Re: Extending Array

$thing is a hash reference, not an array. See perldata and perlreftut for some more information.

Two ways to extend it:

$thing = {
    %$thing,
   'C' => 'cutie',
   'D' => 'pie'
};

# OR

$thing->{C} = 'cutie';
$thing->{D} = 'pie';
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The first one generates a new hash and a new reference to it, the second adds items to the existing hash. Which one you should use depends on your use case (and is only relevant if there are other references to the same hash somewhere).

Perl 6 - second systems done right

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Re^2: Extending Array by sriharsha.sm (Initiate) on Mar 21, 2012 at 18:53 UTC
Why do you want to create a new hash? Just adding to the old hash ref won't work? Can you give a scenario where do you use 1st case?	[reply]
Re^3: Extending Array by moritz (Cardinal) on Mar 21, 2012 at 19:07 UTC
Can you give a scenario where do you use 1st case? If I write a subroutine that receives a hash reference as an argument, I wouldn't modify it in-place, because the caller might want to re-use it. Simplified example: `# caller: my %h = (a => 1, b => 2); something(\%h); say $_ for keys %h; # callee sub something { my $arg = shift; $arg = { %arg, c => 3, d => 4, }; # do more stuff with $arg here }` [download] If you use `$arg->{c} = 3` in sub `something`, `%h` changes in the caller's code, which is very likely to cause confusion. Perl 6 - second systems done right	[reply] [d/l] [select]