Basicaly this is what I am trying to do.
A trait that has a member and another trait that inherits that member and initialize it.
Then a class that takes a generic type that must be inherited from the first trait and accesses that member
object main extends App{
val z = new c[b]()
z.p()
}
trait a{
val x : String
}
trait b extends a {
val x = 1
}
class c [A <: a] {
self: A =>
def p(): Unit ={
print(x)
}
}
but in the line
val z = new c[b]()
c[b] is underlined and the compiler complains that
"Illegal inheritance, self-type c[b] does not conform to A"
I want to define multiple traits that inherit from a father trait and then define a class that takes one of the fathers child traits and accesses its values
I found the answer, I simply had to extend the object while creating it, like this:
object main extends App{
val z = new c[b] with b
z.p()
}
trait a{
val x : String
}
trait b extends a {
val x = "1"
}
class c[A <: a]{
this: A =>
def p(): Unit ={
print(x)
}
}
Related
I have the following trait and classes:
trait A{
def func1: String
def func2: Int
}
class F extends A{
def func1: String = ???
def func2: Int = ???
}
class G extends A{
def func1: String = ???
def func2: Int = ???
}
And I want to create the following trait and class:
trait X{
val a: ***
}
class Z extends X{
override val a = new G
}
What do I need to write on '***' in trait X so that I can assign val a with a class F or G ?
Thank you.
EDIT
Sorry, I forgot to mention that trait A is A[Any] and traits F and G can be 'F extends A[String]' and 'G extends A[Int]'. In other words, the classes can extends the trait A with different types of parameters.
Re. your edit, your code fails because trait A only takes type parameter of a fixed type. If you declare base class having a A[Any], you cannot override it with a A[Int] or A[String]. There are two ways to make your code works.
One is to make trait A type argument covariant, so that you can override with subclass of the type argument you declared in base class (note the + in trait A definition):
trait A[+T]
class F extends A[String]
class G extends A[Int]
trait X {
val a: A[Any]
}
class Z extends X {
val a = new G
}
See here: Sample
Another way to work is to declare a as a A[_] (_ means it takes any type as a parameter, which is different from Any):
trait A[T]
class F extends A[String]
class G extends A[Int]
trait X {
val a: A[_]
}
class Z extends X {
val a = new G
}
See here: Sample
I need to provide type class instances for a bunch of case classes all derived from a single trait, but as far as I understand Scala compiler expects an instance for a specific class and doesn't go up the inheritance hierarchy. So this code:
trait Base
sealed trait Child extends Base
case class Concrete() extends Child
trait Printable[A] {
def print(value: A): String
}
object WtfTrait {
def print[A](x: A)(implicit ev: Printable[A]) = {
println(ev.print(x))
}
implicit val printableBase = new Printable[Base] {
override def print(value: Base): String = value.toString
}
val x = Concrete()
print(x)
}
doesn't compile with an error reading could not find implicit value for parameter ev: Printable[Impl]. Is there a way to define a single type class instance for the base trait and avoid repitition maybe by using Shapeless or something.
Guess you mean Printable[Concrete] (that's to say a Show typeclass instance).
Need to update to printableBase definition as bellow:
trait Base
sealed trait Child extends Base
case class Concrete() extends Child
trait Printable[A] {
def print(value: A): String
}
object WtfTrait {
def print[A](x: A)(implicit ev: Printable[A]) = {
println(ev.print(x))
}
// HERE
implicit def printableBase[T <: Base] = new Printable[T] {
override def print(value: T): String = value.toString
}
val x = Concrete()
print(x)
}
Printable can be made contravariant by adding a - sign:
trait Printable[-A]
This makes Printable[X] a subtype of Printable[Y] if Y is a subtype of X. In particular, Printable[Base] is a subtype of Printable[Concrete] and can be used when the compiler looks for an implicit of that type.
Let's say I have some trait:
trait A[T] { def foo: T }
A class which extends it:
class B[T](t: T) extends A[T] { def foo = t }
And a subtrait of the parent trait:
trait C[T] extends A[T]
I want to mix C in with B.
val foo = new B("foo") with C[String]
This works fine, but I'd rather not need to specify the type parameter again, since B is already of type A[String]. However, I recognize Scala doesn't support the following:
val foo = new B("foo") with C
My question is where there's some other mechanism in the type system to support not having to specify the type parameters when C is mixed in. What I was thinking of was something as follows:
trait C {
self: A[T] => ...
}
One would think this kind of thing would fix what C could be mixed into. However, it isn't valid Scala. Something like:
trait C {
type T
self: A[T] =>
}
does not work either.
You can do this using an abstract type:
trait A {
type AT
def foo: AT
}
class B[T](t: T) extends A {
type AT = T
def foo = t
}
trait C extends A
val foo = new B("foo") with C
The definition is a bit more verbose, but your requirement of not having to type T again is satisfied.
How about this:
scala> trait C {
| self: A[_] =>
| }
defined trait C
scala> val foo = new B("foo") with C
foo: B[String] with C = $anon$1#f2df380
scala> foo.foo
res16: String = foo
I need overriding implicit in object res defing one from trait. The purpose is to define custom implicits in one place (trait B). Trait a is defined in external library. Is it possible?
trait t {
}
object m extends t
object q extends t
trait a {
implicit val o: t = m
}
trait b {
implicit val o: t = q
}
trait c {
def action(implicit v: t): Unit = {}
}
object res extends c with a with b {
//i need smth like override val o = super[b].o
val ololo= action
}
It is not possible to mix-in two unrelated traits that both contain a member with the same identifier. The Scala compiler has no way to resolve which one would take precedence in this scenario. Since types A and B are related, o does not even need to have the same type in both of them. Trait B needs to extend A and override o. There isn't any other way to override a member without using inheritance.
trait T
case object M extends T
case object Q extends T
trait A {
implicit val o: T = M
}
trait B extends A {
override implicit val o: T = Q
}
trait C {
def action(implicit v: T): Unit = println(v)
}
object Res extends C with B {
def call() = action
}
scala> Res.call()
Q
Since you are already mixing A into Res anyway, it is assumed that A is not difficult to extend, since Res must implement any other unmentioned abstract members.
I'm currently working on a A* implementation in Scala. To accomplish a clean structure I would like to use a nested case class structure which implements a self-bounded trait. However, I experience some issues when implementing this in the Scala IDE. The following code will not compile:
trait A[T <: A[T]]
class B {
case class C(int: Int) extends A[C] // A[this.C] won't work either
def make = C(5)
}
object D {
def foo[T <: A[T]](some: T) = {}
val c = new B().make
foo(c) // this does not compile
}
Is there any way I can make this structure work?
Not sure why you want this, but here's why it won't work as is:
The type of D.c is B#C. It is a path-dependent type where we don't know what instance of B it belongs to. However, C extends A[C], which is the already the same as saying A[this.C] in that context, which is bound to a specific instance of B. foo sees the type parameter T as B#C, which is not the same as b.C for some b.
You have two options to make this compile.
Relax the constraints of A to B#C:
trait A[T <: A[T]]
class B {
case class C(int: Int) extends A[B#C]
def make = C(5)
}
object D {
def foo[T <: A[T]](some: A[T]) = {}
val c = new B().make
foo(c)
}
Or handle the path-dependent type, so that c has type b.C:
trait A[T <: A[T]]
class B {
case class C(int: Int) extends A[C]
def make = C(5)
}
object D {
def foo[T <: A[T]](some: A[T]) = {}
val b = new B
val c: b.C = b.make
foo(c)
}