I have a abstract entity class that 3 slightly different entities implements. In my 3 sub classes I have overriden the equals and has methods but the question is, should I also do this in the abstract entity? If I dont I will not be able to compare entities that are only defined by abstract entity unless i cast them. If i do a equals will I risk to compare to different sub entities and get that they are alike?
Example:
abstract class Log{}
SystemLog extends Log{}
UserLog extends Log{}
public void test(Log log){
Log myInner = new SystemLog();
if(log.equals(myInner)){
//do random stuff
}
}
I cannot see problem with casting. Type of argument to equals is Object, so you have to cast anyway to have access to attributes.
If you define equals method in each subclasses, when comes the situation where equals in abstract superclass is called?
If i do a equals will I risk to compare to different sub entities and
get that they are alike
You are in the risk of comparing different subentities to each others anyway. Just imagine Set with superclass as type populated with objects that are two instances of two different subclasses. It has not too much to do with do you override equals in superclass or not.
In your example equals method possibly implemented in abstract class Log will not be called, if we have implementation already in actual subclass:
Assuming:
UserLog extends Log{
public boolean equals(Object o) {
//I do override equals method so I am one who is called.
//and here you go and check type with something like
if (this == o) return true;
if (!(o instanceof UserLog)) return false;//accepts subclasses of UserLog
....
}
...
}
//And then somewhere else
Log ul = new UserLog();
test(ul);
Related
I have written the following code:
class a {
object c {
var a = "STATIC"
def m() = print("STATIC METHOD")
var f = () => print("STATIC FUNCTION")
}
}
object m {
def main(args: Array[String]) = {
var o = new a()
o.c.m()
}
}
Can I say that the variables, functions and methods that are declared in object c can be static?
If I change name of object c with a then will the object becomes a companion object?
Scala has no true meaning of 'static' that Java does.
The fact that objects have a backing on the JVM that uses static methods / fields is a leaking implementation detail that you only need to deal with if using Java/JVM interop.
Unless you explicitly need that interop, you need to stop thinking of declared objects as 'static' and instead think of them as singletons within their given scope.
An inner object nested under a class, means that there is only ever going to be 1 instance of that object, for each class instance, unlike inner classes which could have multiple instances.
This applies at the top level as well, except that Scala can do additional compatibility with other JVM languages, and mark some of the methods/members as static.
Fields and methods in an object are how Scala declares things that you would have used static for in Java. I guess you can, for intuition sake, say that those fields are usually static (as in only one of those in a JVM at once).
However, in your example, you put an object inside a class, making it no longer static. You can easily check this with a few lines of code (that you can find here on Scastie).
class MyClass {
object Embedded {
val a = "field"
def m = println("method")
}
}
val a = new MyClass().Embedded
val b = new MyClass().Embedded
// prints "a and b are different objects"
if (a eq b)
println("a and b are the same object")
else
println("a and b are different objects")
Regarding your second question: no, class and object must be in the same scope in order for it to be a companion object. You can find more details on the Scala Language Specification.
I quote from there:
Generally, a companion module of a class is an object which has the same name as the class and is defined in the same scope and compilation unit. Conversely, the class is called the companion class of the module.
To answer you questions:
The methods and fields in a.c are not globally static because they need an instance of a to exist. If a were an object, a.c would be static too.
If you want to have a companion object with static fields and methods for your class a it has to be defined outside of a's code block, like this:
class a {
/* non-static stuff goes here */
}
object a {
/* static stuff goes there */
def m() = print("STATIC METHOD")
}
You must keep both in the same file, defining the object or the
class first doesn't matter, so it generally depend on a convention or what makes most sense depending on use case.
If you want to call the static method a.m inside the class a, you will still need to call it a.m and not just m. But the class a will be able to use private fields and methods of object a, because they are companions.
As others already said, static doesn't really exist in Scala, but the concept transpires from Java since Scala is in most cases compiled into java bytecode.
Last advice, the convention is usually the same in Scala and in Java for classes and object: the first-letter of their name should be uppercase (except in some advanced Scala cases)
Well I am new to Scala programming. I struck ed with a question, what is the reason Each Auxiliary Constructor should call one of the previous defined constructor in Scala?
Considering following two situation written in Java:
public class Random{
public int a;
public int b;
public Random(int _a, int _b){
this.a = _a;
this.b = _b;
}
public Random(int _a){
this(_a, 0);
}
}
and
public class Random{
public int a;
public int b;
public Random(int _a, int _b){
this.a = _a;
this.b = _b;
}
public Random(int _a){
this.a = _a;
this.b = 0;
}
}
Since both case are valid in Java, however the design of first pattern would be better than second example because of better maintainability
Considering following situation:
You want to change the name of either attributes
In first example, you only need to change once, which is the this.a to this.newname, and you don't need to go through all the overloaded constructor to change it.
However in second example, you need to go through all the overloaded constructor.
Because of this design reason, in Scala you must call back the default constructor for the auxilliary constructor.
In Scala, each class a Primary constructor, this constructor must be called to construct the object. This is unlike Java which has no concept of primary and auxiliary constructors and all constructors have equal priority.
The primary constructor is special in that fields and parameters declared within it become field on the class usable by other methods in the class called later or even directly accessible if declared public. If the primary constructor isn't called you can't guarantee that these fields are properly initialized.
So then to enforce that the primary constructor gets called, auxiliary constructors must call this primary constructor, or call another constructor as long as the chain of constructor calls eventually leads to the primary constructor.
we have two types of constructors in Scala.
i) Primary Constructor
ii) Secondary Constructor
We can instantiate the class in scala through primary constructor only, even though we declare any object using auxiliary constructor.
Since the auxiliary constructor is the part of primary constructor(because we define methods, variables ,auxiliary constructors inside primary constructor) and each auxiliary constructor should call the previous defined constructor(auxiliary or primary) and this call finally reaches to primary constructor.
From this we can conclude that primary constructor is the main entry point to instantiate the class and this is the reason each auxiliary constructor must and should call another constructor(predefined auxiliary or primary).
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 needed two instances that has access to each other privates. i naturaly thought of a companion object that grants access to a one and only instance of it's companion class. the class itself i made private, so users cannot just create instances using new.
object A {
def apply = dual
lazy val dual = new A
}
private class A {
//some irrelevant logic...
}
this code does not compile. i get: class A escapes its defining scope as part of type A error, which i don't really understand. my current workaround was to define a trait with every method declaration the class should have and make class A extend that trait, while dual is of the trait type, and not class A type.
what's the theoretic problem i'm missing here? why is this forbiden?
Paolo's solution is good (+1), but he didn't explain the error message, so let me try that. The problem stems from the fact that every method needs a return type. Your original definition of apply and dual returned an object of class A, thus the implicit return type of both was A. That implies that A must be visible to clients - how else could they call the function or access the val? Moreover, as both - and their parent object too - are public, they are globally visible. However, you declared A private which means it must not be visible outside its package. So there is a conflict which can't be resolved by the compiler.
The general rule is that all parameter and return type of functions / members must have (at least) the same scope of visibility as the referring member itself*. Thus one trivial way to solve this problem would be to reduce the visibility of apply and dual to private. This would satisfy the compiler, but not you :-)
Your solution gets around the problem by changing the static return type to a public trait, which thus has the same visibility as the members referring to it. The dynamic type of the returned object is still class A, however, this need not be visible to clients. This is a classic example of the principle "program to interfaces, not implementations".
Note that to apply this principle to the full extent, one could turn class A into a private inner class of object A, thus making it innaccessible even for other classes within the same package:
trait A {
//...
}
object A {
def apply: A = dual
lazy val dual: A = new AImpl
private class AImpl extends A {
//some irrelevant logic...
}
}
* To be pedantic, the enclosing class / object may reduce the visibility of its members, like here:
private class Holder {
def member = new Hidden
}
private class Hidden
where member is public but its enclosing class is private, effectively hiding its members from the external world. So the compiler emits no complaints here.
I think you don't want a private class, but a class with a private constructor.
class A private()
object A {
def apply = dual
lazy val dual = new A
}
Now your class is "visible" to outside code, but only your companion object can create instances of it.
I have the following grails domain objects
abstract class A {
String name
}
class B extends A {
String propertySpecificToB
}
class C extends A {
String propertySpecificToC
}
and I can successfully save them to my database (which in this case is MongoDB). However, I would like to list all the names of the rows in my database, so I do something like:
A.list()
But, that throws an InstantiationException as it tries to create instances of the abstract class A. How can I list off all rows (regardless of which class it is). I could make A non-abstract, but it would never be valid to have a A. Also, I'd like to place some abstract methods in A.
I'd really just like it to return a list of A, with the list actually containing Bs and Cs.
I found a related post, but that didn't solve this problem.