原始使用参数化 class 'ABC'

Question

我有以下界面：

public interface AsynchronousJobRunner<T extends AsynchronousJob> extends Runnable {
    public void kill();
    public void addJobExecutionListener(JobExecutionListener listener);
    public void removeJobExecutionListener(JobExecutionListener listener);
    public AsynchronousJobRunner withJob(T job);
}

AsynchronousJob 是一个抽象 class ，可以使用以下抽象方法扩展以表示其他作业：

public abstract class AsynchronousJob implements JSONSerializable, HasId {...}

    /**
     * Returns the class of the {@link AsynchronousJobRunner} that runs this type of job.
     * @return The appropriate class of the job runner for this class.
     */
    public abstract Class<? extends AsynchronousJobRunner> jobRunnerClass();

我还有以下 ExportJob 扩展基础 class 并具有以下签名和方法：

public class ExportJobRunner extends BaseJobRunner<ExportJob> { ...}

    @Override
    public Class<? extends AsynchronousJobRunner> jobRunnerClass() {
        return ExportJobRunner.class;
    }

两种 jobRunnerClass() 方法都有 Raw use of parameterized class 'AsynchronousJobRunner' 警告。

使警告消失的简单解决方案是：

public abstract Class<? extends AsynchronousJobRunner<?>> jobRunnerClass();


@Override
public Class<? extends AsynchronousJobRunner<?> jobRunnerClass() {
    return ExportJobRunner.class;
}

但是正确的解决方案是什么 why/how？

编辑:

我最终只是将界面代码更改为：

public interface AsynchronousJobRunner<T extends AsynchronousJob<T>> extends Runnable {
    void kill();
    void addJobExecutionListener(JobExecutionListener listener);
    void removeJobExecutionListener(JobExecutionListener listener);
    AsynchronousJobRunner<T> withJob(T job);
}

更改为 AsynchronousJob class：

public abstract class AsynchronousJob<T> implements JSONSerializable, HasId { ...
 /**
     * Returns the class of the {@link AsynchronousJobRunner} that runs this type of job.
     * @return The appropriate class of the job runner for this class.
     */
    public abstract Class<? extends AsynchronousJobRunner<? extends AsynchronousJob<T>> jobRunnerClass();

}

ExportJob class:

public public class ExportJob extends AsynchronousJob<ExportJob> {...
    @Override
    public Class<? extends AsynchronousJobRunner<? extends AsynchronousJob<ExportJob>>> jobRunnerClass() {
        return ExportJobRunner.class;
    }
}

而ExportJobRunnerclass保持不变。

我还忘了说，由于作业被序列化到数据库，所以正在进行一些注入魔法：

    /**
     * Instantiates an {@link AsynchronousJobRunner} instance for the provided job.
     * <p>
     * In order for the creation of the runner to succeed, the {@link AsynchronousJob#jobRunnerClass()}
     * method of the job must specify the appropriate class for its runner.
     *
     * @param job job to create runner for
     * @return job runner configured for the specified {@code job} parameter
     */
    private <T extends AsynchronousJob<T>> AsynchronousJobRunner<T> createJobRunner(T job) {
        return ((AsynchronousJobRunner<T>)injector.getInstance(job.jobRunnerClass())).withJob(job);
    }

我接受了@Andrew Vershinin 的回答，因为他让我走上了正确的思路。

Answer 1

我认为，您可以使用 AsynchronousJob 的递归声明：

public abstract class AsynchronousJob<T extends AsynchronousJob<T>> 
    implements JSONSerializable, HasId {

    public abstract Class<? extends AsynchronousJobRunner<T>> jobRunnerClass();
//...
}

不要忘记相应地更新 AsynchronousJobRunner 定义：

public interface AsynchronousJobRunner<T extends AsynchronousJob<T>> //...

然后，在你的工作 class 中，你可以在 return 类型中使用 class 本身：

public class ExportJob extends AsynchronousJob<ExportJob> {
    @Override
    public Class<? extends AsynchronousJobRunner<ExportJob>> jobRunnerClass() {
        return ExportJobRunner.class;
    }
}

这样可以确保类型一致性：T returns 运行ner 类型的工作，运行这种类型的工作 - 你无法确保这一点在 jobRunnerClass 签名中使用 ? extends AsynchronousJob，因为你可以 return 任何运行ner 类型并且它会编译。

原始使用参数化 class 'ABC'

Raw use of parameterized class 'ABC'

java

generics

type-erasure

raw-types

suppress-warnings