多个异步任务

Question

I have two async mehotds that not directly related to one another (they do both write out though).. and i don't want either to block the efficiency of the other..

If i place one of the activities in a set of tasks like this:

var list = new List<Task>();

foreach (var t in things)
{
    list.Add(My.MethodAsync(t)); // method returns Task
}

await Task.WhenAll(t);

I know that it will not block, and potentially run in parallel..

If i another async method though though:

var list = new List<Task>();

foreach (var t in things)
{
    list.Add(My.MethodAsync(t)); // method returns Task
}

foreach (var t in things)
{
    await My.OtherAsync(t)); // method returns Task
}

await Task.WhenAll(t);

The Task.WhenAll() needs to wait for the second foreach to complete...

Therefore, is something like this possible / recommend?

var allTasks = new List<Task>();

foreach (var t in things)
{
    allTasks.Add(My.MethodAsync(t)); // method returns Task
    allTasks.Add(My.OtherAsync(t)); // method returns Task
}

await Task.WhenAll(t);

Having typed this out, even though the tasks themselves are independent, it would be nice i guess if an error was raised if both do not output.. hmm

question still stands though ;)

Answer 1

var allTasks = new List<Task>();

foreach (var t in things)
{
    allTasks.Add(My.MethodAsync(t); // method returns Task
    allTasks.Add(My.OtherAsync(t); // method returns Task
}

await Task.WhenAll(t).ConfigureAwait(false);

This will work just fine. As was pointed out in the comments, if your methods return a Task that was started already (see here ), then all Task.WhenAll will do is wait asynchronously for the tasks to complete. These tasks may run in parallel if their run-time is long enough that you can start one before the previous finishes. Note that I added ConfigureAwait(false) to your code. Technically, it doesn't actually change what your code is doing as await is done with false as the default but it does clarify your intent so I prefer it when I write asynchronous method calls.