Scala Constructors, Named Arguments, and Implicit Getters/Setters - scala

Is it possible to use named arguments in a Scala constructor, and later on override getters and setters without breaking the constructor interface or making the code extremely ugly?
Take the following bit of scala code
class Person( var FirstName: String, var LastName: String )
Nice and clean. This would create a simple class called person, which we could use in the following way
val john = new Person( FirstName="John", LastName="Doe" )
john.FirstName = "Joe"
println( john.FirstName )
Later, we decide we want to add some validation to the FirstName setter. As such, we create a new private local variable and override the getter and setter methods
class Person( var _FirstName: String, var _LastName: String ) {
def FirstName = _FirstName
def FirstName_= (value:String) = _FirstName = value
}
Still somewhat clean, however in order to do this, we've had to change the constructor argument names, thus breaking the external interface.
The first solution to this problem I came up with was
class Person {
var _FirstName:String = null
var LastName:String = null
def FirstName = _FirstName
def FirstName_= (value:String) = _FirstName = value
def this( FirstName: String, LastName: String ){
this()
this._FirstName = FirstName
this.LastName = LastName
}
}
Which is somewhat ugly and inelegant, and removes most of the nice reasons I was using scala in the first place.
Is there a better way of doing this?
tl;dr How to override getters/setters for members defined in the default constructor without making the code ugly or changing the public interface?

Did you consider using an companion object?
class Person private (f: String, l: String ) {
var FirstName = f
var LastName = l
}
object Person {
def apply(FirstName:String, LastName:String) =
new Person(FirstName, LastName)
}

If you're not already using implicit conversions to create the arguments, you can do something like this:
def validateName(s: String) = {
if (s.length>0 && s(0).isUpper) s
else throw new IllegalArgumentException(s+" is not a name!")
}
object Example {
private[Example] class ValidatedName(val s: String) { }
class Person(var firstName: ValidatedName, var lastName: String) { }
implicit def string2valid(s: String) = new ValidatedName(validateName(s))
implicit def valid2string(v: ValidatedName) = v.s
}
scala> new Example.Person("Joe","Schmoe")
res17: Example.Person = Example$Person#51887dd5
scala> new Example.Person("ee","cummings")
java.lang.IllegalArgumentException: ee is not a name!
It's not binary compatible, but it is source compatible (again, if the names weren't already relying upon implicit conversions).
Another slightly longer possibility is to create a stealth ancestor:
class CheckedPerson(private var first: String, var lastName: String) {
def firstName = first
def firstName_=(s: String) { first = validateName(s) }
}
class Person(firstName: String, lastName: String) extends
CheckedPerson(validateName(firstName),lastName) { }
for which I'm not sure about binary compatibility, but will definitely give source compatibility.

No. There is currently no way to do that, it's currently not the focus of research.
It is one of my major pet peeves I have with the language: There is no sensible way to combine constructor arguments and self-defined getter/setter methods.
If you're not happy with the functionality class Person( var FirstName: String, var LastName: String ) provides, it basically means "back to Java's verboseness".

Related

How to instance a class using array | Kotlin

I need to create an instance of a class using an array of values.
I know how much parameters the class have
All class parameters are string
I tried:
class Person(val name: String, val lastName: String)
{
}
fun main()
{
val values= listOf<String>("James", "Smith")
val myPerson = Person(values);
}
Is possibly do something like that?
You could create a custom constructor that takes a list and uses it to instantiate your class:
class Person(val name: String, val lastName: String) {
constructor(values: List<String>) : this(values[0], values[1])
}
However, I would say you should avoid this since it is very error-prone (what if the provided values list is empty or have only one element?).

Modify one value in a Scala class constructor in a concise way

If I want to modify one single parameter in a constructor.
In the Scala case class, the apply method will be overridden twice. Unless apply applies ( no pun ) to auxiliary constructor.
Related to
Modifying case class constructor parameter before setting value
How to override apply in a case class companion
How one can modify one single input from a constructor ?
Criteria :
The class must hold immutable data. All data must be accessible.
Note it doesn't have to be case class.
Additionally , no need to use the apply method.
no extra unused parameters. In the example below, _fistName is still accessible but unused.
case class Person(
lastName: String,
_fistName: String, ... )
{ lazy val fistName = _fistName.toLowerCase }
Here are two simple approaches.
Using class:
class Person(first: String, last: String) {
val firstName = first.toLowerCase
val lastName = last.toLowerCase()
}
val person = new Person("Adam", "Smith")
println(person.firstName + " " + person.lastName) // adam smith
Using trait + object's apply():
trait Person {
val firstName: String
val lastName: String
}
object Person {
def apply(first: String, last: String) = new Person {
override val firstName: String = first.toLowerCase
override val lastName: String = last.toLowerCase
}
}
val person = Person("Adam", "Smith")
println(person.firstName + " " + person.lastName) // adam smith
Note that classes must be instantiated with new, while traits (created with apply()) don't.
Why no case classes? Well, they are designed to serve as ADTs (abstract data types). Basically, they are thought of as containers for some data without any logic. That's why they get apply() out-of-the-box. If you want to override it, then means your class doesn't have the semantics of a case class.
I can see that #prayag took the effort of trying to help you force it into a case class, but seriously, if it doesn't have to be one (and you said so in the question), then don't make it one.
The reference you had posted seems to have lot of answers as well.
Two simple ways I could think of
make it abstract case class and define companion object which would mutate the value you want
define the member of case class as var and mutate it.
eg. (using scalatest)
class CaseClassSpecs extends FunSpec {
describe("case class modify") {
it("APPROACH 1 : modifies abstract case class member") {
object Item {
def apply(itemId: String, itemName: String) :Item = new Item(itemId.toLowerCase, itemName) {}
}
abstract case class Item private (val itemId: String, itemName: String)
val item1 = Item("SKU-ONE", "Shirts")
assert(item1.itemId == "sku-one")
assert(item1.itemName == "Shirts")
}
it("APPROACH 2 : modifies case class member which is var") {
case class Item (var itemId: String, itemName: String) {
itemId = itemId.toLowerCase()
}
val item1 = Item("SKU-ONE", "Shirts")
assert(item1.itemId == "sku-one")
assert(item1.itemName == "Shirts")
}
}
}
Class parameters are not necessarily class members. You can have class parameters that do not become class members.
Method 1
class Foo(bar0: String) {
val bar = bar0.toLowerCase()
}
#main
def main(): Unit = {
println(Foo("AsDfGh").bar)
}
prints:
asdfgh
and the decompiled code is:
public class Foo {
private final String bar;
public Foo(final String bar0) {
this.bar = bar0.toLowerCase();
}
public String bar() {
return this.bar;
}
}
You see, bar0 is a "temporary" value, it does not become a field because it is not referenced.
So if you want to change a value, just do not use the original value in the methods.
Method 2
For case classes, it does not seem to work in 2022, but here is another trick:
case class Point (x: Double, y: Double)
class PolarPoint(r: Double, alpha: Double) extends Point(r*Math.cos(alpha), r*Math.sin(alpha))
Here r and alpha do not become members of PolarPoint.
If you don't need two types, you can make the 1st constructor protected:
case class Foo protected(x:Int)
class FooImpl(x0:Int) extends Foo(myFunc(x0))
You will reference objects as Foos but create FooImpls.
Method 3
Your class can have multiple parameter lists and implicits:
class Qux(x:String)(implicit val y:String = x.toLowerCase())
is converted to:
public class Qux {
private final String y;
public static String $lessinit$greater$default$2(String var0) {
return Qux$.MODULE$.$lessinit$greater$default$2(var0);
}
public Qux(final String x, final String y) {
this.y = y;
}
public String y() {
return this.y;
}
}
You see that here only y becomes a field.

How can one use Amazon's DynamoDBMapper in Scala?

I'm having trouble using Amazon's DynamoDBMapper in Scala code. The main sticking point is getting the JVM to recognize #DynamoDBHashkey when it is used in a case class, like:
case class MyCoolCaseClass(#DynamoDBHashKey(attributeName = "my_id") myId: String) {}
Any pointers from someone who has integrated this client library into a Scala project? (I'm hoping to not simply fallback to the low-level API, though that may be a decent decision once exhausting my options with the Mapper).
I had to do this:
import annotation.meta.beanGetter
import beans.BeanProperty
import com.amazonaws.services.dynamodbv2.datamodeling._
#DynamoDBTable(tableName="DEMOTAB")
case class DemoItem( // it's a case class for the free stuff, but can be ordinary class
#(DynamoDBHashKey #beanGetter) // would not work without meta annotation
#BeanProperty var id:String, // must be var or mapper can't instantiate one
#BeanProperty var number:Integer
) {
def this() = this(null, null) // needed by DynamoDB Mapper to instantiate
}
The DynamoDB mapper uses reflection to find the getters and setters. The SDK assumes Java-style conventions, i.e. that your getters and setters start with "get" or "is", and setters start with "set". You can see the reflection code on github.
I've been able to get it working, but it feels just like writing Java :(
#DynamoDBTable(tableName = "awesome_table")
class TheBestClass {
private var id : Integer = _
#DynamoDBHashKey
def getId() = id
def setId(_id: Integer) = id = _id
}
This works for me, including the boolean
#DynamoDBTable(tableName = "User")
case class User(
#(DynamoDBHashKey #field)
#(DynamoDBAutoGeneratedKey #field)
#BeanProperty var id: String,
#(DynamoDBAttribute #field)
#BeanProperty var firstName: String,
#(DynamoDBAttribute #field)
#BeanProperty var lastName: String,
#(DynamoDBAttribute #field)
#BeanProperty var active: Boolean
)
{
def this() = this(null, null, null, false)
}

Scala: to put constraints on the fields of a class

In Scala how to put constraints on the fields of a class?
In a package I have the domain of my model, in another package I have dsl to instantiate my model.
The basic form of the model is this:
abstract class Element {
var name: String
var description: String
var types : Set[Type]
}
class SAComponent (var name :String,
var description : String,
var properties : Set[Property] = Set(),
var types : Set[Type] = Set(),
) extends Component
Element is the root of my model.
I want to put constraints on the fields of Element, so that each class that inherits name and description and types of Element respects these constraints.
In other words I need to define the I get for these fields. Right?
How should I do?
I tried that, but the constraints are not respected:
abstract class Element {
def name: String
def name_= (value: String): Unit = {if (isBadValue(value)throw new IllegalArgumentException
name = value
}
var description : String,
var types : Set[Type] = Set }
class Component (override var name : String, var description: String) extends Element
The problem is that some fields that must respect the constraints,
in the constructor of the concrete classes must be initialized to a null value. So the "require" for me is not a good solution.
Thank you all.
Checking at initialization does not work for you because you want stateful objects, which can be avoided by using case classes. Instead of mutating the state of an object, you may want to create new objects using copy(field=value), which is automatically generated for case classes.
If you still want to go with stateful objects, I guess you want something like
abstract class Element {
private var _name: String = null
def name_= (value: String) {
require(!isBadValue(value),"Bad Value")
_name = value
}
def name = _name
def isBadValue(value: String): Boolean
}
class Component (initialName : String) extends Element {
name = initialName
def isBadValue(value: String) = value=="name"
}
val c = new Component(null) // works
c.name = "name" // exception
Another thing to point out: the setter generated by override var name in your code overrides your name_= method, which you may already know.
Abstract val:
trait Init {
val name: String
require(!isBadName(name))
def isBadName(name: String) = true
}
Creation:
new { val name = "init" } with Init

Scala: Copying case classes with trait

I'm fairly new to Scala and I have a question about the best way to copy a case class while preserving data that comes from traits. For example, let's say I have the following:
trait Auditing {
var createTime: Timestamp = new Timestamp(System.currentTimeMillis)
}
case class User(val userName: String, val email: String) extends Auditing
val user = User("Joe", "joe#blah.com")
Then I want to make a new copy with one parameter changed:
val user2 = user.copy(email = "joe#newemail.com")
Now, in the example above, the property createTime does not get copied over because it is not defined in the constructor of the User case class. So my question is: assuming that moving createTime into the constructor is not an option, what is the best way for getting a copy of the User object that includes the value from the trait?
I'm using Scala 2.9.1
Thanks in advance!
Joe
You can override the copy method with that behavior.
case class User(val userName: String, val email: String) extends Auditing
{
def copy(userName = this.userName, email = this.email) {
val copiedUser = User(userName, email)
copiedUser.createTime = createTime
copiedUser
}
}
While I see no other solution than Reuben's, I don't understand the requirement to leave the constructor args untouched. This would be the most natural solution:
case class User(userName: String, email: String,
override val createTime:Timestamp = new Timestamp(System.currentTimeMillis))
extends Auditing
If you don't want the user to be able to overwrite createTime, you can still use:
case class User private (userName: String, email: String,
override val createTime:Timestamp) extends Auditing {
def this(userName: String, email: String) =
this(userName, email, new Timestamp(System.currentTimeMillis))
}
The only drawback is that you need to write new User("Joe", "joe#blah.com"), as the primary constructor is now private.
You might be better of not using a case class. You can easily implement the
functionality you need yourself. The below code implements the copy method you wanted, a constructor without new, hides the original constructor, and creates an extractor so that you can use User in case statements.
class User private(val userName: String,
val email: String,
val timeStamp: Timestamp =
new Timestamp(System.currentTimeMillis)) {
def copy(uName: String = userName,
eMail: String = email) =
new User(uName, eMail, timeStamp)
}
object User {
def apply(userName: String, email: String) =
new User(userName, email)
def unapply(u: User) = Some((u.userName, u.email, u.timeStamp))
}