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Home/ Questions/Q 8239287
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Editorial Team
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Editorial Team
Asked: June 7, 20262026-06-07T20:05:43+00:00 2026-06-07T20:05:43+00:00

Why is the Tags property of Book empty after this code runs? class Program

  • 0

Why is the Tags property of Book empty after this code runs?

class Program
{
    static void Main(string[] args)
    {
        List<Book> books = new List<Book>();
        List<String> tags = new List<String> {"tag1", "tag2", "tag3"};
        String title = "a title";
        books.Add(new Book
        {
            Title = title,
            Author = "an author",
            Tags = tags
        });
        Console.WriteLine("(" + title + ")");
        Console.WriteLine((books[0]).Tags.Count());

        title = String.Empty;
        tags.Clear();

        Console.WriteLine("(" + title + ")");
        Console.WriteLine((books[0]).Tags.Count());
    }
}

The code for Book:

public class Book
{
    public String Title { get; set; }
    public String Author { get; set; }
    public List<String> Tags { get; set; }
}

Running this code outputs

("a title")
3
()
0

Are tags and title being passed by reference here? Renaming the respective properties produces the same output.

EDIT:

I just realised that I meant for every Console.WriteLine statement to refer to the object, not just the tags one. I meant this:

Book aBook = books[0];
Console.WriteLine("(" + aBook.Title + ")");
Console.WriteLine(aBook.Tags.Count());

title = String.Empty;
tags.Clear();

Console.WriteLine("(" + aBook.Title + ")");
Console.WriteLine(aBook.Tags.Count());

which as expected, outputs:

("a title")
3
("a title")
0

but since I made a mistake in my initial question, I’m leaving it as is since the parts of the answers that refer to title are referencing the original code.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-07T20:05:46+00:00Added an answer on June 7, 2026 at 8:05 pm

    List<T> is a reference type, so yes, you get reference semantics here.

    You need to assign a copy of tags to the property if you want them to be independent.

    Tags = new List<string>(tags);
    
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