Question

我有一种名为 adapt 的方法，我不确切知道T和？之间的区别。

对我来说， adapt 必须接受任何类。

public void getRequestor(Long requestorId,  ResponseHandler<RequestorResponse> handler){
    String url = this.serviceUrl;

   //Error happens in this.adapt...
    RestClient.url(url).get(null).onResponse(this.adapt(handler), RequestorResponse.class); 
}

private <T> RestClient.ResponseHandler<T> adapt(ResponseHandler<T> handler){
    return new RestClient.ResponseHandler<T>() {
        @Override
        public void onResponse(ResponseEntity<T> response) {
            if(response.getStatusCode().is2xxSuccessful()) {
                handler.onResponse(response.getBody());
                return;
            }
        handler.onError(new com.sensedia.api.interfaces.Error(String.valueOf(response.getStatusCodeValue())));
        }
    };
 }

RestClient类

public class RestClient {
   private final String url;

   private RestClient(String url) {
     this.url = url;
   }

   public static RestClient url(String url) {
      return new RestClient(url);
   }
    //Others methods 
    @FunctionalInterface
    public interface ResponseHandler<T> {
       void onResponse(ResponseEntity<T> var1);
    }

    public static class PagedOperation<T> extends AbstractOperation<T, RestClient.PagedOperation<T>> {
        private final RestClient.Limit limit;
        private final RestClient.Offset offset;

        public PagedOperation(String url, RestClient.Limit limit, RestClient.Offset offset) {
            super(url);
            this.limit = limit;
            this.offset = offset;
        }

        public void onResponse(RestClient.ResponseHandler<T> handler, Class<T> type) {
            UriComponentsBuilder builder = UriComponentsBuilder.fromHttpUrl(this.url()).queryParam("_limit", new Object[]{this.limit.get()}).queryParam("_offset", new Object[]{this.offset.get()});
            ResponseEntity<T> response = this.client().getForEntity(builder.build().encode().toUri(), type);
            handler.onResponse(response);
        }
    }
}

但我无法通过此处理程序。为什么会这样？

错误

不兼容的类型：推断类型不符合相等性 constraint（s）[ERROR]推断：java.lang.Object [ERROR]相等 constraints（s）：java.lang.Object，com.daniela.infra.RequestorResponse

Answer 1

好像你试图将定义的类RequestorResponse传递给泛型类型this.adapt(handler)

ResponseHandler＆lt; RequestorResponse＆gt;处理

Java不允许您这样做，因为您的通用类型T可能与RequestorResponse不同，

我建议你这样做

public void getRequestor(Long requestorId,  ResponseHandler<T> handler){
    String url = this.serviceUrl;
    RestClient.url(url).get(null).onResponse(this.adapt(handler), RequestorResponse.class);
}

或者将this.adapt更改为

private <T> RestClient.ResponseHandler<T> adapt(ResponseHandler<?> handler){

这里有一些参考文献：

<T>通用类型：https://docs.oracle.com/javase/tutorial/java/generics/types.html
<?>通配符：https://docs.oracle.com/javase/tutorial/extra/generics/wildcards.html

与T不兼容的类型

1 个答案: