Geek Answers Handbook

C # - How to convert Lazy <List <T>> to Lazy <List <U>>?

I have Lazy<List<T>>where T is a class that contains a huge string and metadata about files. Let us call them property HugeStringand propertyMetadata

I have this class U, which has the same property HugeString, by the way. I need to convert Lazy<List<T>>to Lazy<List<U>>without loading everything.

Is it possible?

Here I create my list, and inside this method I get information about the file and the file itself:

entity.VersionedItems =
    new Lazy<List<VersionedItemEntity>>(
        () => VersionedItemEntity.GetFromTFSChanges(entity,chng.Changes));

This is what I want to do (commented)

ChangesetList.Add(
    new HistoryLogEntryModel()
    {
        Revision = changeset.Changeset.ToString(),
        Author = changeset.User,
        Date = changeset.Date.ToString("dd/MM/yyyy"),
        Message = changeset.Comment,
        //VersionedItems = changeset.VersionedItems
    }

But HistoryLogEntryModel has a different version of VersionedItems. And I need to convert some variables. If I converted one thing to another, it would load everything, and that would be unnecessary and slow.

Is this the right approach? How else can I achieve this?

adv.

~

+5
2

Lazy<List<T>> Lazy<List<U>>.

var uLazy = new Lazy<List<U>>(() => tLazy.Value.Select(t => (U)t).ToList());
+4

- :

public static class ExtensionMethods
{
    public static Lazy<U> Convert<T,U>(this Lazy<T> source, Func<Lazy<T>, Lazy<U>> convert)
    {
        return convert(source);
    }
}

Lazy<List<int>> source = new Lazy<List<int>>();

Lazy<List<string>> converted = source.Convert(x => 
{ 
 return new Lazy<List<string>>()
            {
                Items = x.Items.ConvertAll<string>(i => i.ToString())
            };
});

, , .

0

More articles:


All Articles