Question

我在CompletableFuture中发生异常时尝试设置默认值。我按handle方法使其工作如下：

private static void testHandle() {
    String name = null;
    CompletableFuture<String> completableFuture
            = CompletableFuture.supplyAsync(() -> {
        if (name == null) {
            throw new RuntimeException("Computation error!");
        }
        return "Hello, " + name;
    }).handle((s, t) -> s != null ? s : "Hello, Stranger!" + t.toString());
    out.println(completableFuture.join());
}

但当我尝试在发生不良事件时使用CompletableFuture停止completeExceptionally并跟踪异常，如下所示我无法抓住异常就像我刚才那样。

private static void testCompleteExceptionally() {
    String name = "Hearen";
    CompletableFuture<String> completableFuture
            = CompletableFuture.supplyAsync(() -> {
        delay(500L);
        if (name == null) {
            throw new RuntimeException("Computation error!");
        }
        return "Hello, " + name;
    }).handle((s, t) -> {
        try {
            throw t.getCause(); 
        } catch (Throwable e) {
            out.println(e.toString()); // I was hoping to record the custom exceptions here;
        }
        return s != null ? s : "Hello, Stranger!" + t.toString();
    });

    if (name != null) {
        completableFuture.completeExceptionally(new RuntimeException("Calculation failed!")); // when bad things happen, I try to complete it by exception;
    }
    out.println(completableFuture.join());

}

更新 2018-06-09感谢您的帮助，@ Daniele

private static void testCompleteExceptionally() {
    String name = "Hearen";
    CompletableFuture<String> completableFuture
            = CompletableFuture.supplyAsync(() -> {
        delay(500L);
        if (name == null) {
            throw new RuntimeException("Computation error!");
        }
        return "Hello, " + name;
    });

    if (name != null) {
        completableFuture.completeExceptionally(new RuntimeException("Calculation failed!"));
    }
    out.println(completableFuture.handle((s, t) ->  s != null ? s : "Hello, Stranger!" + t.toString()).join());

}

join()之前的句柄按预期工作。但在这种情况下，returned value将为null。

基于handle API

返回 new CompletionStage，当此阶段正常或异常完成时，将使用此阶段的结果和异常作为所提供函数的参数执行。

Answer 1

您正在构建一个future，并使用handle（以便获得另一个未来），然后完成handle返回的特殊未来。

你应该特别完成内部future本身，而不是handle。

这里的要点是handle返回另一个未来;并且你不应该特别完成“外部”未来，因为这样做会绕过处理行为。

代码下方;

package stackOv;

import java.util.concurrent.CompletableFuture;
import java.util.function.BiFunction;
import java.util.function.Supplier;

public class TestHandle {
  BiFunction<String, Throwable, String> handle2 = new BiFunction<String, Throwable, String>() {
    @Override
    public String apply(String s, Throwable t) {
      try {
        throw t.getCause(); 
      } catch (Throwable e) {
        // I was hoping to record the custom exceptions here;
        System.out.println(e.toString());
      }
      return s != null ? s : "Hello, Stranger!" + t.toString();
    }
  };

  private void testCompleteExceptionally() {
    String name = "Hearen";
    Supplier<String> supplier2 = () -> {
      delay(500L);
      if (name == null) {
        throw new RuntimeException("Computation error!");
      }
      return "Hello, " + name;
    };
    CompletableFuture<String> completableFuture = CompletableFuture.supplyAsync(supplier2);

    if (name != null) {
      // when bad things happen, I try to complete it by exception;
      completableFuture.completeExceptionally(new RuntimeException("Calculation failed!"));      
    }
    System.out.println(completableFuture.handle(handle2).join());
  }

  public static void main(String[] args) {
    TestHandle th = new TestHandle();
    th.testCompleteExceptionally();
  }

  private static void delay(long milli) {
    try { Thread.sleep(milli); } catch (InterruptedException e) {}    
  }
}

CompletableFuture句柄和完成异常无法一起工作？

1 个答案: