Avi*_*ash 6 java generics scala scala-java-interop
我有以下等级java为我的interface
public interface Identifiable<T extends Comparable<T>> extends Serializable {
public T getId();
}
public interface Function extends Identifiable {
public String getId();
}
public abstract class Adapter implements Function {
public abstract String getId();
}
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当我尝试按如下方式实施Adapter时scala
class MultiGetFunction extends Adapter {
def getId() : String = this.getClass.getName
}
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我收到了以下错误
Multiple markers at this line
- overriding method getId in trait Identifiable of type ()T; method getId has incompatible
type
- overrides Adapter.getId
- implements Function.getId
- implements Identifiable.getId
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Jat*_*tin 16
一般来说,使用Scala的java代码中的原始类型是一种痛苦.
试试以下内容:
public interface Function extends Identifiable<String> {
public String getId();
}
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该错误可能是由于编译器无法确定类型,T因为在声明时没有提到类型Function extends Identifiable.这是从错误中解释的:
:17:错误:在trait中覆盖方法getId可识别type()T; 方法getId具有不兼容的类型
Scala与Java 1.5及更高版本兼容.对于以前的版本,您需要破解.如果您无法更改Adapter,那么您可以在Java中创建Scala包装器:
public abstract class ScalaAdapter extends Adapter {
@Override
public String getId() {
// TODO Auto-generated method stub
return getScalaId();
}
public abstract String getScalaId();
}
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然后在Scala中使用它:
scala> class Multi extends ScalaAdapter {
| def getScalaId():String = "!2"
| }
defined class Multi
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