Scala: referencing companion object from a child class - class

I'm thinking of a following Scala class layout. I have a basic trait that represents an Item - an interface of what ought to be an immutable object that we can query for name, weight and do some object-specific stuff by invoking methods like equip:
trait Item {
def name: String
def weight: Int
def equip // ... more abstract methods
}
I can create Item implementations, creating case classes by hand, but I'd like to have some sort of "dictionary"-based items - i.e. a static map holds mapping from type ID to values, name and weight methods just query the dictionary with a stored type ID:
object Weapon {
final val NameDict = Map(1 -> "short sword", 2 -> "long sword")
final val WeightDict = Map(1 -> 15, 2 -> 30)
}
case class Weapon(typ: Int) extends Item {
import Weapon._
def name = NameDict(typ)
def weight = WeightDict(typ)
def equip = ... // implementation for weapon
}
So far, so good. I can use Weapon(1) to reference an item object for "short sword". However, the same basic principle applies to any other item types, such as Armor, which uses exactly the same name and weight implementation, but completely different equip and other abstract methods implementation, for example:
object Armor {
final val NameDict = Map(3 -> "bronze armor", 4 -> "iron armor")
final val WeightDict = Map(3 -> 100, 4 -> 200)
}
case class Armor(typ: Int) extends Item {
import Armor._
def name = NameDict(typ)
def weight = WeightDict(typ)
def equip = ... // implementation for armor
}
Looks pretty similar to Weapon, isn't it? I'd like to factor out the common pattern (i.e. implementation of lookups in companion object dictionaries and common typ value) in something like that:
abstract class DictionaryItem(typ: Int) extends Item {
def name = ???.NameDict(typ)
def weight = ???.WeightDict(typ)
}
object Weapon {
final val NameDict = Map(1 -> "sword", 2 -> "bow")
final val WeightDict = Map(1 -> 15, 2 -> 30)
}
case class Weapon(typ: Int) extends DictionaryItem(typ) {
def equip = ... // implementation for weapon
}
But how do I reference a children's companion object (like Weapon.NameDict) from parent class - i.e. what should I use instead of ???.NameDict(typ) and how do I explain to the compiler that children's companion object must include these dictionaries? Is there a better, more Scala-esque approach to such a problem?

Instead of looking up the names and weights of items in maps, you could make use of instances of your case classes. In fact there are some good reasons you might want to take this approach:
Avoid Complexity: It's simple enough, you might not have needed to post this
question about how inheritance works with companion classes.
Avoid Repetition: Using case classes, you would only have to
define your items once.
Avoid Errors: Using synthetic keys, would require you to
use the correct key in both the name and weight dictionaries. See
item 2.
Write Idiomatic Code: Scala is both object-oriented and functional. In
this case, orient yourself around the Item.
Avoid Overhead: Get all the item's properties with a single lookup.
Here's a different take in which Item plays the starring role:
package object items {
val weapons = Weapon("halbred", 25) :: Weapon("pike", 35) :: Nil
val armor = Armor("plate", 65) :: Armor("chainmail", 40) :: Nil
val dictionary = (weapons ::: armor) map(item => item.name -> item) toMap
}
Notice that the names and weights are packaged in the instances instead of packaged as values in a map. And there are no indexes to manage. However, if you needed an index for some reason you could easily use the item's position in the list of equipment or add a property to the Item trait.
And, you might use it like this:
> items dictionary("halbred") // Weapon(halbred, 25)
> items dictionary("plate") weight // 65
One final note, the final keyword in the Scala word is kind of anachronistic. When you define a val in Scala, it's already immutable: Why are `private val` and `private final val` different?

som-snytt is right - just make the Armor and Weapon objects extend a trait, say, Dictionary which has NameDict and WeightDict members, then have DictionaryItem(typ: Dictionary).
It seems like a pretty convoluted way of doing things though, and it's easy to make mistakes and hard to refactor if you make all your properties lookups by number. The more natural way would be to put the actual objects in the map, something like:
case class Weapon(name: String, weight: Int, equip: ...) extends Item
case class Armor(name: String, weight: Int, equip: ...) extends Item
val ShortSword = Weapon(name = "short sword",
weight = 15,
equip = ...)
val LongSword = Weapon("long sword", 20, ...)
val Weapons = Map(
1 -> ShortSword,
2 -> LongSword
)
so you'd write for example Weapons(1).name rather than Weapon.NameDict(1). Or, you could just write ShortSword.name. Or, if you only need to refer to them via their index, you could define the items directly in the Map.

I think you want object Weapon extends Dictionary. I once called it an Armory, to mean a factory for arms. But any -ory or -ary will do. I would also change the integral typ to an Enumeration. Then define Weapon.apply(kind: Kind) to return a Sword when you Weapon(Kinds.Sword).
This pattern of factory methods on traits to be implemented by companion objects is used by the collections framework, which is educational.

Related

HList/KList from class values

I want to be able to create a class/trait that behaves somewhat like an enumeration (HEnum in the first snippet below). I can't use a plain enumeration because each enum value could have a different type (though the container class will be the same): Key[A]. I'd like to be able to construct the enum roughly like this:
class Key[A](val name: String)
object A extends HEnum {
val a = new Key[String]("a")
val b = new Key[Int]("b")
val c = new Key[Float]("c")
}
And then I'd like to be able to perform more or less basic HList operations like:
A.find[String] // returns the first element with type Key[String]
A.find("b") // returns the first element with name "b", type should (hopefully) be Key[Int]
So far I've been playing with an HList as the underlying data structure, but constructing one with the proper type has proven difficult. My most successful attempt looks like this:
class Key[A](val name: String)
object Key {
def apply[A, L <: HList](name: String, l: L): (Key[A], Key[A] :: L) = {
val key = new Key[A](name)
(key, key :: l)
}
}
object A {
val (a, akeys) = Key[String, HNil]("a", HNil)
val (b, bkeys) = Key[Int, Key[String] :: HList]("b", akeys)
val (c, ckeys) = Key[Float, Key[Int] :: HList]("c", bkeys)
val values = ckeys // use this for lookups, etc
def find[A]: Key[A] = values.select[A]
def find[A](name: String): Key[A] = ...
}
The problem here is that the interface is clunky. Adding a new value anywhere besides the end of the list of values is error prone and no matter what, you have to manually update values any time a new value is introduced. My solution without HList involved a List[Key[_]] and error prone/unsafe casting to the proper type when needed.
EDIT
I should also mention that the enum example found here is not particularly helpful to me (although, if that can be adapted, then great). The added compiler checks for exhaustive pattern matches are nice (and I would ultimately want that) but this enum still only allows a homogeneous collection of enum values.

Easy idiomatic way to define Ordering for a simple case class

I have a list of simple scala case class instances and I want to print them in predictable, lexicographical order using list.sorted, but receive "No implicit Ordering defined for ...".
Is there exist an implicit that provides lexicographical ordering for case classes?
Is there simple idiomatic way to mix-in lexicographical ordering into case class?
scala> case class A(tag:String, load:Int)
scala> val l = List(A("words",50),A("article",2),A("lines",7))
scala> l.sorted.foreach(println)
<console>:11: error: No implicit Ordering defined for A.
l.sorted.foreach(println)
^
I am not happy with a 'hack':
scala> l.map(_.toString).sorted.foreach(println)
A(article,2)
A(lines,7)
A(words,50)
My personal favorite method is to make use of the provided implicit ordering for Tuples, as it is clear, concise, and correct:
case class A(tag: String, load: Int) extends Ordered[A] {
// Required as of Scala 2.11 for reasons unknown - the companion to Ordered
// should already be in implicit scope
import scala.math.Ordered.orderingToOrdered
def compare(that: A): Int = (this.tag, this.load) compare (that.tag, that.load)
}
This works because the companion to Ordered defines an implicit conversion from Ordering[T] to Ordered[T] which is in scope for any class implementing Ordered. The existence of implicit Orderings for Tuples enables a conversion from TupleN[...] to Ordered[TupleN[...]] provided an implicit Ordering[TN] exists for all elements T1, ..., TN of the tuple, which should always be the case because it makes no sense to sort on a data type with no Ordering.
The implicit ordering for Tuples is your go-to for any sorting scenario involving a composite sort key:
as.sortBy(a => (a.tag, a.load))
As this answer has proven popular I would like to expand on it, noting that a solution resembling the following could under some circumstances be considered enterprise-grade™:
case class Employee(id: Int, firstName: String, lastName: String)
object Employee {
// Note that because `Ordering[A]` is not contravariant, the declaration
// must be type-parametrized in the event that you want the implicit
// ordering to apply to subclasses of `Employee`.
implicit def orderingByName[A <: Employee]: Ordering[A] =
Ordering.by(e => (e.lastName, e.firstName))
val orderingById: Ordering[Employee] = Ordering.by(e => e.id)
}
Given es: SeqLike[Employee], es.sorted() will sort by name, and es.sorted(Employee.orderingById) will sort by id. This has a few benefits:
The sorts are defined in a single location as visible code artifacts. This is useful if you have complex sorts on many fields.
Most sorting functionality implemented in the scala library operates using instances of Ordering, so providing an ordering directly eliminates an implicit conversion in most cases.
object A {
implicit val ord = Ordering.by(unapply)
}
This has the benefit that it is updated automatically whenever A changes. But, A's fields need to be placed in the order by which the ordering will use them.
To summarize, there are three ways to do this:
For one-off sorting use .sortBy method, as #Shadowlands have showed
For reusing of sorting extend case class with Ordered trait, as #Keith said.
Define a custom ordering. The benefit of this solution is that you can reuse orderings and have multiple ways to sort instances of the same class:
case class A(tag:String, load:Int)
object A {
val lexicographicalOrdering = Ordering.by { foo: A =>
foo.tag
}
val loadOrdering = Ordering.by { foo: A =>
foo.load
}
}
implicit val ord = A.lexicographicalOrdering
val l = List(A("words",1), A("article",2), A("lines",3)).sorted
// List(A(article,2), A(lines,3), A(words,1))
// now in some other scope
implicit val ord = A.loadOrdering
val l = List(A("words",1), A("article",2), A("lines",3)).sorted
// List(A(words,1), A(article,2), A(lines,3))
Answering your question Is there any standard function included into the Scala that can do magic like List((2,1),(1,2)).sorted
There is a set of predefined orderings, e.g. for String, tuples up to 9 arity and so on.
No such thing exists for case classes, since it is not easy thing to roll off, given that field names are not known a-priori (at least without macros magic) and you can't access case class fields in a way other than by name/using product iterator.
The unapply method of the companion object provides a conversion from your case class to an Option[Tuple], where the Tuple is the tuple corresponding to the first argument list of the case class. In other words:
case class Person(name : String, age : Int, email : String)
def sortPeople(people : List[Person]) =
people.sortBy(Person.unapply)
The sortBy method would be one typical way of doing this, eg (sort on tag field):
scala> l.sortBy(_.tag)foreach(println)
A(article,2)
A(lines,7)
A(words,50)
Since you used a case class you could extend with Ordered like such:
case class A(tag:String, load:Int) extends Ordered[A] {
def compare( a:A ) = tag.compareTo(a.tag)
}
val ls = List( A("words",50), A("article",2), A("lines",7) )
ls.sorted
My personal favorite method is using the SAM(Single abstraction method) with 2.12 as mentioned over the below example:
case class Team(city:String, mascot:String)
//Create two choices to sort by, city and mascot
object MyPredef3 {
// Below used in 2.11
implicit val teamsSortedByCity: Ordering[Team] = new Ordering[Team] {
override def compare(x: Team, y: Team) = x.city compare y.city
}
implicit val teamsSortedByMascot: Ordering[Team] = new Ordering[Team] {
override def compare(x: Team, y: Team) = x.mascot compare y.mascot
}
/*
Below used in 2.12
implicit val teamsSortedByCity: Ordering[Team] =
(x: Team, y: Team) => x.city compare y.city
implicit val teamsSortedByMascot: Ordering[Team] =
(x: Team, y: Team) => x.mascot compare y.mascot
*/
}
object _6OrderingAList extends App {
//Create some sports teams
val teams = List(Team("Cincinnati", "Bengals"),
Team("Madrid", "Real Madrid"),
Team("Las Vegas", "Golden Knights"),
Team("Houston", "Astros"),
Team("Cleveland", "Cavaliers"),
Team("Arizona", "Diamondbacks"))
//import the implicit rule we want, in this case city
import MyPredef3.teamsSortedByCity
//min finds the minimum, since we are sorting
//by city, Arizona wins.
println(teams.min.city)
}

Scala "update" immutable object best practices

With a mutable object I can write something like
var user = DAO.getUser(id)
user.name = "John"
user.email ="john#doe.com"
// logic on user
If user is immutable then I need to clone\copy it on every change operation.
I know a few ways to perform this
case class copy
method (like changeName) that creates a new object with the new property
What is the best practice?
And one more question. Is there any existing technique to get "changes" relative to the original object(for example to generate update statement)?
Both ways you've mentioned belongs to functional and OO paradigms respectively. If you prefer functional decomposition with abstract data type, which, in Scala, is represented by case classes, then choose copy method. Using mutators is not a good practice in my option, cause that will pull you back to Java/C#/C++ way of life.
On the other hand making ADT case class like
case class Person(name: String, age: String)
is more consise then:
class Person(_name: String, _age: String) {
var name = _name
var age = _a
def changeName(newName: String): Unit = { name = newName }
// ... and so on
}
(not the best imperative code, can be shorter, but clear).
Of cause there is another way with mutators, just to return a new object on each call:
class Person(val name: String,
val age: String) {
def changeName(newName: String): Unit = new Person(newName, age)
// ... and so on
}
But still case class way is more consise.
And if you go futher, to concurrent/parallel programming, you'll see that functional consept with immutable value are much better, then tring to guess in what state your object currently are.
Update
Thanks to the senia, forgot to mention two things.
Lenses
At the most basic level, lenses are sort of getters and setters for immutable data and looks like this:
case class Lens[A,B](get: A => B, set: (A,B) => A) {
def apply(a: A) = get(a)
// ...
}
That is it. A lens is a an object that contains two functions: get and set. get takes an A and returns a B. set takes an A and B and returns a new A. It’s easy to see that the type B is a value contained in A. When we pass an instance to get we return that value. When we pass an A and a B to set we update the value B in A and return a new A reflecting the change. For convenience the get is aliased to apply. There is a good intro to Scalaz Lens case class
Records
This one, ofcause, comes from the shapeless library and called Records. An implementation of extensible records modelled as HLists of associations. Keys are encoded using singleton types and fully determine the types of their corresponding values (ex from github):
object author extends Field[String]
object title extends Field[String]
object price extends Field[Double]
object inPrint extends Field[Boolean]
val book =
(author -> "Benjamin Pierce") ::
(title -> "Types and Programming Languages") ::
(price -> 44.11) ::
HNil
// Read price field
val currentPrice = book.get(price) // Inferred type is Double
currentPrice == 44.11
// Update price field, relying on static type of currentPrice
val updated = book + (price -> (currentPrice+2.0))
// Add a new field
val extended = updated + (inPrint -> true)

What would be an appropriate collection type for storing few elements?

Of the different collection types supported in Scala (lists, maps, hashmaps, set etc) what would be an appropriate collection type for implementing something that can be done by C code below
typedef enum { GOOD BAD MAX_QUALITY } quality
struct student_data s_data[MAX_QUALITY];
The collection size is small... 2 or 3 elements, but having a collection helps to keep the code elegant, when performing similar operations on the data .
Thanks!
List or Seq of different case classes should do the trick. When I say different case classes, I really mean:
case class CaseClass1(arg1: String, arg2: Int, arg3: OtherCaseClass)
case class OtherCaseClass(arg1: String, arg2: String)
val foo: List[CaseClass1] = ...
Then instances of CaseClass1 are composed and stored in your list.
If your goal is to create an enum-like data structure which provides a fast index lookup, I would go for:
sealed trait Quality { val index: Int }
case class BAD() extends Quality { val index = 0 }
case class GOOD() extends Quality { val index = 1 }
case class MAX_QUALITY() extends Quality { val index = 2 }
This allows to use pattern matching on an arbitrary quality: Quality and the verbose syntax quality.index makes it explicit that quality is used as an Int index at that point.
Can you please elaborate further ?
If your intention is to create structure in which you associate to something of type quality a data structure of type student_data then you can use a Map
Depending on whether you need mutable access, or you just create the structure and then the access is read-only you would use mutable or immutable maps.
I suggest, in your case, to use immutable Maps. The Scala Collection Library provides specialized versions for small maps, up to 5 inserted key-value pairs, which provide better performance and memory footprint than the unspecialized types of map.
val x = Map(1 -> 2, 2 -> 3)
println(x.getClass)
>> class scala.collection.immutable.Map$Map2
You can also define an alias, to work more easily with your structures
type StudentStruct = Map[Quality, StudentData]
Usage:
val studentStruct = Map(Bad -> studentData)
val studentStruct2 = studentStruct + (Good -> studentData2)
Hope it helps.

Does Scala have record update syntax for making modified clones of immutable data structures?

In Mercury I can use:
A = B^some_field := SomeValue
to bind A to a copy of B, except that some_field is SomeValue instead of whatever it was in B. I believe the Haskell equivalent is something like:
a = b { some_field = some_value }
Does Scala have something like this for "modifying" immutable values. The alternative seems to be to have a constructor that directly sets every field in the instance, which isn't always ideal (if there are invarients the constructor should be maintaining). Plus it would be really clunky and much more fragile if I had to explicitly pass every other value in the instance I want to have a modified copy of.
I couldn't find anything about this by googling, or in a brief survey of the language reference manual or "Scala By Example" (which I have read start-to-finish, but haven't absorbed all of yet, so it may well be in there).
I can see that this feature could have some weird interactions with Java-style access protection and subclasses though...
If you define your class as a case class, a convenient copy method is generated, and calling it you can specify with named parameters new values for certain fields.
scala> case class Sample(str: String, int: Int)
defined class Sample
scala> val s = Sample("text", 42)
s: Sample = Sample(text,42)
scala> val s2 = s.copy(str = "newText")
s2: Sample = Sample(newText,42)
It even works with polymorphic case classes:
scala> case class Sample[T](t: T, int: Int)
defined class Sample
scala> val s = Sample("text", 42)
s: Sample[java.lang.String] = Sample(text,42)
scala> val s2 = s.copy(t = List(1,2,3), 42)
s2: Sample[List[Int]] = Sample(List(1, 2, 3),42)
Note that s2 has a different type than s.
You can use case classes for this, but you don't have to. Case classes are nothing magical - the modifier case just saves you a lot of typing.
The copy method is realized by the use of named and default parameters. The names are the same as the fields and the defaults are the current values of the fields. Here's an example:
class ClassWithCopy(val field1:String, val field2:Int) {
def copy(field1:String = this.field1, field2:Int = this.field2) = {
new ClassWithCopy(field1,field2);
}
}
You can use this just like the copy method on case classes. Named and default parameters are a very useful feature, and not only for copy methods.
If the object you're planning on modifying is a case class then you can use the autogenerated copy method:
scala> val user = User(2, "Sen")
user: User = User(2,Sen)
scala> val corrected = user.copy(name = "Sean")
corrected: User = User(2,Sean)