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in reply to Why does this core? or How am I being a bonehead

Here's an attempt at a more elegant approach. I can't test it without sample data. 
foreach my $record ( @dataArr ) {
    my (@date) = &ParseRecur($record->[3], $base, $start, $stop);
    foreach my $line (@date) {
        @bigArr = ( @$record[0, 1, 2], $line, @$record[4] );
    }
}

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You might want to put it into a subroutine, and have it return @bigArr... not sure where that's being dealt with...

stephen

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Re: Re: Why does this core? or How am I being a bonehead
by SparkeyG (Curate) on Mar 16, 2001 at 01:46 UTC
    Brother Stephan,

    Your solution in much more elegant than mine, and it works on my data-set.
    The one thing that I'd left out was that @bigArr will be a list-of-lists. The following references to $place, a temp var, and making the assignment into a anon array fixes that problem. 

    my $place=0;

foreach my $record ( @dataArr ) {
    my (@date) = &ParseRecur($record->[3], $base, $start, $stop);
    foreach my $line (@date) {
        $bigArr[$place] = [ @$record[0, 1, 2], $line, @$record[4] ];
        $place++;
    }
}

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    --SparkeyG

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[select]
      Brother SparkeyG...

      How about this? (Minor change) 

      my @bigArr = ();

foreach my $record ( @dataArr ) {
    foreach my $line ( &ParseRecur($record->[3], $base, $start, $stop)
+ ) {
        push(@bigArr, [ @$record[0, 1, 2], $line, @$record[4] ]);
    }
}

      [download]
      <PEDANTIC>
      As a general principle: using counting variables like $place can increase program complexity unnecessarily... if you're building a list it's best to use push.

      In the same spirit, I eliminated @date, since it's only used in one place.
      </PEDANTIC>

      stephen

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