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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):

    use strict; use warnings; 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 1;

    In reply to Re^4: pack and unpack multiple arrays with one common repeat prefix by hexcoder
    in thread pack and unpack multiple arrays with one common repeat prefix by hexcoder

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