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    <p>This is one of the reasons existential types (the Scala equivalent of wildcard types) are a Bad Thing (TM) that is best avoided when not doing Java interop: the compiler cannot (or just isn't smart enough to) reason normally about which types are equal to which, because they are all gone... </p> <p>To make the compiler understand that those types are the same, you need to give that type a name somehow. </p> <p>A type parameter is a possibility: you define a parametrized method with the content of your for comprehension, so that within the method, the type is well-known.</p> <pre><code>def callHandler(url: String) { def call[T](r: Route[T]) = { val args = r.matcher.matchUrl(url) if (args.isDefined) r.handler(args.get) // or args.foreach(r.handler) } for (r &lt;- routes) call(r) // or routes.foreach(call) } </code></pre> <p><em>Note</em>: in simpler cases, you could also use variance to have a <code>List[Route[Any]]</code>, and your problem goes away, the type is well-known again. Here I am not sure you could make <code>Route[A]</code> covariant.</p> <p><a href="https://stackoverflow.com/questions/1031042/scalas-existential-types">Existential types</a> mostly exist to represent Java wildcards, Java raw types and the JVM's view of types (reflection and stuff), even though <strong>they are more powerful than those three constructs</strong>. <strike>If you can design things to avoid using them, you will save yourself a lot of pain.</strike> This was a bit controversial, but a least, there are a lot of limitations with the way they interact with type inference, so one has to be careful.</p>
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    1. POHow to describe and fix this Scala type mismatch error?
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    1. USgourlaysama
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    1. COI understand the sentiment here, but it's just not true that Scala's existential types are _only_ for Java interoperability (Scala spends a whole keyword—and it doesn't have all that many—on `forSome`, for example). There are lots of completely legitimate (although sure, more advanced) uses that have nothing to do with Java.
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      1. USTravis Brown
    2. COYes, they are strictly more powerful that Java's wildcards and such (and much older), but I remember Martin Odersky saying somewhere that they wouldn't be in Scala if it were not for compatibility with erasure, wildcards and raw types.
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    3. COI don't understand existential types -- like the meaning of `[T <: Foo]` vs `[T forSome { type T <: Foo }]` -- but I updated the question with another subquestion. I'll probably use your suggestion here and define the helper method. Thanks.
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      1. USRob N
 

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