Return all enumerables with yield return at once; without looping through

c# ienumerable yield yield-return
John Oxley picture John Oxley · Aug 13, 2009 · Viewed 38.8k times · Source

I have the following function to get validation errors for a card. My question relates to dealing with GetErrors. Both methods have the same return type IEnumerable<ErrorInfo>.

private static IEnumerable<ErrorInfo> GetErrors(Card card)
{
    var errors = GetMoreErrors(card);
    foreach (var e in errors)
        yield return e;

    // further yield returns for more validation errors
}

Is it possible to return all the errors in GetMoreErrors without having to enumerate through them?

Thinking about it this is probably a stupid question, but I want to make sure I'm not going wrong.

Answer

Jon Skeet picture Jon Skeet · Aug 13, 2009

It's definitely not a stupid question, and it's something that F# supports with yield! for a whole collection vs yield for a single item. (That can be very useful in terms of tail recursion...)

Unfortunately it's not supported in C#.

However, if you have several methods each returning an IEnumerable<ErrorInfo>, you can use Enumerable.Concat to make your code simpler:

private static IEnumerable<ErrorInfo> GetErrors(Card card)
{
    return GetMoreErrors(card).Concat(GetOtherErrors())
                              .Concat(GetValidationErrors())
                              .Concat(AnyMoreErrors())
                              .Concat(ICantBelieveHowManyErrorsYouHave());
}

There's one very important difference between the two implementations though: this one will call all of the methods immediately, even though it will only use the returned iterators one at a time. Your existing code will wait until it's looped through everything in GetMoreErrors() before it even asks about the next errors.

Usually this isn't important, but it's worth understanding what's going to happen when.

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