Why Would Anyone Use Perl 6?

Re:Why Would Anyone Use Perl 6? by awwaiid on Wednesday January 13, 2016 @08:13AM (#10893987)

The | junctions seem a lot like sets to me, though I guess there is a lot of fancy you can use. One that comes to mind is that you can pretend that it is a single value, even if it is a complex junction of & and | things:

    > (5|4) + 1
    any(6, 5)
    > (5&4) + 1
    all(6, 5)
    > my $x = 5|4|3|2
    any(5, 4, 3, 2)
    > say "yep!" if $x + 2 ~~ 7
    yep!
    > ((2|3|4|5|6|7) & (3|4)) + 1 ~~ 5
    True
    > ((2|3|4|5|6|7) & (3|4)) + 1 ~~ 6
    False

Kinda weird, especially & junctions. The docs[1] say "Junctions are meant to be used as matchers in boolean context; introspection of junctions is not supported. If you feel the urge to introspect a junction, use a Set or a related type instead."

So far in my Perl 6 exploration I don't think I've used/needed junctions -- strikes me as a specific sort of use-case, but maybe familiarity would help me spot more (useful) opportunities.

[1] http://doc.perl6.org/type/Junction, though like all P6 docs needs more contributions

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Re:Why Would Anyone Use Perl 6? by akavi on Wednesday January 13, 2016 @10:25AM (#10894936)

If I understand correctly, it's a nice literal syntax for a vector/set hybrid data structure?

If you wanted, you could get pretty close in ruby with only a sprinkle of ill-advised metaprogramming.

    > junc{5|4} + 1
    any(6, 5)
    > junc{5&4} + 1
    all(6, 5)
    > x = junc{5|4|3|2}
    any(5, 4, 3, 2)
    > puts "yep!" if x + 2 =~ 7
    yep!
    > junc{(2|3|4|5|6|7) & (3|4)} + 1 =~ 5
    True
    > junc{(2|3|4|5|6|7) & (3|4)} + 1 =~ 6
    False

would be possible.

I guess I'm just curious if this is a situation that comes up often enough to need a built in literal syntax.

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