java - 与 T 不兼容的类型

Question

我有一个名为 adapt 的方法，但我不知道 T 和 ? 之间的确切区别。

对我来说，适应必须接受任何类(class)。

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);
        }
    }
}

但是我无法传递这个处理程序。为什么会出现这种情况？

错误

incompatible types: inferred type does not conform to equality constraint(s) [ERROR] inferred: java.lang.Object [ERROR] equality constraints(s): java.lang.Object,com.daniela.infra.RequestorResponse

Answer 1

似乎您正在尝试传递定义的类 RequestorResponse转换为泛型类型 this.adapt(handler)

ResponseHandler < RequestorResponse > handler

Java 不会允许你这样做，因为你的通用类型 T可能与 RequestorResponse 的类别不同

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){

这里有一些引用:

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