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In Perl 5 it would be 'sub { }', isn't it? (Score:1)
If the Counter / LazyIterator example was written in Perl 5, one would use '
sub { $count++ }', isn't it? Why the same method, i.e. 'sub { }' as circumfix operator for block in Perl 6?P.S. Would you resubmit this blog post to http://perlgeek.de/blog-en [perlgeek.de] for it to be easier to find?
Re:In Perl 5 it would be 'sub { }', isn't it? (Score:1)
If the Counter / LazyIterator example was written in Perl 5, one would use 'sub { $count++ }', isn't it? Why the same method, i.e. 'sub { }' as circumfix operator for block in Perl 6?
Yeah,
sub { }would be the nearest equivalent in Perl 5. It creates a reference to a closure. You can write that in Perl 6 as well, but thesubkeyword gives the closure a bit more of a "shell", as outlined in this post [perl.org].P.S. Would you resubmit this blog post to http://perlgeek.de/blog-en [perlgeek.de] for it to be easier to find?
If you want to read Moritz's and my blog posts in the same feed, I suggest you check out Planet Perl 6 [perl.org].
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