How to create an instances for typeclass with dependent type using shapeless

Question

I\'m trying to derive a tuple instance for a type class with dependent type. I\'m using shapeless to create summon the type class for the tuple elements. I\'m having trouble mat

Answer 1

There are several mistakes in your code.

Firstly, if you return (k1, k2) then k1, k2 should be the[Identifiable[K1]].identify(id._1), the[Identifiable[K2]].identify(id._2) correspondingly and not vice versa as you defined them. (Typo is fixed.)

Secondly, you forgot type refinement. You declare return type of identifiableTuple to be Identifiable[(K1,K2)] instead of correct Identifiable[(K1,K2)] { type K = (a.K, b.K)} (aka Identifiable.Aux[(K1,K2), (a.K, b.K)]). If you keep Identifiable[(K1,K2)] you actually upcast right hand side

new Identifiable[(K1,K2)]{
  ...
  type K = (a.K, b.K)   
  ...
}

and information that for this implicit instance type K = (a.K, b.K) will be lost.

Since you have to restore type refinement you can't write identifiableTuple with context bounds, you have to write it with implicit block

implicit def identifiableTuple[K1, K2](implicit
  a: Identifiable[K1],
  b: Identifiable[K2]
): Identifiable[(K1, K2)] {type K = (a.K, b.K)} = new Identifiable[(K1, K2)] {
  type K = (a.K, b.K)
  override def identify(id: (K1, K2)): K = {
    val k1 = a.identify(id._1)
    val k2 = b.identify(id._2)
    (k1, k2)
  }
}

You can test your code at compile time

implicit val int: Identifiable[Int] { type K = Double } = null
implicit val str: Identifiable[String] { type K = Char } = null
implicitly[Identifiable[(Int, String)] { type K = (Double, Char)}]

You can rewrite this with Aux pattern type Aux[M, K0] = Identifiable[M] { type K = K0 }

implicit def identifiableTuple[K1, K2](implicit
  a: Identifiable[K1],
  b: Identifiable[K2]
): Identifiable.Aux[(K1, K2), (a.K, b.K)] = new Identifiable[(K1, K2)] {
  type K = (a.K, b.K)
  override def identify(id: (K1, K2)): K = {
    val k1 = a.identify(id._1)
    val k2 = b.identify(id._2)
    (k1, k2)
  }
} // (*)

and

implicit val int: Identifiable.Aux[Int, Double] = null
implicit val str: Identifiable.Aux[String, Char] = null
implicitly[Identifiable.Aux[(Int, String), (Double, Char)]]

This is similar to @MateuszKubuszok's answer

implicit def identifiableTuple[M1, M2, K1, K2](implicit
  a: Identifiable.Aux[M1, K1],
  b: Identifiable.Aux[M2, K2]
): Identifiable.Aux[(M1, M2), (K1, K2)] = new Identifiable[(M1, M2)] {
  type K = (K1, K2)
  override def identify(id: (M1, M2)): K = {
    val k1 = a.identify(id._1)
    val k2 = b.identify(id._2)
    (k1, k2)
  }
} // (**)

although the latter needs extra inferrence of two type parameters.

And thirdly, you can't write (*) with implicitly or even the inside like

implicit def identifiableTuple[K1, K2](implicit
  a: Identifiable[K1],
  b: Identifiable[K2]
): Identifiable.Aux[(K1, K2), (a.K, b.K)] = new Identifiable[(K1, K2)] {
  type K = (a.K, b.K)
  override def identify(id: (K1, K2)): K = {
    val k1 = the[Identifiable[K1]].identify(id._1)
    val k2 = the[Identifiable[K2]].identify(id._2)
    (k1, k2)
  }
}

The thing is that path-dependent types are defined in Scala so that even when a == a1, b == b1 types a.K and a1.K, b.K and b1.K are different (a1, b1 are the[Identifiable[K1]], the[Identifiable[K2]]). So you return (k1, k2) of wrong type (a1.K,b1.K).

But if you write it in (**) style

implicit def identifiableTuple[M1, M2, K1, K2](implicit
  a: Identifiable.Aux[M1, K1],
  b: Identifiable.Aux[M2, K2]
): Identifiable.Aux[(M1, M2), (K1, K2)] = new Identifiable[(M1, M2)] {
  type K = (K1, K2)
  override def identify(id: (M1, M2)): K = {
    val k1 = the[Identifiable[M1]].identify(id._1)
    val k2 = the[Identifiable[M2]].identify(id._2)
    (k1, k2)
  }
}

then it will be ok (with the but not with implicitly) because compiler infers that (k1,k2) has type (K1,K2).