CS8328 question:out ref is avaliable

[XNTEABDSC]

ref type is type,so we can use "out ref T value" to do thing like return ref

[Rekkonnect]

No, currently you cannot do out ref T value. The real type of the parameter is still T; ref is a modifier showing that the parameter is passed by reference.

To do that you need to wrap your T onto another object, like a Box<T> or something that basically acts like a reference to T. Your box could be either of the two, depending on how you lay your structure:

Class

public class Box<T>
{
    public T Value { get; set; }
}

Struct - Unsafe

public ref struct Ref<T>
{
    private ref T value;

    public ref T Value
    {
        get => ref value;
    }
    
    public unsafe void SetValue(ref T value)
    {
        this.value = ref value;
    }
}

Usually, the simplest and safest option is by doing the Box<T> class, the Ref<T> struct I defined above could be unsafe and harder to handle if you aren't too familiar with unsafe constructs.

[XNTEABDSC]

public ref T Get(int i);
public bool Has(int i);
public bool TryGet(int i,out ref T value)

I think my needs are reasonable.

[Rekkonnect]

In that case, a simple box would be your solution; you also avoid technicalities around using the ref you're providing, and it's clearer that you intend the ref be modified.