[erlang-questions] Re. How to decode octet to septets using erlang (prasantha kumara)

[Hynek Vychodil]

Well, GSM using weird coding.
This similar to your solution is:

unpack(UD) ->
    << <<0:1,A:1,B:1,C:1,D:1,E:1,F:1,G:1>> ||
        <<G:1,F:1,E:1,D:1,C:1,B:1,A:1>> <=
            << <<A:1,B:1,C:1,D:1,E:1,F:1,G:1,H:1>> ||
<<H:1,G:1,F:1,E:1,D:1,C:1,B:1,A:1>> <= UD >>
    >>.

It is 2-10 times faster on my laptop (depend of GC hit) in byte code and 4
times in native.

2009/3/5 Tony Seebregts <tony.seebregts@REDACTED>

> Hmmm , the short answer is that the code I suggested is taken from a module
> to convert a GSM octet stream into a septet stream and it gives a different
> answer when using your more compact code :-) :
> UD:      << 203,176,184,204,86,191,1 >>
> Mine:    [75,97,98,101,108,106,111,0]
> Yours:  << 101,108,23,12,98,90,126,1>>
>
> I remember the GSM octet-septet conversion being quite awkward to get right
> because the bytes were the 'wrong way round'  which is probably why I ended
> up with the code I did i.e. having to use flip(...).
>
>
> 2009/3/5 Hynek Vychodil <vychodil.hynek@REDACTED>
>
>> Why not?
>>
>>
>> unpack(UD) ->
>>   << <<0:1,X/bits>> || <<X:7/bits>> <= UD >>.
>>
>> 2009/3/5 Tony Seebregts <tony.seebregts@REDACTED>
>>
>>> This is an extract from the code I use:
>>>
>>>> unpack(UD) ->
>>>>    septets(list_to_binary([ flip(X) || X <- binary_to_list(UD) ])).
>>>>
>>>> septets(T) ->
>>>>     [ flipx(X) || <<X:7>> <= T].
>>>>
>>>> % Flips the bits in an octet/septet for easy unpacking
>>>>
>>>> flip(Octet) ->
>>>>    <<B8:1,B7:1,B6:1,B5:1,B4:1,B3:1,B2:1,B1:1>>  = <<Octet>>,
>>>>    <<B1:1,B2:1,B3:1,B4:1,B5:1,B6:1,B7:1,B8:1>>.
>>>>
>>>
>>> There is probably a better way of doing it but I was in a rush at the
>>> time :-)
>>>
>>>
>>>
>>>
>>>
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>>>
>>
>>
>>
>> --
>> --Hynek (Pichi) Vychodil
>>
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>
>
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-- 
--Hynek (Pichi) Vychodil

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