If you run this in scala with the -Xlog-implicits
argument passed, you get more information:
scala.this.Prefed.conforms is not a valid implicit value for (T) => Ordered[T] because:
type mismatch:
found : <:<[T,T]
required: (T) => Ordered[T]
scala.this.predef.conforms is not a valid implicit value for (Ordered[T]) => Ordered[Ordered[T]] because:
type mismatch:
found : <:<[Ordered[T], Ordered[T]]
required : (Ordered[T]) => Ordered[Ordered[T]]
math.this.Ordering.ordered is not a valid implicit value for Ordering[T] because:
type arguments [T] do not conform to method ordered's type parameter bounds [A <: scala.math.Ordered[A]]
This is mostly speculation, but would seem to make some sense. I will try to investigate further:
This seems to suggest that there are three implicits that are being considered here. Ultimately, the signature of sorted
requires it to find something of type Ordering[T]
. So it's trying to construct your implicit function ordering
. Firstly, it's trying to fill in conv
by finding an implicit of type (T) => Ordered[T]
, where it's searching in Predef - which seems like barking up the wrong tree. It's then trying to find an implicit for (Ordered[T]) => Ordered[Ordered[T]]
in the same place, since by
takes an implicit parameter of type Ordering[S]
, where S
is Ordered[T]
by virtue of conv
. So it can't construct ordering
.
It then tries to use ordering
in math.Ordering, but this also doesn't fit. However, I think this is what's giving the somewhat confusing 'diverging implicits' message. The problem isn't that they're diverging, it's that there isn't a suitable one in scope, but it's being confused by the fact that there are two paths to go down. If one tries to define def foo[T <% Ordered[T]](s : Seq[T]) = s.sorted
without the implicit ordered function, then it fails with just a nice message saying that it can't find a suitable implicit.
val seq: Seq[Int] = in.readData()
but by helping the type inferencer by explicitly stating the type parameters of the method, then it worked:val seq = in.readData[Seq[Int]]()
. However, stating type parameters ofsorted
does not seem to help here for the OT.