CompletableFuture handle和CompleteExceptionly不能一起工作？

Question

当异常发生在CompletableFuture中时，我试图设置一个默认值，我通过如下方法使其工作：

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());
}

但是，当我试图在不好的事情发生时异常地使用complete来停止CompletableFuture，并按如下方式跟踪异常时，我无法像刚才那样捕捉到异常。

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());

}

更新日期2018-06-09谢谢您的帮助，@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());

}

在join（）之前封闭的句柄按预期工作。但在这种情况下，返回值将为null。

基于handle API

返回一个新的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提供的函数的参数来执行该阶段。




Answer 1

您正在构建一个未来，用一个句柄进行管道连接（从而获得另一个未来），然后异常地完成句柄返回的未来。

您应该异常地完成内部未来本身，而不是句柄。

这里的重点是句柄返回另一个未来；您不应该例外地完成“外部”未来，因为这样做会绕过处理行为。

低于代码；

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) {}    
  }
}