[erlang-questions] double to float

[Sten Kvamme]

Thanks, that made it!

/Sten

On Oct 18, 2007, at 20:08 , Tony Rogvall wrote:

> Erlang defaults to big endian (historical reasons) numbers that is  
> high order bytes first,
> for intel like machines they use low order bytes first.
>
> use  X/little-float
>
> /Tony
>
> On 18 okt 2007, at 19.18, Sten Kvamme wrote:
>
>> Vlad pointed in the right direction, so it's working now (thanks
>> Vlad!). But it still puzzles me that I have to mirror the 8 bytes to
>> have erl accepting them as a float. In C I just do like this:
>> fread(Xrealp, sizeof(Xreal), 1, fpdxf);
>>
>> Here's the erlang code snippet:
>>
>> ac_double(Thefile) ->
>> 	<<X:8/binary, B/binary>> = Thefile,
>>     <<X1,X2,X3,X4,X5,X6,X7,X8>> = X,
>>     <<Y/float>> = <<X8,X7,X6,X5,X4,X3,X2,X1>>,
>> 	{B,Y}.
>>
>> Here is the whole thing (remember I have just bought the book ;-)
>> http://www.kvamme.se/pub/dxf.erl.txt
>>
>> And here's a small binary dxf file to run on it.
>> http://www.kvamme.se/pub/b.dxf
>>
>> I'm learning fast and it's real fun. I can truly recommend the book.
>>
>> /Sten Kvamme
>>
>> On Oct 16, 2007, at 20:43 , Vlad Dumitrescu wrote:
>>
>>> Hi and welcome,
>>>
>>> On 10/16/07, Sten Kvamme <sten@REDACTED> wrote:
>>>> I'm reading a binary file with lots of double data types. Have
>>>> someone already written some code to convert the double 8 bytes,
>>>> <<184,162,71,68,75,65,104,64>> to a float?
>>>
>>> Do you mean
>>>    <<X/float>> = <<184,162,71,68,75,65,104,64>>
>>> ?
>>>
>>> regards,
>>> Vlad
>>
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