The question is what do I need to do to replicate the functionality (for other class cooked on my own) of the scala enumerations:
val MY_ENUM_1, MY_ENUM_2 = Value
this creates 2 instances of Value. How can I do this for something like:
object App extends App {
class Foo {}
val foo1, foo2 = Foo
}
You need to add the new keyword:
scala> class Foo
defined class Foo
scala> val foo1, foo2 = new Foo
foo1: Foo = Foo#470e2030
foo2: Foo = Foo#3fb4f649
val MY_ENUM_1, MY_ENUM_2 = Value
works because it actually calls the method Value from Enumeration, which is:
/** Creates a fresh value, part of this enumeration. */
protected final def Value: Value = Value(nextId)
which will end up calling
protected final def Value(i: Int, name: String): Value = new Val(i, name)
which in turn will return a Val (and also increment nextId).
You can see the whole process in the source code of Enumeration (starts line #128, the Val class is defined line #209).
Related
Is the below Scala class is mutable or immutable?
I believe that its immutable as I can't edit the variables or access them once its created but whats making me doubt myself is the fact that it returns the current instance of a variable using its functions. It also does not have final in front of it which is further making me doubt myself.
class person(name:String, dob:String){
def getName = name
def getDob = dob
def display() = {
println("Name "+name+" dob: "+dob)
}
}
Thanks,
You have a misconception with the term Immutable:
I believe that its immutable as I can't edit the variables or access
them once its created
That's the definition of a private thing (method, variable, ...). Immutability refers to the fact that you cannot mutate state, that is, you can't change the value of something unless you create a new instance of it.
Let's see it with an example:
trait Foo{
def myMutableValue: Int
}
class Clazz extends Foo{
var myMutableValue = 1
def changeState(): Int = {
myMutableValue += 1
myMutableValue
}
}
val bar = new Clazz
bar.changeState() // myMutableValue = 2
bar.changeState() // myMutableValue = 3
bar.changeState() // myMutableValue = 4
bar.myMutableValue // myMutableValue = 4
With that example, in your instance of Clazz (bar) you're changing the state of a class attribute, in this case myMutableValue is changing its value every time I invoke changeState.
Please note that the class is public by default and changeState is also public and that doesn't means that is immutable.
Now, let's see an immutable approach:
trait Foo{
def myMutableValue: Int
}
class Clazz extends Foo{
val myMutableValue = 1
def changeState(): Int = myMutableValue + 1
}
val instance = new Clazz
instance.changeState() // myMutableValue = 2
instance.changeState() // myMutableValue = 2
instance.changeState() // myMutableValue = 2
instance.myMutableValue // 1
With this approach, every call to changeState will evaluate to 2, no matter how many times I call the function. That is, because we're dealing with an immutable value (val myMutableValue = 1). Every invocation of changeState will perform the evaluation and return a copy of that value. You're not modifying in any way the value of myMutableValue.
Please take a look to this and this.
Also, please take a look at your code, you have some errors:
By convention, class name should be capitalized (Person instead of person).
You don't need to reassign your class values with def (def getNameand def getDob). You can use class values as is.
Lastly:
It also does not have final in front of it which is further making me
doubt myself.
Again, you're talking about different things. final, as in Java, is a modifier to prevent your class to be extended. It doesn't relate in any way to immutability In adition, if you want to prevent mutability in your subclass you have to make all their members final (see this).
Since your example is coded in Scala you have all the tools that the language itself offers at your disposal (e.g. val, sealed, final)
Please note that I've used a trait to explain the possible use of def.
EDIT: about final modifier and immutability
Thanks to #Silvio Mayolo and #puhlen for the comments and clarification about final
I found an interesting scala implementation of Builder pattern, but I can't understand what a few lines mean:
case class Built(a:Int, b:String){}
trait Status
trait Done extends Status
trait Need extends Status
class Builder[A<:Status,B<:Status] private(){
private var built = Built(0,"")
def setA(a0:Int)={
built = built.copy(a=a0)
this.asInstanceOf[Builder[Done,B]]
}
def setB(b0: String) = {
built = built.copy(b = b0)
this.asInstanceOf[Builder[A,Done]]
}
def result(implicit ev: Builder[A,B] <:< Builder[Done,Done]) = built
}
object Builder{
def apply() = new Builder[Need,Need]
}
1) What does private() mean in class Builder[A<:Status,B<:Status] private() class declaration?
2) What is the meaning of implicit ev: Builder[A,B] <:< Builder[Done,Done] in result function?
1)
The private means that the primary constructor for Builder can not be accessed from outside.
Since there are no other constructors, the only way to get an instance is through the companion object with the apply method.
Example:
val builder = Builder()
2)
You have methods in Builder to set both parameters for the Built case-class.
The method result gives you the constructed Built-instance. The evidence makes sure that you have set both parameters and will not allow you to create an instance if you didn't do it.
Example (I did not test this, so please correct me if I am wrong):
val builderA = Builder().setA(3)
val resultA = builderA.result //should not compile because this is Builder[Done, Need]
val builderAB = builderA.setB("hello") //now this is Builder[Done, Done]
val resultAB = builderAB.result //should compile and yield Built(3, "hello")
For your first question, the keyword private in this position means the constructor for the class is private.
Search results so far have led me to believe this is impossible without either a non-primary constructor
class Foo { // NOT OK: 2 extra lines--doesn't leverage Scala's conciseness
private var _x = 0
def this(x: Int) { this(); _x = x }
def x = _x
}
val f = new Foo(x = 123) // OK: named parameter is 'x'
or sacrificing the name of the parameter in the primary constructor (making calls using named parameters ugly)
class Foo(private var _x: Int) { // OK: concise
def x = _x
}
val f = new Foo(_x = 123) // NOT OK: named parameter should be 'x' not '_x'
ideally, one could do something like this:
class Foo(private var x: Int) { // OK: concise
// make just the getter public
public x
}
val f = new Foo(x = 123) // OK: named parameter is 'x'
I know named parameters are a new thing in the Java world, so it's probably not that important to most, but coming from a language where named parameters are more popular (Python), this issue immediately pops up.
So my question is: is this possible? (probably not), and if not, why is such an (in my opinion) important use case left uncovered by the language design? By that, I mean that the code either has to sacrifice clean naming or concise definitions, which is a hallmark of Scala.
P.S. Consider the case where a public field needs suddenly to be made private, while keeping the getter public, in which case the developer has to change 1 line and add 3 lines to achieve the effect while keeping the interface identical:
class Foo(var x: Int) {} // no boilerplate
->
class Foo { // lots of boilerplate
private var _x: Int = 0
def this(x: Int) { this(); _x = x }
def x = _x
}
Whether this is indeed a design flaw is rather debatable. One would consider that complicating the syntax to allow this particular use case is not worthwhile.
Also, Scala is after all a predominantly functional language, so the presence of vars in your program should not be that frequent, again raising the question if this particular use case needs to be handled in a special way.
However, it seems that a simple solution to your problem would be to use an apply method in the companion object:
class Foo private(private var _x: Int) {
def x = _x
}
object Foo {
def apply(x: Int): Foo = new Foo(x)
}
Usage:
val f = Foo(x = 3)
println(f.x)
LATER EDIT:
Here is a solution similar to what you originally requested, but that changes the naming a bit:
class Foo(initialX: Int) {
private var _x = initialX
def x = _x
}
Usage:
val f = new Foo(initialX = 3)
The concept you are trying to express, which is an object whose state is mutable from within the object and yet immutable from the perspective of other objects ... that would probably be expressed as an Akka actor within the context of an actor system. Outside the context of an actor system, it would seem to be a Java conception of what it means to be an object, transplanted to Scala.
import akka.actor.Actor
class Foo(var x: Int) extends Actor {
import Foo._
def receive = {
case WhatIsX => sender ! x
}
}
object Foo {
object WhatIsX
}
Not sure about earlier versions, but In Scala 3 it can easily be implemented like follows:
// class with no argument constructor
class Foo {
// prive field
private var _x: Int = 0
// public getter
def x: Int = _x
// public setter
def x_=(newValue: Int): Unit =
_x = newValue
//auxiliary constructor
def this(value: Int) =
this()
_x = value
}
Note
Any definition within the primary constructor makes the definition public, unless you prepend it with private modifier
Append _= after a method name with Unit return type to make it a setter
Prepending a constructor parameter neither with val nor with var, makes it private
Then it follows:
val noArgFoo = Foo() // no argument case
println(noArgFoo.x) // the public getter prints 0
val withArgFoo = Foo(5) // with argument case
println(withArgFoo.x) // the public getter prints 5
noArgFoo.x = 100 // use the public setter to update x value
println(noArgFoo.x) // the public getter prints 100
withArgFoo.x = 1000 // use the public setter to update x value
println(withArgFoo.x) // the public getter prints 1000
This solution is exactly what you asked; in a principled way and without any ad hoc workaround e.g. using companion objects and the apply method.
What is the difference between defining an object using the new operator vs defining a standalone object by extending the class?
More specifically, given the type class GenericType { ... }, what is the difference between val a = new GenericType and object a extends GenericType?
As a practical matter, object declarations are initialized with the same mechanism as new in the bytecode. However, there are quite a few differences:
object as singletons -- each belongs to a class of which only one instance exists;
object is lazily initialized -- they'll only be created/initialized when first referred to;
an object and a class (or trait) of the same name are companions;
methods defined on object generate static forwarders on the companion class;
members of the object can access private members of the companion class;
when searching for implicits, companion objects of relevant* classes or traits are looked into.
These are just some of the differences that I can think of right of the bat. There are probably others.
* What are the "relevant" classes or traits is a longer story -- look up questions on Stack Overflow that explain it if you are interested. Look at the wiki for the scala tag if you have trouble finding them.
object definition (whether it extends something or not) means singleton object creation.
scala> class GenericType
defined class GenericType
scala> val a = new GenericType
a: GenericType = GenericType#2d581156
scala> val a = new GenericType
a: GenericType = GenericType#71e7c512
scala> object genericObject extends GenericType
defined module genericObject
scala> val a = genericObject
a: genericObject.type = genericObject$#5549fe36
scala> val a = genericObject
a: genericObject.type = genericObject$#5549fe36
While object declarations have a different semantic than a new expression, a local object declaration is for all intents and purpose the same thing as a lazy val of the same name. Consider:
class Foo( name: String ) {
println(name+".new")
def doSomething( arg: Int ) {
println(name+".doSomething("+arg+")")
}
}
def bar( x: => Foo ) {
x.doSomething(1)
x.doSomething(2)
}
def test1() {
lazy val a = new Foo("a")
bar( a )
}
def test2() {
object b extends Foo("b")
bar( b )
}
test1 defines a as a lazy val initialized with a new instance of Foo, while test2 defines b as an object extending Foo.
In essence, both lazily create a new instance of Foo and give it a name (a/b).
You can try it in the REPL and verify that they both behave the same:
scala> test1()
a.new
a.doSomething(1)
a.doSomething(2)
scala> test2()
b.new
b.doSomething(1)
b.doSomething(2)
So despite the semantic differences between object and a lazy val (in particular the special treatment of object's by the language, as outlined by Daniel C. Sobral),
a lazy val can always be substituted with a corresponding object (not that it's a very good practice), and the same goes for a lazy val/object that is a member of a class/trait.
The main practical difference I can think of will be that the object has a more specific static type: b is of type b.type (which extends Foo) while a has exactly the type Foo.
Are there public instance variables anymore in Scala? I'm reading Programming in Scala, which covers Scala 2.8. If I'm understanding it correctly, it claims that vars in 2.8 are by default public.
I'm trying to write code for 2.9.1.final now, and instance variables are now by default private? But there's no public keyword that I'm aware of. (Interestingly enough, it appears it used to exist sometime in the 2.x series, but it mysteriously disappeared somewhere along the line.)
Am I missing something obvious?
Also, by extension, is there an easy way to declare a variable passed to a class constructor to be public (since it appears that those also have default private visibility now too)?
Example:
class Instance(label: String, attributes: Array[Int]){
val f = 0
}
Eclipse claims that label, attributes, and f are all private. Scala 2.9.1.final is being used as the library.
In scala, if you omit the modifer, then instance fields are public by default:
scala> class Foo { var foo = 1 }
defined class Foo
scala> class Bar { def bar() = { val f = new Foo; f.foo = 5; }}
defined class Bar
No worries there. However, when you use a variable in a constructor, the variable is not necessarily turned into a field:
scala> class Foo(foo: Int)
defined class Foo
scala> class Bar { def bar() = { val f = new Foo(5); println(f.foo) }}
<console>:8: error: value foo is not a member of Foo
class Bar { def bar() = { val f = new Foo(5); println(f.foo) }}
^
so you can declare it as a val or var to have it available:
scala> class Foo(val foo: Int)
defined class Foo
scala> class Bar { def bar() = { val f = new Foo(5); println(f.foo) }}
defined class Bar
Note that all fields are actually private, but scala exposes accessor methods (foo() and foo_=(t: Int)) to enable you to access the fields), which is why scala-ide says that the fields are private (assuming you mean when you hover over the variable).