OpenJDK 1.6做嵌套类型推断的Bug?

时间:2012-06-17 06:54:42

标签: java generics type-inference

我有以下示例Java程序,它可以在Oracle JDK下编译,但不能在OpenJDK上编译:

public class GenericsBug {

  public static void main(String[] args) {
      GenericsBug bug = new GenericsBug();
      // This line causes the error for not finding matching types:
      AResp resp = bug.execute(new AReq());
  }

  public <T extends Request,R extends Response<T>> R execute(T request) {
      return null;
  }
}

class Request { }
class Response<T extends Request> {}

class AReq extends Request {}
class AResp extends Response<AReq> {}

使用

进行编译
java version "1.6.0_31"
Java(TM) SE Runtime Environment (build 1.6.0_31-b04-415-11M3635)
Java HotSpot(TM) 64-Bit Server VM (build 20.6-b01-415, mixed mode)

java version "1.7.0_03"
OpenJDK Runtime Environment (IcedTea7 2.1.1pre) (7~u3-2.1.1~pre1-1ubuntu3)
OpenJDK Client VM (build 22.0-b10, mixed mode, sharing)

工作正常,但

失败
java version "1.6.0_18"
OpenJDK Runtime Environment (IcedTea6 1.8.13) (6b18-1.8.13-0+squeeze1)
OpenJDK 64-Bit Server VM (build 14.0-b16, mixed mode)

我在这里得到以下编译错误:

GenericsBug.java:9: type parameters of <R>R cannot be determined; no unique maximal instance exists for type variable R with upper bounds AResp,Response<T>
    AResp resp = bug.execute(new AReq());
                            ^
 1 error

所以我的问题是,如果这是OpenJDK中的错误,或者我是否在使用泛型类型推断做错了什么?


由于在下面的评论中,对于此代码的有用性存在一些疑问(虽然这个问题纯粹是语法相关的;-),这里给出了更多内容的给定示例。它使用双分派来让请求自己创建响应,并使用参数化的execute来尽可能使用类型,以便只允许相同的请求和响应类型对。使用多个重载execute()方法也可能完成相同的方案。无论这段代码是否有用,关于此用例的Oracle JDK和OpenJDK 1.6之间的不同语法处理的问题仍然存在。

 public class Client {

   public static void main(String[] args) {
     Client client = new Client();

     // AReq and AResp must match, otherwise an compile error will happen
     AResp respA = client.execute(new AReq());
     System.out.println(respA.getAContent());

     // Same for BReq and BResp
     BResp respB = client.execute(new BReq());
     System.out.println(respB.getBContent());
   }

   public <T extends Request,R extends Response<T>> R execute(T request) {
     // Fetch the response somehow, eg. by using a HttpClient:
     String responseBody = "....";
     // Let the request itself create the response
     return request.createResponse(responseBody);
 }

}
// ==================================================================================

// Abstract definition of a requests
abstract class Request {
    abstract <R extends Response<? extends Request>> R createResponse(String content); 
}
class Response<T extends Request> {}

// Two request/response pairs with specific request/response specific members
class AReq extends Request {
    AResp createResponse(String content) {
        return new AResp(content);
    }
}
class AResp extends Response<AReq> {
    private String aContent;

    public AResp(String pContent) {
        aContent = "AResp: " + pContent;
    }

    public String getAContent() {
        return aContent;
    }
} 

class BReq extends Request {
    BResp createResponse(String content) {
        return new BResp(content);
    }
}
class BResp extends Response<BReq> {
    private String bContent;

    public BResp(String pContent) {
        bContent = "BResp: " + pContent;
    }

    public String getBContent() {
        return bContent;
    }
}

2 个答案:

答案 0 :(得分:1)

想象你也

class AResp2 extends Response<AReq> {}

然后两个人的合法性都是合法的:

  

AResp resp = bug.execute(new AReq());

     

AResp2 resp = bug.execute(new AReq());

所以我相信OpenJDK在这里是正确的。

可以采取哪些措施来解决问题:

  

公共响应执行(T请求){}

然后手动将响应转换为AResp。

或者将请求紧密连接到响应,以便编译器可以按请求类型明确地确定响应类型:

class Request<R extends Response> { }
class Response {}

class AReq extends Request<AResp> {}
class AResp extends Response {}

public <T extends Request<R>, R extends Response> R execute(T request) {  }

答案 1 :(得分:1)

你的问题是execute 泛型,正如阿列克谢的答案所示。你怎么能实现一个方法,可以为任何R返回Response< T >的任何所需子类T(尽管return null在技术上应该有效,这就是为什么我认为Oracle JDK实际上是正确的。)

GenericsBug该课程,而非GenericsBug.execute方法,应以TR作为参数。