Yeah the key part is not whether something is a _stream_ (functions in Stream
module tend to return %Stream{} or fun/2) but whether something is an
_enumerable_. There’s no benefit turning, say, list into a stream, list is
already _enumerable_!
> On 24 Jul 2024, at 09:38, Robert Dober <robert.do...@gmail.com> wrote:
>
> Oh I think I get it
> Why create just a wrapper around an enum with `Stream.map(...;&(&1))` when
> you can call all the transformations on the enum anyway.
> In other words instead of `x|>Stream.new|>Stream.something|>... I just get
> rid of the Stream.new
>
> Correct?
>
> On Wednesday, July 24, 2024 at 9:22:45 AM UTC+2 Robert Dober wrote:
>> I am afraid I do not get the message José, did you mean that
>> `Stream.map(enum, &(&1))` is an antipattern?
>> Well my use case is very simple, I have a list and I want to make it lazy,
>> why should I not use an idiom like the above?
>> IIUC I should not use Stream at all for this, right? But is there a
>> different way? There is no Enum.to_stream for example.
>>
>> Hmm maybe Stream and List do not play in the same class as Stream does not
>> implement Collectable either?
>>
>> Can you help me out with this, please
>> Robert
>>
>> On Wednesday, July 24, 2024 at 12:17:22 AM UTC+2 José Valim wrote:
>>> While some functions in the Stream module may return the `Stream` struct,
>>> you must never explicitly check for the `Stream` struct, as streams
>>> may come in several shapes, such as `IO.Stream`, `File.Stream`, or
>>> even `Range`s.
>>>
>>> The functions in the Stream module only guarantee to return enumerables
>>> and their implementation (structs, anonymous functions, etc) may
>>> change at any time. For example, a function that returns an anonymous
>>> function today may return a struct in future releases.
>>>
>>> Instead of checking for a particular type, you must instead write
>>> assertive code that assumes you have an enumerable, using the functions
>>> in the `Enume` or `Stream` module accordingly.
>
>
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