Question

错误：（65,52）java：不兼容类型：推理变量U具有不兼容的边界等式约束：akka.http.javadsl.model.HttpResponse 下界：com.myactors.ChatActor.ChatMessage

以下行编译显示错误：

 CompletionStage<HttpResponse> httpResponse =
                        postChat(url,
                                context().system(), chatData)
                                .thenApplyAsync(httpResponse -> new ChatActor.ChatMessage(httpResponse,"1234"));


public static class ChatMessage{
   HttpResponse httpResponse;
    String name;

    public ChatMessage( HttpResponse httpResponse, String name) {
        this.httpResponse = httpResponse;
        this.name = name;
    }

    public HttpResponse getHttpResponse() {
        return httpResponse;
    }

    public String getname() {
        return name;
    }
}

这是HttpResponse是Akka Http。

我不知道它在说什么。应该怎样解决它？

Answer 1

方法.thenApplyAsync(httpResponse -> new ChatActor.ChatMessage(httpResponse,"1234"))返回CompletionStage<ChatActor.ChatMessage>而非CompletionStage<HttpResponse>，与函数httpResponse -> new ChatActor.ChatMessage(httpResponse,"1234")一样，您将类型HttpResponse的实例转换为实例类型为ChatActor.ChatMessage。

只需更改变量的类型：

CompletionStage<ChatActor.ChatMessage> httpResponse =
     postChat(url, context().system(), chatData)
    .thenApplyAsync(httpResponse -> new ChatActor.ChatMessage(httpResponse,"1234"));

编译器之所以不简单地说“无法将CompletionStage<ChatActor.ChatMessage>类型的实例分配给CompletionStage<HttpResponse> 类型的变量”的原因在于应用的新类型推断通用方法thenApplyAsync。编译器尝试查找方法的类型参数U的类型，该类型可以使调用有效，即如果ChatActor.ChatMessage和HttpResponse之间存在子类型关系，则可能。由于此操作失败，编译器会告诉您它无法找到U的有效类型。

java：不兼容的类型：推理变量U具有不兼容的边界

1 个答案: