Rogach - 6 months ago 31

Scala Question

I'm trying to print peano numbers, something like this:

`sealed trait Nat`

trait _0 extends Nat

trait Succ[N <: Nat] extends Nat

type _1 = Succ[_0]

type _2 = Succ[_1]

class RepNat[T <: Nat](val value: Int)

def rep[T <: Nat](implicit r: RepNat[T]) = r.value

implicit val repZero = new RepNat[_0](0)

implicit def repSucc[A <: Succ[B], B <: Nat](implicit r: RepNat[B]): RepNat[A] = new RepNat[A](r.value + 1)

println(rep[_0])

println(rep[_1])

// does not work, implicits do not resolve recursively:

// implicitly[RepNat[_2]]

// println(rep[_2])

// but explicit instantiation works:

println(rep[_2](repSucc(implicitly[RepNat[_1]])))

Answer

Recursive implicits do work. The following definition works fine for `_2`

:

```
implicit def repSucc[A <: Nat, B <: Nat](implicit
ev: A <:< Succ[B],
r: RepNat[B]
): RepNat[A] =
new RepNat[A](r.value + 1)
```

The reason, I believe, is that
`repSucc`

gets a single actual type argument `A`

, and needs to calculate `B`

from that. With your definition it tries to assign `A`

and `B`

at the same time, and thus `B`

gets assigned effectively to `Nothing`

.

This a common problem with type inference in Scala, and the usual solution is to move the type bound `A <: M[B]`

to a generalised type constraint `A <:< M[B]`

.

Also, note that the order of implicit parameters matters: first the compiler calculates `B`

from `A`

with `ev: A <:< Succ[B]`

, and then finds the `RepNat`

implementation for `B`

, possibly recursively.