Write while loop equivalent in elixir

aadeshere1

I have a another noob question about loop. Since elixir is immutable, while loop is not directly possible.

total = 10
while total != 0
   puts "hello"
   total -= 1
end

The question and answer in this example use list for loop. Convert ruby while loop into elixir

36 comments

#while #loop

3 29716 36

2023-10-19 09:05:29 UTC

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sribe

I just want to comment on something, at first glance all the solutions posted on this page look kind of complicated for such a simple thing. But you almost never need to write such code in the real world.

In actual apps, you want to iterate over some collection of data items. The for or while loop is how you do that in an imperative language: you have some “iterator” kind of value–a simple index for an array, something else for a map or set–increment that, check if it’s out of range, use it to access your data collection.

In functional languages, you pass the operation you want to perform on the data as an argument to one of the collection’s functions. In Elixir, Enum.map or Enum.reduce. This actually results in more compact code because it eliminates the whole “get an iterator, increment, check it, use it” dance.

So while this particular exercise can help you understand some low-level mechanics of Elixir, it could also be misleading. Please don’t think you have to jump through the hoops of all these solutions in order to do something with an array of values you get back from your database or submitted from a web form

That said, here is my solution:

Enum.each(10..1, fn(i) -> IO.puts(i) end)

or, using some shorthand:

Enum.each(10..1, &(IO.puts("#{&1}")))

or, if you want to prove you actually understand passing around functions:

Enum.each(10..1, &IO.puts/1)

But even that could be misleading, because in a functional language you don’t often just “run a loop” over a collection for side effects, you usually produce a value, either a set of values produce from each source value (Enum.map) or a single value derived from the whole set (Enum.reduce).

Post #6

LostKobrakai

I found Stream.unfold to be quite useful for places where I don’t know how many iterations are needed to get to the end.

stream = Stream.unfold(10, fn x -> 
  if x != 0 do
    IO.puts "Hello"
    {:ok, x - 1}
  else
    nil
  end 
end)

Stream.run(stream)

There’s also Enum.reduce_while, but it needs an enumerable as input to start with.

Post #1

benwilson512

Author of Craft GraphQL APIs in Elixir with Absinthe

@silverdr OK sure, here’s a non manual loop version:

Stream.repeatedly(&generate_name/0) |> Stream.take(16) |> Enum.find(&not_taken?/1)

This will return the first generated name that isn’t taken, OR quit after 16 tries. It will only generate as many names as are needed. Enumerable works great here too. Of course, you can always use manual recursion.

Post #28

Last Post!

thbar

An equivalent with potentially infinite running time (useful when polling an API for results in a quick Mix.install/2 script):

Stream.unfold(1, fn
  acc ->
    :timer.sleep(1_000)
    IO.puts("Waiting...")
    # TODO here: poll API and decide based on output status whether to leave or not
    if acc >= 5, do: nil, else: {acc, acc + 1}
end)
|> Stream.run()

Thanks @LostKobrakai for the inspiration, I wouldn’t have thought about Stream.unfold for this!

Post #36