Generics “上的Scala类型推断失败?”;?扩展;在Java代码中
我有以下简单的Java代码:Generics “上的Scala类型推断失败?”;?扩展;在Java代码中,generics,scala,type-inference,scala-2.7,Generics,Scala,Type Inference,Scala 2.7,我有以下简单的Java代码: 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<O
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;
}
}
我得到以下错误:
错误:类型不匹配找到:lang.this.class[scala.this.Predef.String]
required:lang.this.class[?0]对于某些{type?0来说,该推理适用于Scala 2.8.0.Beta1 对于早期版本,如果您更改了以下内容,则它将起作用:
public <S extends T> Query<S> refine(Class<? extends S> clazz)
被改进的
def refine[S>:_root\uuu.scala.Nothing:scala.Nothing:Nothing:scala.Nothing是的-我意识到更改是有效的,但在我看来这是不必要的限制。例如,我应该说refine[Collection](classOf[LinkedList])
或类似的话
val sq = Query.newQuery().refine(classOf[String])
val sq = Query.newQuery().refine[String](classOf[String])
public <S extends T> Query<S> refine(Class<? extends S> clazz)
public <S extends T> Query<S> refine(Class<S> clazz)
def refine[S >: _root_.scala.Nothing <: T](clazz: Class[_$1] forSome {
type _$1 >: Nothing <: S
}): Query[S] = _;
def refine[S >: _root_.scala.Nothing <: T](clazz: Class[S]): Query[S] = _;
def refine[S >: scala.Nothing <: T](clazz: Class[_$1] forSome {
type _$1 >: Nothing <: S
} = _): Query[S] = _;
def refine[S >: scala.Nothing <: T](clazz: Class[S] = _): Query[S] = _;