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问题:
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(Left(0), Right("a")).iterator.toArray xs: Array[Product with Serializable with Either[Int,java.lang.String]] = Array(Left(0), Right(a)) scala> f(xs) res39: Int = 0 scala> f(List(Left(0), Right("a")).iterator.toArray) :9: error: polymorphic expression cannot be instantiated to expected type; 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) ^
Update: Debilski suggests a better example (not 100% sure this is demonstrating the same underlying phenomenon):
Seq(0).toArray : Seq[Int] // compiles Seq(Some(0)).toArray : Seq[Option[Int]] // doesn't
回答1:
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.
回答2:
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]])Int scala> f(xs) res5: Int = 0
BTW, no need to get an iterator before calling toArray.
回答3:
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.
