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Scalac does not check type equality for existential types #9899

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scabug opened this Issue Aug 23, 2016 · 5 comments

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scabug commented Aug 23, 2016

Welcome to Scala 2.11.8 (Java HotSpot(TM) 64-Bit Server VM, Java 1.8.0_77).
Type in expressions for evaluation. Or try :help.

scala> import language.existentials
import language.existentials

scala> val pair: (A => String, A) forSome { type A } = ( { a: Int => a.toString }, 19 )
pair: (A => String, A) forSome { type A } = (<function1>,19)

scala> pair._1(pair._2)
<console>:13: error: type mismatch;
 found   : pair._2.type (with underlying type A)
 required: A
       pair._1(pair._2)
                    ^
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scabug commented Aug 23, 2016

Imported From: https://issues.scala-lang.org/browse/SI-9899?orig=1
Reporter: @Atry
Affected Versions: 2.11.8

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scabug commented Aug 23, 2016

@SethTisue said:
I think this is working as designed as per SLS 6.1: "The following skolemization rule is applied universally for every expression: If the type of an expression would be an existential type T, then the type of the expression is assumed instead to be a skolemization of T." pair._1 and pair._2 get two different skolems which then don't match.

You can pattern match instead; this compiles:

pair match { case (f, x) => f(x) }

You might also be interested in http://stackoverflow.com/a/39061519/86485 which is similar.

@scabug scabug closed this Aug 23, 2016

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scabug commented Aug 23, 2016

@SethTisue said:
Closing as "Not a bug" rather than leaving it open as a feature request, because it's not clear exactly what the feature would be or how it would work.

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scabug commented Aug 23, 2016

@SethTisue said:
Note that this compiles as well:

def applyPair[A](p: (A => String, A)) = p._1(p._2)
applyPair(pair)

so pattern-matching isn't the only workaround.

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scabug commented Aug 23, 2016

@adriaanm said:
Seth & I pulled on this thread a bit, and it turns out it revealed a different bug in the type checker, tracked as scala/scala-dev#205

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