Why does the following occur in Scala 2.9.0.1?
scala> def f(xs: Seq[Either[Int,String]]) = 0
f: (xs: Seq[Either[Int,String]])Int
scala> val xs = List(
I believe this is because you cannot turn an Array into an Array, but you can convert a Sequence into an Array. The method wants a Sequence that it can use to make an Array.
Bottom line is to check the method signatures and not guess what they are based on the method name.
The important part is this:
found : [B >: Product with Serializable with Either[Int,java.lang.String]]Array[B]
required: Seq[Either[Int,String]]
f(List(Left(0), Right("a")).iterator.toArray)
The toArray
method wants a Seq (so a List would be fine) and it returns an Array. You have passed it an Array and it does not know what to do. Either make the Array into a Seq first, or just skip the toArray
method entirely.
If you go back one step it is clear that the iterator
method takes your List and returns an Array. Each method invocation is a function call.
The best person to explain this is Adriaan Moors, and he already did that here on Stack Overflow -- lookup answers from him and you'll find it.
Anyway, the problem is that the type of List(Left(0), Right("a")).iterator.toArray
cannot be inferred within the boundaries expected by f
. It does not conform to Seq[Either[Int, String]]
without an implicit conversion, and no implicit conversion can be applied because it (the type) cannot be determined. It's like an egg&chicken problem.
If you use <%
or assign it to a val, you break the cycle in the inference.
This has nothing to do with Either
but rather with Array
handling. If you convert it manually to a Seq
it works:
scala> f(xs.toSeq)
res4: Int = 0
A Scala Array
is not a Seq
(because it is in fact a Java array). But a WappedArray
is. You could also redefine your function f
has:
scala> def f[A <% Seq[Either[Int,String]]](xs: A) = 0
f: [A](xs: A)(implicit evidence$1: (A) => Seq[Either[Int,String]])Int
scala> f(xs)
res5: Int = 0
BTW, no need to get an iterator before calling toArray
.