Trouble understanding octet string (binary) iteration

The bit/binary syntax usable with for is a feature of for with it being a special form. Binaries/bitstrings do not implement the Enumerable protocol. The issue is that there’s no clear cadance as to where a binary or bitstring should be split at. for knowing what syntax is on the matching side can adjust that automatically. With Enumerable you however don’t know anything about the consumer of the individual pieces. You therefore need to manually split the binary or bitstring up.

I see. So it’s not only syntactic sugar; in certain cases it actually does new things?

Is there another way to get at the enumeration-of-bitstrings? (Of course, I understand I’ll need to specify the desired element “width” in any case.)

The function I’m trying to build will need early-exit; I want to iterate through the octet string only until I find a non-0 octet, then return. Is there any “lazy” way to go iterate through an octet-string — that is, without first “greedily” slurping the entire string up into a List like for seems to do?

Hmm, I think I figured it out, now.

def binary_to_stream(s) do
  Stream.unfold(s, fn
    <<>> -> nil
    <<next::8, rem::binary>> -> {next, rem}
  end)
end

You can use Stream.unfold to turn a binary into an enumerable of “pieces”:

<<1, 2, 3>>
|> Stream.unfold(fn 
    <<i::8, r::binary>> -> {i, r}
    <<>> -> nil
end)
|> Enum.into([])
# [1, 2, 3]

You can, but it’s worth noting that for is going to produce more optimal code than wrapping this in a stream will do.

If the early exit really makes a difference, you could also do explicit recursion.