List of inner classes/objects during compilation - scala

This is the simplified code of a class(object) I'm working on:
object T {
val default = A
val options = List(A,B,C)
sealed trait T
object A extends T {
override def toString = "A"
}
object B extends T {
override def toString = "B"
}
object C extends T {
override def toString = "C"
}
}
This hierarchy maps directly to GUI element which requires a options = List(A,B,C) to build.
Problem with current approach:
Every time I add an extra object I have to change the option manually. This isn't too much work however the code isn't too elegant this way.
My question is:
Can I generate a list of inner objects during compile time? I wouldn't like to do this during runtime, it would be an overkill.

To add to #Samar's comment, to make it clear. The following is what you need:
import scala.reflect.runtime.universe._
case object K {
val default = A
val options = getObjects[this.type]
sealed trait T
object A extends T {
override def toString = "A"
}
object B extends T {
override def toString = "B"
}
object C extends T {
override def toString = "C"
}
def getObjects[T: TypeTag] = typeOf[T].members.collect {
case m: ModuleSymbol if m.isModule => m
}.toList
}

Related

Syntax extension using type class methods in Scala

I want to bind a check method to the Test in such a way that the implementation does not contain an argument (look at the last line). It is necessary to use type classes here, but I'm new in Scala, so I have problems.
Object Checker is my attempts to solve the problem. Perhaps it is enough to make changes to it...
trait Test[+T] extends Iterable[T]
class TestIterable[+T](iterable: Iterable[T]) extends Test[T] {
override def iterator: Iterator[T] = iterable.iterator
}
object Test {
def apply[T](iterable: Iterable[T]): Test[T] = new TestIterable[T](iterable)
}
trait Check[M] {
def check(m: M): M
}
object Checker {
def apply[M](implicit instance: Check[M]): Check[M] = instance
implicit def checkSyntax[M: Check](m: M): CheckOps[M] = new CheckOps[M](m)
private implicit def test[T](m: Test[T]) : Check[Test[T]] = {
new Check[Test[T]] {
override def check(m: Test[T]) = m
}
}
final class CheckOps[M: Check](m: M) {
def x2: M = Checker[M].check(m)
}
}
import Checker._
val test123 = Test(Seq(1, 2, 3))
Test(test123).check

how to copy instance and override value field declared in trait

Suppose I have some abstract value field defined in a trait:
trait Base {
val toBeOverride: String
}
case class Impl(other:Int) extends Base {
override val toBeOverride = "some value"
}
How can I write a function that I can easily get a cloned instance only overriding the toBeOverride value, like this:
// copy only available to case class instance
// v does not have method 'copy'
def overrideBaseValue[T <: Base](v: Base) =
v.copy(toBeOverride = "prefix" + v.toBeOverride)
?
Edit
#som-snytt, I don't think this is a duplicate, just like a Trait is not the same as an Abstract Class. And the answers of that question do not satisfy me, see below.
#Blaisorblade, yes, it is a problem. For instances of each sub case class, the toBeOverride field are the same, so it should not appear in the constructor.
For now all the suggestions are to define an customized copy method in each(!) sub case class and that in my opinion is ugly and shows the incapability of the language.
The simplest solution is to just add the method you want to Base:
trait Base {
val toBeOverride: String
def copyBase(newToBeOverridden: String): Base
}
case class Impl(other:Int, override val toBeOverride: String = "some value") extends Base {
def copyBase(newToBeOverridden: String) = copy(toBeOverride = newToBeOverridden)
}
This also allows to directly create an instance of Impl while specifying the value of toBeOverride (which wasn't possible). The only disadvantage is that now pattern matches using Impl have to change syntax - please update your question and add a comment if that's a problem.
BTW, if you just want to add a prefix (as in your example), that's no problem:
case class Impl(other:Int, override val toBeOverride: String = "some value") extends Base {
def copyBase(newToBeOverridden: String) = copy(toBeOverride = toBeOverride + newToBeOverridden)
}
Here are two mechanisms.
Apparently, in the near future you'll be able to write a macro that can emit the anonymous subclass, but until then, I think this typeclass is not arduous.
Just kicking the tires on Dynamic here.
import scala.language.dynamics
import scala.reflect._
import scala.reflect.runtime.{ currentMirror => cm }
import scala.reflect.runtime.universe._
trait Base {
def m: String
}
case class Impl(p: Int) extends Base {
override val m = "some value"
}
trait Basic extends Dynamic {
protected def m: String
def selectDynamic(f: String): Any =
if ("m" == f) m else reflecting(this, f)
protected def reflecting(b: Basic, f: String) = {
val im = cm.reflect(b)
val member = im.symbol.typeSignature member newTermName(f)
require(member != NoSymbol, s"No such member $f")
(im reflectMethod member.asMethod)()
}
}
case class Implic(p: Int) extends Basic {
override protected val m = "some value"
}
object Test extends App {
implicit class Copy[A <: Base](val b: A) {
def overriding(overm: String): A = (b match {
case impl: Impl => new Impl(impl.p) { override val m = overm }
case b: Base => new Base { override val m = overm }
}).asInstanceOf[A]
}
implicit class Proxy[A <: Basic : ClassTag](val b: A) {
def proximately(overm: String): Basic = new Basic {
override val m = overm
override def selectDynamic(f: String): Any =
if ("m" == f) overm else reflecting(b, f)
override def toString = b.toString
}
}
// asked for this
//def overriding[T <: Base](v: Base) = v.copy(m = "prefix" + v.m)
/* want something like this
def overriding[T <: Base](v: Base) = new Impl(v.p) {
override val m = "some value"
} */
val a = Impl(5)
val b = a overriding "bee good"
Console println s"$a with ${a.m} ~> $b with ${b.m}"
// or
val c = Implic(7)
val d = c proximately "dynomite"
Console println s"$c with ${c.m} ~> $d with ${d.m}"
}
Since traits don't get copy methods automatically, you can try using a Base case class instead:
case class Base(toBeOverride: String)
case class Impl(other: Int, someVal: String = "some value") extends Base(someVal)
def overrideBaseValue[T <: Base](v: Base) =
v.copy(toBeOverride = "prefix" + v.toBeOverride)
The problem that you're going to run into though, is that copy returns an instance of Base and I don't think that you can convert it back to your original Impl class. For instance, this won't compile:
def overrideBaseValue[T <: Base](v: T): T =
v.copy(toBeOverride = "prefix" + v.toBeOverride)

How to avoid awful type casts working with path dependent types?

I am new to Scala and dont know why i have to do an (unintuitive for me) type cast related to path dependent types in the following code.
(I don't like getters, setters nor nulls, they are here to separate operations and disambiguate the source of errors)
// Module A public API
class ModA {
trait A
}
// Module B public API that depends on types defined in Module A
class ModB(val modA: ModA) {
trait B {
def getA: modA.A;
def setA(anA: modA.A);
}
}
// One implementation of Module A
class ModAImpl extends ModA {
class AImpl extends A
}
// One implementation of Module B
class ModBImpl(mod: ModA) extends ModB(mod) {
class BImpl extends B {
private[this] var privA: modA.A = _;
override def getA = privA;
override def setA(anA: modA.A) = privA = anA;
}
}
object Main {
def main(args: Array[String]): Unit = {
// wiring the modules
val modAImpl = new ModAImpl;
val modBImpl = new ModBImpl(modAImpl);
// wiring objects
val a = new modAImpl.AImpl;
val b = new modBImpl.BImpl;
b.setA(a); //don't compile and complain: type mismatch; found: modAImpl.A required: modBImpl.modA.A
//i have to do this horrible and coutnerintuitive cast to workaround it
b.setA(a.asInstanceOf[modBImpl.modA.A]);
var someA: modAImpl.A = null;
someA = b.getA; // don't compile with same reason
someA = b.getA.asInstanceOf[modAImpl.A]; // horrible cast to workaround
println(a == b.getA); // however this prints true
println(a eq b.getA); // this prints true too
}
}
I have read about singleton types to inform the compiler when two types are the same, but I don't know how to apply this here.
Thanks in advance.
You can stick a type parameter on the ModB types:
class ModA { trait A }
class ModB[AA](val modA: ModA { type A = AA }) {
trait B {
def getA: AA
def setA(anA: AA)
}
}
class ModAImpl extends ModA { class AImpl extends A }
class ModBImpl[AA](
mod: ModA { type A = AA }) extends ModB(mod) {
class BImpl extends B {
private[this] var privA: AA = _
override def getA = privA
override def setA(anA: AA) = privA = anA
}
}
And the type inference all works out as desired:
scala> val modAImpl = new ModAImpl
modAImpl: ModAImpl = ModAImpl#7139edf6
scala> val modBImpl = new ModBImpl(modAImpl)
modBImpl: ModBImpl[modAImpl.A] = ModBImpl#1dd7b098
scala> val a = new modAImpl.AImpl
a: modAImpl.AImpl = ModAImpl$AImpl#4cbde97a
scala> val b = new modBImpl.BImpl
b: modBImpl.BImpl = ModBImpl$BImpl#63dfafd6
scala> b.setA(a)
Let's start by simplifying your code ridding out the unnecessary complexity.
class Aout {
class Ain
}
class Bout(val link: Aout) {
class Bin(val field: link.Ain)
}
object Main {
def main(args: Array[String]): Unit = {
// wiring outer object
val aout: Aout = new Aout;
val bout: Bout = new Bout(aout);
// wiring inner object
val ain: aout.Ain = new aout.Ain;
val bin: bout.Bin = new bout.Bin(ain); //don't compile and complain: type mismatch; found: aout.Ain required: bout.link.Ain
}
}
Answer:
The compiler complains with a type mismatch error because he compares the paths of the two declared types involved in the assignment, and they are different.
The compiler is not intelligent enough to notice that at that point aout eq bout.link. You have to tell him.
So, replacing the line
val ain: aout.Ain = new aout.Ain;
with
val ain: bout.link.Ain = new bout.link.Ain;
solves the path-dependent type mismatch.
Anyway, correcting the type's path in your original code is not enough because there is also an inheritance problem.
One solution to that is to make the class ModBImpl know the ModAImpl class like this:
class ModA {
trait A
}
class ModB[M <: ModA](val modA: M) { // type parameter added
trait B {
def getA: modA.A;
def setA(anA: modA.A);
}
}
class ModAImpl extends ModA {
class AImpl extends A
}
class ModBImpl(mod: ModAImpl) extends ModB(mod) { // changed type of `mod` parameter from `ModA` to `ModAImpl`
class BImpl extends B {
private[this] var privA: modA.A = _;
override def getA: modA.A = privA;
override def setA(anA: modA.A): Unit = privA = anA;
}
}
object Main {
def main(args: Array[String]): Unit = {
val modAImpl = new ModAImpl;
val modBImpl = new ModBImpl(modAImpl);
val a: modBImpl.modA.AImpl = new modBImpl.modA.AImpl; // changed the path of the type
val b: modBImpl.BImpl = new modBImpl.BImpl;
b.setA(a); // here IntellijIde complains with a type mismatch error, but the ScalaIDE (eclipse) and the scala compiler work fine.
}
}
If the rules of your business don't allow that the ModBImpl class has knowledge of the ModAImpl class, tell me so we can find another solution.

How to specify the return type of a function to be a (arbitrary) monad?

In short, I want to declare a trait like this:
trait Test {
def test(amount: Int): A[Int] // where A must be a Monad
}
so that I can use it without knowing what monad that A is, like:
class Usecase {
def someFun(t: Test) = for { i <- t.test(3) } yield i+1
}
more details...
essentially, I want to do something like this:
class MonadResultA extends SomeUnknownType {
// the base function
def test(s: String): Option[Int] = Some(3)
}
class MonadResultB(a: MonadResultA) extends SomeUnknownType {
// added a layer of Writer on top of base function
def test(s: String): WriterT[Option, String, Int] = WriterT.put(a.test(s))("the log")
}
class Process {
def work(x: SomeUnknownType) {
for {
i <- x.test("key")
} yield i+1
}
}
I wanted to be able to pass any instances of MonadResultA or MonadResultB without making any changes to the function work.
The missing piece is that SomeUnknowType, which I guess should have a test like below to make the work function compiles.
trait SomeUnknowType {
def test(s: String): T[Int] // where T must be some Monad
}
As I've said, I'm still learning this monad thing... if you find my code is not the right way to do it, you're more than welcomed to point it out~
thanks a lot~~
Assuming you have a type class called Monad you can just write
def test[A:Monad](amount: Int): A[Int]
The compiler will require that there is an implicit of type Monad[A] in scope when test is called.
EDIT:
I'm still not sure what you're looking for, but you could package up a monad value with its corresponding type class in a trait like this:
//trait that holds value and monad
trait ValueWithMonad[E] {
type A[+E]
type M <: Monad[A]
val v:A[E]
val m:M
}
object M {
//example implementation of test method
def test(amount:Int):ValueWithMonad[Int] = new ValueWithMonad[Int] {
type A[+E] = Option[E]
type M = Monad[Option]
override val v = Option(amount)
override val m = OptionMonad
}
//test can now be used like this
def t {
val vwm = test(1)
vwm.m.bind(vwm.v, (x:Int) => {
println(x)
vwm.m.ret(x)
})
}
}
trait Monad[A[_]] {
def bind[E,E2](m:A[E], f:E=>A[E2]):A[E2]
def ret[E](e:E):A[E]
}
object OptionMonad extends Monad[Option] {
override def bind[E,E2](m:Option[E], f:E=>Option[E2]) = m.flatMap(f)
override def ret[E](e:E) = Some(e)
}

Method polymorphism

I am trying to write a generic method f[T](id:String) that is something like this:
case class A(x:String)
case class B(y:String)
case class C(z:String)
def f[T](id:String): T = { /* equivalent to T(id) */ }
val result1:A = f[A]("123") // returns A("123")
val result2:B = f{B]("345") // returns B("345")
val result3:C = f[C]("567") // returns C("567")
Unfortunately I cannot figure out how to work with the type T inside the method, besides using reflection. By "working with the type T" i mean for example being able to do something like the following, which I know doesn't work (for illustration purposes only):
T match {
case A => A(id)
case B => B(id)
}
or simply invoke T(ID) to create a new object of whatever type T is.
I can of course break up this into three methods:
def f1(id:String): A = { A(id) }
def f2(id:String): B = { B(id) }
def f3(id:String): C = { C(id) }
val result1:A = f1("123") // returns A("123")
val result2:B = f2("345") // returns B("345")
val result3:C = f3("567") // returns C("567")
but I'm hoping there is a way to keep it as one generic method to avoid some ugly boilerplate code duplication, and still be nearl as fast as the tree method version.
If you do not want to use reflection (ClassTag or TypeTag), you could use a Factory type class to achieve the desired functionality (unless it defeats the purpose of your generic function by generating a lot of duplicated simple code ;)).
case class A(s: String)
case class B(s: String)
case class C(s: String)
trait Factory[T] extends ((String) => T) {
def apply(arg: String): T
}
object Factory {
implicit object AFactory extends Factory[A] {
override def apply(arg: String): A = A(arg)
}
implicit object BFactory extends Factory[B] {
override def apply(arg: String): B = B(arg)
}
implicit object CFactory extends Factory[C] {
override def apply(arg: String): C = C(arg)
}
}
def create[T : Factory](arg: String): T = implicitly[Factory[T]].apply(arg)
create[A]("foo") | -> res0: A = A(foo)