Why is a Dictionary "not ordered"?

fearofawhackplanet picture fearofawhackplanet · Jun 17, 2011 · Viewed 11.7k times · Source

I have read this in answer to many questions on here. But what exactly does it mean?

var test = new Dictionary<int, string>();
test.Add(0, "zero");
test.Add(1, "one");
test.Add(2, "two");
test.Add(3, "three");

Assert(test.ElementAt(2).Value == "two");

The above code seems to work as expected. So in what manner is a dictionary considered unordered? Under what circumstances could the above code fail?

Answer

Jon Skeet picture Jon Skeet · Jun 17, 2011

Well, for one thing it's not clear whether you expect this to be insertion-order or key-order. For example, what would you expect the result to be if you wrote:

var test = new Dictionary<int, string>();
test.Add(3, "three");
test.Add(2, "two");
test.Add(1, "one");
test.Add(0, "zero");

Console.WriteLine(test.ElementAt(0).Value);

Would you expect "three" or "zero"?

As it happens, I think the current implementation preserves insertion ordering so long as you never delete anything - but you must not rely on this. It's an implementation detail, and that could change in the future.

Deletions also affect this. For example, what would you expect the result of this program to be?

using System;
using System.Collections.Generic;

class Test
{ 
    static void Main() 
    {
        var test = new Dictionary<int, string>();
        test.Add(3, "three");
        test.Add(2, "two");
        test.Add(1, "one");
        test.Add(0, "zero");

        test.Remove(2);
        test.Add(5, "five");

        foreach (var pair in test)
        {
            Console.WriteLine(pair.Key);
        }
    }     
}

It's actually (on my box) 3, 5, 1, 0. The new entry for 5 has used the vacated entry previously used by 2. That's not going to be guaranteed either though.

Rehashing (when the dictionary's underlying storage needs to be expanded) could affect things... all kinds of things do.

Just don't treat it as an ordered collection. It's not designed for that. Even if it happens to work now, you're relying on undocumented behaviour which goes against the purpose of the class.