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Re^3: How to return a list using recursion?

by Anonymous Monk
on Aug 07, 2012 at 15:04 UTC ( #986016=note: print w/replies, xml ) Need Help??


in reply to Re^2: How to return a list using recursion?
in thread How to return a list using recursion?

With older versions of Perl, it can be rewritten as a proper iterator :-)

use strict; use warnings; my @list = ('a' .. 'z'); my $size = 5; my $chunker = _array_chunk_it($size, @list); while (my @chunk = $chunker->()) { print "Array chunk of $size =\n", join("\n", @chunk), "\n\n"; } sub _array_chunk_it { my ($count, @array) = @_; return sub { my @chunk; for (1 .. $count) { push(@chunk, shift @array) if @array; } return @chunk; } }

(Using a closure we can actually call _array_chunk_it multiple times, in an overlapping manner. Have multiple iterators running simultaneously. Plus it looks quite a bit nicer.)

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[choroba]: originally, you sent a pull request to someone by email, i.e. you asked them to pull from your repo
[Discipulus]: if is the author of a patch, is more like a 'request to push' into rather than a pull request..
[choroba]: as it's considered unpolite to push to someone else's repo
[choroba]: and in fact, you don't push into the target repo, you push into your branch of your fork
[choroba]: the maintainer of the upstream repo than "merges" the pull request, i.e. they pull from your fork into the upstream
[Discipulus]: ' i.e. you asked them to pull from your repo' =~ I (subj) want to push
[Discipulus]: chorobayour words are reasonable
[choroba]: I'm just repeating some else's words as I remembered them after having asked the same question
[choroba]: s/some/someone/
Discipulus what a pity all people do not speak only in eatalian..

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