No subject

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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