data. On closer inspection (should have done that before, sorry),
the procedure uninterleave that you gave is a direct translation of
that code, and it also uses little endian (the high byte is at
address i*4+1, the low byte at i*4, hence the little byte first,
little endian). Procedure uninterleave_little_endian uninterleaves
as well, but assumes that the machine it's running on is little
endian, so it can get away with casting two bytes to a 16-bit
integer, assuming that the input is little endian. The "official"
encoder example must work on big endian as well as little endian
machines, so it can't use the second trick. Delphi only works on
x86 anyway, so you may as well use the faster version with casts.
Conclusion, your input file is little endian. Since you're running
on an x86 machine, both procedures are equivalent. If you would
compile and run your code on a big endian machine with the same
little-endian file, the procedure uninterleave would give correct
results, procedure uninterleave_little_endian would not.
HTH,
Lourens
--
GPG public key: http://home.student.utwente.nl/l.e.veen/lourens.key
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