获取Func<Task<IResponse>>在超时检查中的结果

Question

我是异步编程的新手，所以不要做太多的事情。拜托，我需要更多的帮助。我照我在另一个问题中再次提出的那样做了。

public async Task<TResponse> SendRequestAsync<TResponse>(Func<Task<TResponse>> sendAsync)
{
        int timeout = 15;
        if (await Task.WhenAny(sendAsync, Task.Delay(timeout) == sendAsync))
        {
            return await sendAsync();
        }
        else
        {
            throw new Exception("time out!!!");
        }
}

但是我需要得到sendAsync()的结果并返回它。我也有疑问：

1)做这件事的最佳方法是什么，以及如何在Task.Delay中使用Func<Task<TResponse>>(或者可能是某种东西而不是它)？我不知道如何将Func转换为任务。

( 2)如果永续请求，则返回await sendAsync()内部。这不太好。如果不知怎么的话，我可以得到我的Func<Task<..>>的结果吗？

Answer 1

既然您是异步编程中的新手--最好不要在一条语句中放太多东西，最好把它分开：

public async Task<TResponse> SendRequestAsync<TResponse>(Func<Task<TResponse>> sendAsync) {
    int timeout = 15;
    // here you create Task which represents ongoing request
    var sendTask = sendAsync();
    // Task which will complete after specified amount of time, in milliseconds
    // which means your timeout should be 15000 (for 15 seconds), not 15
    var delay = Task.Delay(timeout);
    // wait for any of those tasks to complete, returns task that completed first
    var taskThatCompletedFirst = await Task.WhenAny(sendTask, delay);
    if (taskThatCompletedFirst == sendTask) {
        // if that's our task and not "delay" task - we are fine
        // await it so that all exceptions if any are thrown here
        // this will _not_ cause it to execute once again
        return await sendTask;
    }
    else {
        // "delay" task completed first, which means 15 seconds has passed
        // but our request has not been completed
        throw new Exception("time out!!!");
    }
}

Answer 2

请求被发送两次，因为sendAsync是一个Func返回任务，每个调用都不同。首先在Task.WhenAny()下调用它，然后在操作符return await sendAsync()中重复。

为了避免这个重复的调用，您应该将一个任务保存为变量，并将该任务传递给两个调用：

public async Task<TResponse> SendRequestAsync<TResponse>(Func<Task<TResponse>> sendAsync)
{
    int timeout = 15;
    var task = sendAsync();
    if (await Task.WhenAny(task, Task.Delay(timeout) == task))
    {
        return await task;
    }
    else
    {
        throw new Exception("time out!!!");
    }
}

已完成任务上的await只返回其结果，而不重新运行任务。