Parallel.Foreach + yield return?

Short version – no, that isn’t possible via an iterator block; the longer version probably involves synchronized queue/dequeue between the caller’s iterator thread (doing the dequeue) and the parallel workers (doing the enqueue); but as a side note – logs are usually IO-bound, and parallelising things that are IO-bound often doesn’t work very well.

If the caller is going to take some time to consume each, then there may be some merit to an approach that only processes one log at a time, but can do that while the caller is consuming the previous log; i.e. it begins a Task for the next item before the yield, and waits for completion after the yield… but that is again, pretty complex. As a simplified example:

static void Main()
{
    foreach(string s in Get())
    {
        Console.WriteLine(s);
    }
}

static IEnumerable<string> Get() {
    var source = new[] {1, 2, 3, 4, 5};
    Task<string> outstandingItem = null;
    Func<object, string> transform = x => ProcessItem((int) x);
    foreach(var item in source)
    {
        var tmp = outstandingItem;

        // note: passed in as "state", not captured, so not a foreach/capture bug
        outstandingItem = new Task<string>(transform, item);
        outstandingItem.Start();

        if (tmp != null) yield return tmp.Result;
    }
    if (outstandingItem != null) yield return outstandingItem.Result;
}
static string ProcessItem(int i)
{
    return i.ToString();
}