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 its signature (the thing that
follows the name and before the block {...} defining its action).
Your declaration of 'sub s {1}' defined a routine with a zero length
signature.
When you wrote '$str ~~ s:g/ b /x/', you were ultimately calling a
subroutine 's' with a signature that has adverbs (:g) and a regex.
But without the ':g', the compiler choose your subroutine 's', which
meant that it needed to decide what 'b' was, but couldn't.
This is a good thing because it means that you can create a subroutine
's' for your own objects without the compiler blowing up and telling you
that you can't use 's'.
On 23/10/18 04:52, Richard Hogaboom wrote:
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:
b used at line 10
results.
Is the sub name s() interfering with the s/// somehow, but s:g/// does
not?
rahogaboom
