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in reply to Re^2: pack and unpack multiple arrays with one common repeat prefix
in thread pack and unpack multiple arrays with one common repeat prefix

So, for clarity, does the following always hold? Because, from the Perl perspective, that's functionally equivalent to: and thus C/C* v* does everything you need.

#11929 First ask yourself `How would I do this without a computer?' Then have the computer do it the same way.

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Re^4: pack and unpack multiple arrays with one common repeat prefix
by hexcoder (Curate) on Jun 15, 2017 at 22:06 UTC
    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;

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