The following code:
use v6;
my $str = 'abc';
sub s {1};
say s;
$str ~~ s:g/ b /x/;
dd $str;
say $/;
outputs:
1
Str $str = "axc"
(｢b｣)
as expected.
But, just remove the :g global flag and:
===SORRY!=== Error while compiling /home/hogaboom/hogaboom/Perl6/p6ex/./t.p6
Undeclared routine:
Short answer: Yes.
Longer: Perl 6 allows you to over-ride the names of routines. 's' is a
routine. You over-rode it.
Perl 6 is different from most other languages because it uses multiple
dispatch. Effectively this means it is not just the name of the
subroutine (s) that matters, but also it
Hi Richard,
I wouldn't call s/// a routine. It's actually (implemented as) a kind of
quote, like Q, q, qq, qw, and also rx, and tr. I consider that an
important distinction, because the syntax with which you call s/// is
very different from how you call a sub. You can't just `say '(' ~ ')'
1234`,
