTransformer("upper").transform("stack-overflow") without making any unnecessary objects.
Any idea what I should be doing differently? Should I stop worrying about premature optimization and learn to love the instance?
The problem isn't visibility, it's that you simply do not define an object named UpperTransformer anywhere - only a class. If you replace class UpperTransformer with object UpperTransformer, your code should work fine.
trait UserRepository {
def findByFirstName(firstName: String): Seq[User]
}
trait UserBusinessDelegate extends UserRepository {
abstract override def findByFirstName(firstName: String) = {
super.findByFirstName(firstName)
}
}
class MockUserRepository extends UserRepository {
override def findByFirstName(firstName: String) = {
// whatever
}
}
val userRepository = new MockUserRepository with UserBusinessDelegate
userRepository.findByFirstName("John") // OK
However, if I change UserBusinessDelegate as follows:
trait UserBusinessDelegate {
self: UserRepository =>
override def findByFirstName(firstName: String): Seq[User] = {
self.findByFirstName(firstName) // requires explicit return type, thinks this is a recursive call
}
}
val userRepository = new MockUserRepository with UserBusinessDelegate
userRepository.findByFirstName("John") // StackOverflow!!!
I understand stackable pattern and hence how the first case works. My question is why the 2nd doesn't.
In the second snippet you have a recursive call without an exit condition:
override def findByFirstName(firstName: String): Seq[User] = {
self.findByFirstName(firstName)
}
This will always call findByFirstName from UserBusinessDelegate (because you're using self, which basically says that this object will have this kind of behaviour at runtime, not that it's parent will have it and therefore we should call parent's method) creating a new stack frame with each call -> stack overflow.
In the second snippet UserBusinessDelegate's findByFirstName will be called and then you call MockUserRepository's method from it using super -> no recursion -> no stack overflow. You can check Scala's stackable trait pattern for more info.
#Edit: to make it more clear, in the snippet that throws a SO exception the findByFirstName method from MockUserRepository won't be called because you are overriding it in UserBusinessDelegate therefore the anonymous class created with new MockUserRepository with UserBusinessDelegate will only contain the overriden method and that's why the SO, is that clear?
Why would you assume that the method from MockUserRepository would get invoked?
#Edit2: the code doesn't compile without override because self: UserRepository => tells the compiler that a method with such a signature will already be there at runtime and you cannot have 2 methods with the same signature. The first example works only because it's a stackable trait, such traits are dynamically bound and can modify the behaviour but have to call super at some point (which normally isn't allowed without the abstract override modifier, I really recommend going through the link I posted about stackable pattern).
Maybe someone else knows a way, from what I know there's no way to call the mock method unless you change the method name in UserBusinessDelegate and drop the override, then you can call self.findByFirstName and it will call the method from MockUserRepository.
I have the following setup:
trait A
{
def doSomething(): Unit;
}
object B extends A
{
override def doSomething(): Unit =
{
// Implementation
}
}
class B(creator: String) extends A
{
override def doSomething(): Unit =
{
B.doSomething() // Now this is just completely unnecessary, but the compiler of course insists upon implementing the method
}
}
Now you may wonder why I even do this, why I let the class extend the trait as well.
The problem is, that somewhere in the Program there is a Collection of A.
So somewhere:
private val aList: ListBuffer[A] = new ListBuffer[A]
and in there, I also have to put Bs (among other derivates, namely C and D)
So I can't just let the B-class not extend it.
As the implementation is the same for all instances, I want to use an Object.
But there is also a reason I really need this Object. Because there is a class:
abstract class Worker
{
def getAType(): A
def do(): Unit =
{
getAType().doSomething()
}
}
class WorkerA
{
def getAType(): A =
{
return B
}
}
Here the singleton/object of B gets returned. This is needed for the implementation of do() in the Worker.
To summarize:
The object B is needed because of the generic implementation in do() (Worker-Class) and also because doSomething() never changes.
The class B is needed because in the collection of the BaseType A there are different instances of B with different authors.
As both the object and the class have to implement the trait for above reasons I'm in kind of a dilemma here. I couldn't find a satisfying solution that looks neater.
So, my question is (It turns out as a non-native-speaker I should've clarified this more)
Is there any way to let a class extend a trait (or class) and say that any abstract-method implementation should be looked up in the object instead of the class, so that I must only implement "doSomething()" (from the trait) once (in the object)? As I said, the trait fulfills two different tasks here.
One being a BaseType so that the collection can get instances of the class. The other being a contract to ensure the doSomething()-method is there in every object.
So the Object B needs to extend the trait, because a trait is like a Java interface and every (!) Object B (or C, or D) needs to have that method. (So the only option I see -> define an interface/trait and make sure the method is there)
edit: In case anyone wonders. How I really solved the problem: I implemented two traits.
Now for one class (where I need it) I extend both and for the other I only extend one. So I actually never have to implement any method that is not absolutely necessary :)
As I wrote in the comment section, it's really unclear to me what you're asking.
However, looking at your code examples, it seems to me that trait A isn't really required.
You can use the types that already come with the Scala SDK:
object B extends (()=>Unit) {
def apply() { /* implementation */ }
}
Or, as a variant:
object B {
val aType:()=>Unit = {() => /* implementation */ }
}
In the first case, you can access the singleton instance with B, in the second case with B.aType.
In the second case, no explicit declaration of the apply method is needed.
Pick what you like.
The essential message is: You don't need a trait if you just define one simple method.
That's what Scala functions are for.
The list type might look like this:
private val aList:ListBuffer[()=>Unit] = ???
(By the way: Why not declare it as Seq[()=>Unit]? Is it important to the caller that it is a ListBuffer and not some other kind of sequence?)
Your worker might then look like this:
abstract class Worker {
def aType:()=>Unit // no need for the `get` prefix here, or the empty parameter list
def do() {aType()}
}
Note that now the Worker type has become a class that offers a method that invokes a function.
So, there is really no need to have a Worker class.
You can just take the function (aType) directly and invoke it, just so.
If you always want to call the implementation in object B, well - just do that then.
There is no need to wrap the call in instances of other types.
Your example class B just forwards the call to the B object, which is really unnecessary.
There is no need to even create an instance of B.
It does have the private member variable creator, but since it's never used, it will never be accessed in any way.
So, I would recommend to completely remove the class B.
All you need is the type ()=>Unit, which is exactly what you need: A function that takes no parameters and returns nothing.
If you get tired of writing ()=>Unit all the time, you can define a type alias, for example inside the package object.
Here is my recommentation:
type SideEffect = ()=>Unit
Then you can use SideEffect as an alias for ()=>Unit.
That's all I can make of it.
It looks to me that this is probably not what you were looking for.
But maybe this will help you a little bit along the way.
If you want to have a more concrete answer, it would be nice if you would clarify the question.
object B doesn't really have much to do with class B aside from some special rules.
If you wish to reuse that doSomething method you should just reuse the implementation from the object:
class B {
def doSomething() = B.doSomething()
}
If you want to specify object B as a specific instance of class B then you should do the following:
object B extends B("some particular creator") {
...
}
You also do not need override modifiers although they can be handy for compiler checks.
The notion of a companion object extending a trait is useful for defining behavior associated with the class itself (e.g. static methods) as opposed to instances of the class. In other words, it allows your static methods to implement interfaces. Here's an example:
import java.nio.ByteBuffer
// a trait to be implemented by the companion object of a class
// to convey the fixed size of any instance of that class
trait Sized { def size: Int }
// create a buffer based on the size information provided by the
// companion object
def createBuffer(sized: Sized): ByteBuffer = ByteBuffer.allocate(sized.size)
class MyClass(x: Long) {
def writeTo(buffer: ByteBuffer) { buffer.putLong(x) }
}
object MyClass extends Sized {
def size = java.lang.Long.SIZE / java.lang.Byte.SIZE
}
// create a buffer with correct sizing for MyClass whose companion
// object implements Sized. Note that we don't need an instance
// of MyClass to obtain sizing information.
val buf = createBuffer(MyClass)
// write an instance of MyClass to the buffer.
val c = new MyClass(42)
c.writeTo(buf)
I'm working on an automatic mapping framework built on top of Dozer. I won't go into specifics as it's not relevant to the question but in general it's supposed to allow easy transformation from class A to class B. I'd like to register the projections from a class's companion object.
Below is a (simplified) example of how I want this to work, and a Specs test that assures that the projection is being registered properly.
Unfortunately, this doesn't work. From what I can gather, this is because nothing initializes the A companion object. And indeed, if I call any method on the A object (like the commented-out hashCode call, the projection is being registered correctly.
My question is - how can I cause the A object to be initialized automatically, as soon as the JVM starts? I don't mind extending a Trait or something, if necessary.
Thanks.
class A {
var data: String = _
}
class B {
var data: String = _
}
object A {
projekt[A].to[B]
}
"dozer projektor" should {
"transform a simple bean" in {
// A.hashCode
val a = new A
a.data = "text"
val b = a.-->[B]
b.data must_== a.data
}
}
Short answer: You can't. Scala objects are lazy, and are not initialized until first reference. You could reference the object, but then you need a way of ensuring the executing code gets executed, reducing the problem back to the original problem.
In ended up doing this:
trait ProjektionAware with DelayedInit
{
private val initCode = new ListBuffer[() => Unit]
override def delayedInit(body: => Unit)
{
initCode += (() => body)
}
def registerProjektions()
{
for (proc <- initCode) proc()
}
}
object A extends ProjektionAware {
projekt[A].to[B]
}
Now I can use a classpath scanning library to initialize all instances of ProjektionAware on application bootstrap. Not ideal, but works for me.
You can force the instantiation of A to involve the companion object by using an apply() method or some other sort of factory method defined in the object instead of directly using the new A() constructor.
This does not cause the object to be initialized when the JVM starts, which I think as noted in another answer can't generally be done.
As Dave Griffith and Don Roby already noted, it cannot be done at JVM startup in general. However maybe this initialization could wait until first use of your framework?
If so, and if you don't mind resorting to fragile reflection tricks, in your --> method you could obtain reference to the companion object and get it initialize itself.
You can start at Getting object instance by string name in scala.
We could use this sort of a way to ensure that companion object gets initialized first and then the class gets instantiated.
object B {
val i = 0
def apply(): B = new B()
}
class B {
// some method that uses i from Object B
def show = println(B.i)
}
// b first references Object B which calls apply()
// then class B is instantiated
val b = B()
I know that objects are treated pretty much like singletons in scala. However, I have been unable to find an elegant way to specify default behavior on initial instantiation. I can accomplish this by just putting code into the body of the object declaration but this seems overly hacky. Using an apply doesn't really work because it can be called multiple times and doesn't really make sense for this use case.
Any ideas on how to do this?
Classes and objects both run the code in their body upon instantiation, by design. Why is this "hacky"? It's how the language is supposed to work. If you like extra braces, you can always use them (and they'll keep local variables from being preserved and world-viewable).
object Initialized {
// Initalization block
{
val someStrings = List("A","Be","Sea")
someStrings.filter(_.contains('e')).foreach(s => println("Contains e: " + s))
}
def doSomething { println("I was initialized before you saw this.") }
}
scala> Initialized.doSomething
Contains e: Be
Contains e: Sea
I was initialized before you saw this.
scala> Initialized.someStrings
<console>:9: error: value someStrings is not a member of object Initialized
Initialized.someStrings
Rex has it right, I just wanted to point out a pattern I use a lot, that saves you from having to use vars, while avoiding namespace pollution by intermediate values.
object Foo {
val somethingFooNeeds = {
val intermediate = expensiveCalculation
val something = transform(intermediate)
something
}
}
If it makes you feel better, you can create some class with protected constructor and object will create singleton of this class:
sealed class MyClass protected (val a: String, b: Int) {
def doStuff = a + b
}
object MyObject extends MyClass("Hello", b = 1)
Also notice, that sealed stops other classes and objects to extend MyClass and protected will not allow creation of other MyClass instances.
But I personally don't see any problems with some code in the body of the object. You can also create some method like init and just call it:
object MyObject {
init()
def init() {
...
}
}
The body of object and class declarations IS the default constructor and any code placed in there will be executed upon first reference, so that is exactly the way to do it.