接口上的Java CompletableFuture链调用

Question

我有一些工作流抽象依赖于接口WorkflowStep:

public interface WorkflowStep {

    public void executeStep();
}

现在我有三个不同的类来实现这个接口：

GetCoordinatesForWaypoints, DisplayDetails, PlaySounds

public class GetCoordinatesForEndpoints implements WorkflowStep {
    @Override
    public void executeStep() {
        new Thread(new Runnable() {
            @Override
            public void run() {
              //download coordinates from open street map
        }).start();

    }
}

public class DetailsDispatchWorkflow implements DispatchWorkflow {
    private List<WorkflowStep> workflowSteps;

    public DetailsDispatchWorkflow() {
        workflowSteps = new LinkedList<>();
    }

    @Override
    public void start() {
        workflowSteps.add(new GetCoordinatesForEndpoints());
        workflowSteps.add(new DisplayDetails());
        workflowSteps.add(new PlaySounds());
        workflowSteps.forEach(WorkflowStep::executeStep);
    }
}

ExecutorService executorService = Executors.newFixedThreadPool(5);
CompletableFuture<WorkflowStep> workflowStepCompletableFuture = 
CompletableFuture.supplyAsync(() -> new 
GetCoordinatesForEndpoints().executeStep(), executorService);

.thenAccept(() -> new DisplayDetails().executeStep(), executorService);

我得到一个错误，编译器不能推断函数接口类型。我的问题是：如何实现以下调用链：

CompletableFuture<WorkflowStep> workflowStepCompletableFuture = 
    CompletableFuture
        .supplyAsync(() -> new GetCoordinatesForEndpoints().executeStep(), executorService)
        .thenAccept(() -> new DisplayDetails().executeStep(), executorService)
        .thenAcceptAsync(() -> new PlaySounds().executeStep(), executorService);

当所有实例化的对象实现相同的接口时？

Answer 1

您的WorkflowStep接口基本上等同于runnable:没有输入，没有输出。因此，在CompletableFutureAPI中，应该使用相应的runasync()和thenrunasync()方法：

CompletableFuture<Void> workflowStepCompletableFuture = 
    CompletableFuture
        .runAsync(() -> new GetCoordinatesForEndpoints().executeStep(), executorService)
        .thenRunAsync(() -> new DisplayDetails().executeStep(), executorService)
        .thenRunAsync(() -> new PlaySounds().executeStep(), executorService);

这将使它们都异步运行，但按顺序运行（就像您试图做的那样）。

当然，您还应该从实现中删除线程创建，以使其更加有用。