use Perl

• array (Score:0)

  by slanning (5049)

  Seems to be returning the the last part:

  DB<13> @r = qw(this that these 17)
  
  DB<14> x {%$r}{@r}
  0  'this'
  1  'that'
  2  'these'
  3  17
  
  DB<15> x {@r}
  0  'this'
  1  'that'
  2  'these'
  3  17
  
  DB<16> x @r
  0  'this'
  1  'that'
  2  'these'
  3  17
  • Re: (Score:2)

    by jdavidb (1361)

    You're right. I'm getting keys, not values. I should pay more attention to detail. :)

    So what about the debugger is making it ignore the first part in braces, I wonder.

    --
    J. David works really hard, has a passion for writing good software, and knows many of the world's best Perl programmers
• My way of rembering it... (Score:1)

  by jarich (4909)

  This might not help, but I remember hash slices as follows. First really it's just a hash:

  my $hashref = {a => 1, b => 2, c=> 3};
  
  # it's just a hash:
  %$hashref;

  Second we do our lookup:

  my @of_interest = qw(a c);
  
  %$hashref{@of_interest};

  Slices return lists, so we need an @ sign instead of the %:

  @$hashref{@of_interest};

  Which gives us:

  use strict;
  use warnings;
  
  my $hashref = {a => 1, b => 2, c=> 3};
  my @of_interest = qw(a c);
  
  print "@$hashref{@of_interest}\n";

  You can go the other di
• Parsed as blocks (Score:1)

  by tonyc (123)

  The expression you're supplying to the debugger is being parsed as a block:

  tony@zeus:~$ perl -MO=Deparse -e '{%$r}{@f}'
{
        %$r;
}
{
        @f;
}
-e syntax OK

  eval()ling the same expression gives the result you saw in the debugger:

  tony@zeus:~$ perl -e '$r = { this => "that", these => 17 }; @f = qw(this these); @foo = eval q"{%$r}{@f}"; use Data::Dumper; print Dumper \@foo'
$VAR1 = [
                    'this',

  • Re: (Score:2)

    by jdavidb (1361)

    Ah! Thank you for the explanation!

    --
    J. David works really hard, has a passion for writing good software, and knows many of the world's best Perl programmers