Add BufferWithCount or an overload to Buffer()

[softlion]

I know this "idea" has already discussions on rx, but i feel those discussions were aside from the real need. So i'm opening a new one.

BufferWithCount(N) should be a rolling (ie: overlapping) buffer emitting the N last (buffered) items each time one new item is added.
Currently, Buffer(N) is only a non overlapping buffer, emitting distinct group of items.

The current behavior of Buffer makes difficult to compare 2 consecutive values. And to emit a new item only when a specific change occurs between those consecutive values.

The idea is to simplify this code:

var queue = new List<AdapterStatus>();
this.WhenAnyValue(x => x.ViewModel.AdapterStatus)
    .StartWith(AdapterStatus.Unknown)
    .Distinct()
    .Do(value =>
    {
        queue.Add(value);
        if(queue.Count > 2)
            queue.RemoveAt(0);
    })
    .Where(s => queue.Count>=2 && queue[^1] == AdapterStatus.Available && queue[^2] != AdapterStatus.Available) 

Into this:

this.WhenAnyValue(x => x.ViewModel.AdapterStatus)
    .StartWith(AdapterStatus.Unknown)
    .Distinct()
    .BufferWithCount(2) //Keep this curious name to be 'standard' with other languages
    .Where(s => queue[^1] == AdapterStatus.Available && queue[^2] != AdapterStatus.Available) 

Does this make sense ?

[akarnokd]

I'm pretty sure there is a standard Buffer overload with count and skip where you get this overlapping pattern:

Observable.Range(1, 10)
            .Buffer(2, 1)
            .Subscribe(v => Console.WriteLine(string.Join(",", v)));

prints

1,2
2,3
3,4
4,5
5,6
6,7
7,8
8,9
9,10
10

[softlion]

You are right. The names of nearly all methods and parameters should be reviewed at some time in all RX. They are all so misleading.

[dwcullop]

The current behavior of Buffer makes difficult to compare 2 consecutive values. And to emit a new item only when a specific change occurs between those consecutive values.

If that is all you need, there is already the DistinctUntilChanged operator that will suppress events if the value is the same as the last one. There is even an overload that will take IEqualityComparer<T> so you can control what "same" means.