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]; }.
First Post!
LostKobrakai
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.
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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
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