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Scala type inference failure on "? extends" in Java code

开发者 https://www.devze.com 2022-12-24 18:13 出处:网络
I have the following simple Java code: package testj; import java.util.*; public class Query<T> { private static List<Object> l = Arrays.<Object>asList(1, \"Hello\", 3.0);

I have the following simple Java code:

package testj;
import java.util.*;

public class Query<T> {

    private static List<Object> l = Arrays.<Object>asList(1, "Hello", 3.0);

    private final Class<? extends T> clazz;

    public static Query<Object> newQuery() { return new Query<Object>(Object.class); }

    public Query(Class<? extends T> clazz) { this.clazz = clazz; }

    public <S extends T> Query<S> refine(Class<? extends S> clazz) {
        return new Query<S>(clazz);
    }

    public List<T> run() {
        List<T> r = new LinkedList<T>();
        for (Object o : l) {
            if (clazz.isInstance(o)) r.add(clazz.cast(o));
        }
        return r;
    }
}

I can call this from Java as follows:

Query<String> sq = Query.newQuery().refine(String.class); //NOTE NO <String>

But if I tr开发者_开发问答y and do the same from Scala:

val sq = Query.newQuery().refine(classOf[String])

I get the following error:

error: type mismatch

found :lang.this.class[scala.this.Predef.String]

required: lang.this.class[?0] forSome{ type ?0 <: ? }

val sq = Query.newQuery().refine(classOf[String])

This is only fixed by the insertion of the correct type parameter!

val sq = Query.newQuery().refine[String](classOf[String])

Why can't scala infer this from my argument? Note I am using Scala 2.7


The inference works against Scala 2.8.0.Beta1.

For earlier versions, it works if you change:

public <S extends T> Query<S> refine(Class<? extends S> clazz)

to:

public <S extends T> Query<S> refine(Class<S> clazz)

scalac -print suggests that the interpretation of the Java signatures hasn't changed across Scala releases. So the difference is probably in the typer/inferencer itself.

Scala 2.7.5

Original

def refine[S >: _root_.scala.Nothing <: T](clazz: Class[_$1] forSome {
  type _$1 >: Nothing <: S
}): Query[S] = _;

Modified

def refine[S >: _root_.scala.Nothing <: T](clazz: Class[S]): Query[S] = _;

Scala 2.8.0.Beta1

Original

def refine[S >: scala.Nothing <: T](clazz: Class[_$1] forSome {
  type _$1 >: Nothing <: S
} = _): Query[S] = _;

Modified

def refine[S >: scala.Nothing <: T](clazz: Class[S] = _): Query[S] = _;
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