In general, no, it does not hold. For the example I extracted only the part that I had problems with. In real life there is more structure before and after the example structure.
I also hoped to learn how flexible and concise the packing templates can be, but of course I need to digest the perlpacktut documentation now (which I did not knew existed).
Thanks for good advice.
Edit: Here is a more general example with three arrays of different sized types.
Now it is
byte: number of members in each of the following arrays
array of bytes
array of unsigned shorts
array of longs
I updated the script and added the one step solution (using the excellent explanation of Eily
my $testinput = pack('C/a* a* a*',
(pack 'C*', 1, 2),
(pack 'v*', 3, 4),
(pack 'l*', 5, 6));
print join(',', unpack('C/C* v2 l2', $testinput)), "\n";
# gives "1,2,3,4,5,6" which is ok,
# but has the repeat factors for 'v' hardcoded
my $repeat = unpack('C', $testinput);
print join(',', unpack("C/C* v$repeat l$repeat", $testinput)), "\n";
# gives "1,2,3,4,5,6" which is ok, but uses two steps
print join(',', unpack('C/C* @0 CXC /(x[C]) xX /v @0 CXC /((x[C])(x[v]
+)) xX /l', $testinput)), "\n";
# gives "1,2,3,4,5,6" uses one step, but is a bit complex