in reply to Re^2: Perl 6, defining a hash with interpolated string
in thread Perl 6, defining a hash with interpolated string

The last two of the following examples give you interpolation, but it is more of the "normal" interpolation you would expect in Perl. 
use v6;


my %kv1 = :LiteralKey<LiteralValue>;

say :%kv1.perl;



my %kv2 = <LiteralKey>,<LiteralValue>;

say :%kv2.perl;



my $var3 = 'Literal';
my %kv3  = "{$var3}key",'LiteralValue';

say :%kv3.perl;



my $var4 = 'LiteralKey';
my %kv4  = "$var4",'LiteralValue';

say :%kv4.perl;

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gives you 
"kv1" => {"LiteralKey" => "LiteralValue"}
"kv2" => {"LiteralKey" => "LiteralValue"}
"kv3" => {"Literalkey" => "LiteralValue"}
"kv4" => {"LiteralKey" => "LiteralValue"}

[download]

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Re^4: Perl 6, defining a hash with interpolated string
by Mr_Micawber (Beadle) on Aug 04, 2010 at 21:23 UTC

    I noticed playing with this code that the output of .perl differs depending on if the colon is in front of the variable:

    say :%kv4.perl; #"kv1" => {"LiteralKey" => "LiteralValue"}
say %kv4.perl;  #{"LiteralKey" => "LiteralValue"}

    [download]

    the colon seems to be plucking the Hash's name and creating a Pair with it.

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[d/l]
[select]
      Yes that is correct. 
      my $name='Larry';
my $pair = :$name;
say $pair.perl;

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      Outputs: 
      "name" => "Larry"

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      UPDATE:
      This is especially useful for named parameters in subroutine declarations.

      For example: 
      use v6;

foo(  B => 'there',A => 'howdy');

sub foo(Str :$A, Str :$B){
        say "Named parameter A is '$A'";
        say "Named parameter B is '$B'";

}

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[reply]
[d/l]
[select]
        Thanks, that clears a lot up!
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