Quote:\032 = #32 = 20 (??)I first wondered if I made a typo, but I didn't. I think "1A" is correct.
I'm not an expert in erlang, but I guess "\032" refers to one byte, which means that this is octal notation (3 digits). Therefore "\032 = #26 = 1A".
Also I found out that I got the erlang documentation wrong: the "real" encoded data section starts after the second header which also includes 4 bytes of data size:
wings header
#!WINGS-1.0 0D 0A 1A 04 [4 bytes (compressed size + erlang header (but without wing3d header))]
erlang header
83 50 [4 bytes (uncompressed size)]
[the "real" data section starts here !]
Now if I replace the wings header and the erlang header with a gzip header and decode it I can get a way more readable format as you can see here:
https://plus.google.com/photos/+CedricWe...6377178597
I still get an error from gzip "unexpected end of file" though. I guess I've forgotten a footer. But the output data is 5808 Bytes long which exactly matches the big endian integer from the erlang header, so everything seems fine.
Thanks for your your link. This will defenitively help me. With the erlang file specification (http://www.erlang.org/doc/apps/erts/erl_ext_dist.html ) I will now try to reverse engineer that file. Shouldn't be too hard (I hope). If I succeed I will post it here.