Re: Sammy + wildcards + type inference = :(

On Fri, 2015-06-26 at 18:05 -0700, Roman Janusz wrote:
> Is there any chance this will work?
> 
> 
> $ scala -Xexperimental
> Welcome to Scala version 2.11.6 (Java HotSpot(TM) 64-Bit Server VM,
> Java 1.8.0_40).
> Type in expressions to have them evaluated.
> Type :help for more information.
> 
> scala> new java.util.ArrayList[String]().stream.map(_.toInt)
> <console>:8: error: no type parameters for method map: (x$1:
> java.util.function.Function[_ >: String, _ <:
> R])java.util.stream.Stream[R] exist so that it can be applied to
> arguments (java.util.function.Function[String,Int] with Serializable)
>  --- because ---
> argument expression's type is not compatible with formal parameter
> type;
>  found   : java.util.function.Function[String,Int] with Serializable
>  required: java.util.function.Function[_ >: String, _ <: ?R]


Not fatal, I would say.


Welcome to Scala version 2.11.7 (OpenJDK 64-Bit Server VM, Java 1.8.0_45).
Type in expressions to have them evaluated.
Type :help for more information.

scala> import java.util.ArrayList, java.util.stream.{Stream => JStream}
import java.util.ArrayList
import java.util.stream.{Stream=>JStream}

scala> new java.util.ArrayList[String]().stream.map[Int](_.toInt)
res1: java.util.stream.Stream[Int] = java.util.stream.ReferencePipeline$3@5b49c202

scala> new java.util.ArrayList[String]().stream.map(_.toInt): JStream[Int]
res2: java.util.stream.Stream[Int] = java.util.stream.ReferencePipeline$3@47bded3c


Not ideal, but maybe something else is going on here.

-- 
Stephen Compall
"^aCollection allSatisfy: [:each|aCondition]": less is better

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