How does Lazy<T> get around needing new() constraint?

Question

Example 1 (does not compile):

void Main()
{
    var c = new C<D>();
    c.M.F();
}

class C<T>
{
    T _m = null;
    public T M { get {
        if(_m == null) _m = new T();
        return _m; 
    } }
}

class D
{
    public void F() { Console.WriteLine ("i was created"); }
}

Result:

Cannot create an instance of the variable type 'T' because it does not have the new() constraint

Example 2 (works):

void Main()
{
    var c = new C<D>();
    c.M.F();
}

class C<T>
{
    Lazy<T> _m = new Lazy<T>();
    public T M { get { return _m.Value; } }
}

class D
{
    public void F() { Console.WriteLine ("i was created"); }
}

Result:

i was created

Answer 1

If you delve into the source code, you'll see that Lazy<T> ultimately uses Activator:

return new Lazy<T>.Boxed((T)Activator.CreateInstance(typeof(T)));

This is just a shortcut for using reflection. Since it's not instantiating the type via the actual generic type argument (new T()) but rather invoking the constructor through reflection, no where T : new() constraint is needed.