C#8 what does "default!" do on generic types?

Question

Just playing around with C# 8.0 Beta and updating some of my code to use nullable reference types.

I have a node style class for a Trie implementation. Each node has a value of type T. The constructor for the root node does not need a value, so I was setting this to default.

Here's a short version:

public class Trie<T>
{
    public readonly bool caseSensitive;
    public readonly char? letter;
    public readonly Dictionary<char, Trie<T>> children;
    public readonly Trie<T>? parent;
    public readonly int depth;
    public bool completesString;
    public T value;

    public Trie(bool caseSensitive = false)
    {
        this.letter = null;
        this.depth = 0;
        this.parent = null;
        this.children = new Dictionary<char, Trie<T>>();
        this.completesString = false;
        this.caseSensitive = caseSensitive;
        this.value = default;
    }
}

if the last line of the ctor is changed to

   this.value = default!;

as I saw in a different question here, then it compiles just fine. But I don't understand what the ! is doing here, and it's pretty hard to google, since google seems to ignore punctuation in most cases.

What does default! do?

Answer 1

The nullable reference types use static flow analysis to figure out if you have a possible null value (thus issuing a warning the variable is being assigned a null value).

The ! is used to allow the developer to suppress the warning generated. Essentially overriding the compiler.