in reply to Non-destructive array processing

Isn't this what local is for? Something like the following;

my @array = 1..10;
{
  local @array = @array;
  while (my @chunk = splice @array, 0, 2) {
    print "Chunk: @chunk\n";
  }
}
print "Original array is still intact! (@array)\n";

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Re^2: Non-destructive array processing
by Aristotle (Chancellor) on Jan 22, 2003 at 16:09 UTC
    Nearly. 
    local *array = \@array;

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    This will work with lexically scoped variables too.

    Makeshifts last the longest.

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      But what will simply alias the lexical @array with the dynamical @array. So I don't see what you've achieved by doing this.

      One problem with this that you probably didn't foresee is that lexicals are resolved before dynamic variables. Example: 

        my @foo = 1..4;
  local *foo = ['a'..'d'];
  print @foo; # 1234

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      The problem is solved through our() since that creates an aliased lexical: 
        my @foo = 1..4;
  our @foo = 'a'..'d';
  print @foo; # abcd

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      ihb
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Re: Re: Non-destructive array processing
by jdporter (Paladin) on Jan 21, 2003 at 22:24 UTC
    Nope, because local() doesn't work on my variables. Nice thought, though.

    jdporter
    The 6th Rule of Perl Club is -- There is no Rule #6.

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Re: Re: Non-destructive array processing
by ihb (Deacon) on Jan 22, 2003 at 19:01 UTC
    No, but you can use my: 
      my @array = 1..10;
  {
    my @array = @array;
    while (my @chunk = splice @array, 0, 2) {
      print "Chunk: @chunk\n";
    }
  }
  print "Original array is still intact! (@array)\n";

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    Note though that modifications of an @array element (i.e. via $array[$n]) will disappear when the scope is left. This is simply because the inner @array simply is another variable with the values copied. The idea of Juerd's routine was that the elements would be aliased but the array different.

    Hope I've helped,
    ihb
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