java - 没有参数的 Lambda 函数 - 无法推断功能接口(interface)类型

Question

我有一个不需要任何参数的 lambda 函数:

Observable.interval(5, TimeUnit.SECONDS)
  .subscribe(x -> new CryptoCompareRxService().getHistoricalDaily("BTC", "USD")
    .subscribe(historicalDaily -> System.out.println("historicalDaily = " + historicalDaily)));

我想用“()”替换“x”，因为它没用:

Observable.interval(5, TimeUnit.SECONDS)
  .subscribe(() -> new CryptoCompareRxService().getHistoricalDaily("BTC", "USD")
    .subscribe(historicalDaily -> System.out.println("historicalDaily = " + historicalDaily)));

但是当我这样做时，我得到一个错误:

Cannot infer functional interface type

为什么？

这是完整的类(class):

package com.tests.retrofitrxjava;

import com.tests.retrofitrxjava.rx.CryptoCompareRxService;
import org.springframework.boot.SpringApplication;
import org.springframework.boot.autoconfigure.SpringBootApplication;
import rx.Observable;

import java.util.concurrent.TimeUnit;

@SpringBootApplication
public class RetrofitRxjavaApplication {

  public static void main(String[] args) {

    SpringApplication.run(RetrofitRxjavaApplication.class, args);

    Observable.interval(5, TimeUnit.SECONDS)
      .subscribe(x -> new CryptoCompareRxService().getHistoricalDaily("BTC", "USD")
        .subscribe(historicalDaily -> System.out.println("historicalDaily = " + historicalDaily)));

  }

}

Answer 1

我认为你的错误是由于 RX 的变化 Observable在版本 2 中，订阅现在接受 Consumer接受参数

public final Disposable subscribe(Consumer<? super T> onNext)

Consumer<T> A functional interface (callback) that accepts a single value.

而不是 Action与版本 1 一样

public final Subscription subscribe(Action1<? super T> onNext)

Action1<T> extends Action A one-argument action.