如何让方法在不阻塞的情况下返回已启动的任务？

Question

在我的 C# 应用程序中，我已经能够启动进行 http 调用的并发任务，然后等待所有任务完成。

var responseTasks = new List<Task<HttpResponseMessage>>();

// PostChanges makes an async http call and returns the task
responseTasks.Add(myServiceClient.PostChanges(json));
responseTasks.Add(myServiceClient.PostChanges(json));
responseTasks.Add(myServiceClient.PostChanges(json));

Task.WaitAll(responseTasks.ToArray());

但是，如果对 PostChanges 的每个调用都依赖于从 http 请求中获取数据的调用呢？创建封装两个调用并返回已启动任务的方法的最简单方法是什么？该方法必须使两个请求都像这样：

public Task<HttpResponseMessage> GetDataAndPostChanges(MyInput input)
{
    var json = myServiceClient.GetData(input); // this call must complete first
    var response = myServiceClient.PostChanges(json);
    return response; // how to actually return a task immediately that does both calls?
}

然后我想对该方法进行并发调用并等待它们全部完成。

var responseTasks = new List<Task<HttpResponseMessage>>();

// PostChanges makes an async http call and returns the task
responseTasks.Add(myServiceClient.GetDataAndPostChanges(input1));
responseTasks.Add(myServiceClient.GetDataAndPostChanges(input2));
responseTasks.Add(myServiceClient.GetDataAndPostChanges(input3));

Task.WaitAll(responseTasks.ToArray());

Answer 1

使函数异步并等待必要的调用

public async Task<HttpResponseMessage> GetDataAndPostChanges(MyInput input) {
    var json = await myServiceClient.GetData(input); // this call must complete first
    var response = await myServiceClient.PostChanges(json);
    return response;
}

上面的假设是GetData是一个异步函数。