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