[destel]

Hi everyone, I’m the author of the article. Happy to answer any questions or discuss concurrency patterns in Go. Curious how others tackle such problems.

[gabesullice]

I might be too late for you to see this, but I'm curious why your final example requires the function "f" to receive the canWrite channel. I must be missing something. Couldn't the Ordered map signature be:

  func OrderedLoop[A, B any](in <-chan A, done chan<- B, n int, f func(a A))

Instead of:

  func OrderedLoop[A, B any](in <-chan A, done chan<- B, n int, f func(a A, canWrite <-chan struct{}))

Or is there a reason that "f" can't be wrapped in an anonymous function to handle the blocking logic?

[destel]

You're not late. Just yesterday I've configured new reply notifications via RSS.

Typically function "f" does two things. 1. Performs calculations (this can be parallelized). 2. Writes results somewhere (this must happen sequentially and in the correct order).

Here's the typical OrderedLoop usage example from the article:

  OrderedLoop(in, out, n, func(a A, canWrite <-chan struct{}) {
   // [Do processing here]
   
   // Everything above this line is executed concurrently,
   // everything below it is executed sequentially and in order
   <-canWrite
   
   // [Write results somewhere]
  })

Without the "canWrite" the entire body of the "f" will be executed either concurrently or sequentially. With it - we can have this dual behavior, similar to critical sections protected by mutexes.

It's worth mentioning that OrderedLoop is a low-level primitive. User-level functions like OrderedMap, OrderedFilter and others do not need the "canWrite" channel to be exposed.

[kunley]

chan chan Foo seems like a cool trick, looking forward to use it in the code; thanks for the idea.

PS. I realize you present even better solution; still, first version seems like a thing nice enough to have in a toolbox

[destel]

Thanks. This replyTo pattern is very similar to promises in other languages.

[Traubenfuchs]

> Curious how others tackle such problems.

What do you think about the order preserving simplicity of Java?

  List<Input> inputs = ...;

  List<Output> results = inputs.parallelStream()
                             .map(this::processTask)
                             .collect(toList());

If you want more control or have more complex use cases, you can use an ExecutorService of your choice, handle the futures yourself or get creative with Javas new structured concurrency.

[Groxx]

Their planned semantics don't allow for that - there's no backpressure in that system, so it might race ahead and process up to e.g. item 100 while still working on item 1.

If everything fits in memory, that's completely fine. And then yeah, this is wildly overcomplicated, just use a waitgroup and a slice and write each result into its slice index and wait for everything to finish - that matches your Java example.

But when it doesn't fit in memory, that means you have unbounded buffer growth that might OOM.

[destel]

I haven’t used Java for about a decade, so I’m not very familiar with streams api.

Your snippet looks good and concise.

One thing I haven’t emphasized enough in the article is that all algorithms there are designed to work with potentially infinite streams

[kamranjon]

Often in go I’ll create some data structure like a map to hold the new value keyed by the original index (basically a for loop with goroutines inside that close over the index value) - then I just reorder them after waiting for all of them to complete.

Is this basically what Java is doing?

I think that maybe the techniques in this article are a little more complex, allowing you to optimize further (basically continue working as soon as possible instead of just waiting for everything to complete and reordering after the fact) but I’d be curious to know if I’ve missed something.

[gleenn]

It's a reasonable solution. The problem with this solution is mentioned in the article, you necessarily have the worst case memory usage because you have to store everything in the map first. If you don't have too much to store, it will work.