James_E

James_E

Trouble understanding octet string (binary) iteration

The Comprehensions article says:

In Elixir, it is common to loop over an Enumerable … Comprehensions are syntactic sugar for such constructs

However, this simple example of Enumerable manipulation:

# [11, 22, 33]
[1, 2, 3] |> Enum.map(fn i -> i+10*i end)

fails when the input is instead an octet string:

# (Protocol.UndefinedError) protocol Enumerable not implemented for <<1, 2, 3>> of type BitString.
<<1, 2, 3::8>> |> Enum.map(fn i -> i+10*i end)

Instead it only seems to work with the allegedly “syntactic sugar” for comprehension:

# [11, 22, 33]
<<1, 2, 3>> |> (fn s -> (for <<i::8 <- s>>, do: i+10*i) end).()

What am I doing wrong?

What do I need to do to Enumerate a BitString by chunks of a given interval, especially 8?


EDIT [X-Y Problem]:

I’m trying to do this as a component of a larger function which counts the “leading zeroes” of an octet string in a rather perverse way — the octets are counted in something like big-endian*, while the bits within each octet are counted little-endian. So I’d require <<0, 255, 0, 0>> -> 8 but <<0, 254, 0, 0>> -> 9.

*That is, in the “usual” iteration order one gets from Python’s for octet in buf or C’s for ( i=0 ; i<buflen ; i++ ){ octet = buf[i]; }.

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James_E

James_E

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
gregvaughn

gregvaughn

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

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