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[Solved][HOP] Working with Iterators, why do I suddenly get a coderef?

by three18ti (Scribe)
on Nov 03, 2013 at 00:29 UTC ( #1060998=perlquestion: print w/ replies, xml ) Need Help??
three18ti has asked for the wisdom of the Perl Monks concerning the following question:

Hello Monks,

I'm working with Higher Order Perl and I'm working with the section on iterators. One of the chapters is a function that takes iterators as arguments and when one iterator runs out of values the function returns the next iterator.

I understand the code, but I'm getting some odd results. most of the time, append returns the next iterator, but sometimes it returns a reference to the iterator. for instance, I have to call $it6->()->() to get the second iterator in "Invalid append 4" below.

Why does the append seem to behave differently under seemingly similar circumstances?

#!/usr/bin/env perl use strict; use warnings; use Data::Dumper; $Data::Dumper::Deparse = 1; # syntactic sugar sub NEXTVAL { $_[0]->() } sub Iterator (&) { return $_[0] } # turn list into iterator sub list_iterator { my @list = @_; return Iterator { shift @list } } # create iterator that returns values from $a to $b sub upto { my $start = shift; my $end = shift; return Iterator { return $start <= $end ? $start++ : undef ; }; } # append iterators together sub append { my @its = @_; return Iterator { while (@its) { my $val = NEXTVAL $its[0]; return $val if defined $val; shift @its; } return; }; } # helper function to print iterator values sub print_it { my $it = shift; while (my $var = NEXTVAL $it) { print Dumper $var; } } print "Valid append 1\n"; my $it1 = append upto(1,3), upto(5,7), upto(9,10); print_it $it1; print "\n\nValid append 2\n"; my $ita = list_iterator qw(foo bar baz); my $itb = list_iterator qw(zab rab oof); my $itc = upto(1,3); my $it2 = append $ita, $itb, $itc; print_it $it2; print "\n\nInvalid append 1\n"; my $it3 = append upto(1,3), list_iterator qw(foo bar baz), upto (7,9); print_it $it3; print "\n\nInvalid append 2\n"; my $it4 = append list_iterator qw(foo bar baz), upto(1,3), upto(7,9); print_it $it4; print "\n\nInvalid append 3\n"; my $it5 = append list_iterator qw(foo bar baz), list_iterator qw( a b +c ); print_it $it5; print "\n\nInvalid append 4\n"; my $it6 = append list_iterator qw(foo bar baz), list_iterator qw( a b +c ); $it6->(); $it6->(); $it6->(); # prints "a" print $it6->()->() . "\n";

Sample output

Valid append 1 $VAR1 = 1; $VAR1 = 2; $VAR1 = 3; $VAR1 = 5; $VAR1 = 6; $VAR1 = 7; $VAR1 = 9; $VAR1 = 10; Valid append 2 $VAR1 = 'foo'; $VAR1 = 'bar'; $VAR1 = 'baz'; $VAR1 = 'zab'; $VAR1 = 'rab'; $VAR1 = 'oof'; $VAR1 = 1; $VAR1 = 2; $VAR1 = 3; Invalid append 1 $VAR1 = 1; $VAR1 = 2; $VAR1 = 3; $VAR1 = 'foo'; $VAR1 = 'bar'; $VAR1 = 'baz'; $VAR1 = sub { use warnings; use strict 'refs'; return $start <= $end ? $start++ : undef; }; Invalid append 2 $VAR1 = 'foo'; $VAR1 = 'bar'; $VAR1 = 'baz'; $VAR1 = sub { use warnings; use strict 'refs'; return $start <= $end ? $start++ : undef; }; $VAR1 = sub { use warnings; use strict 'refs'; return $start <= $end ? $start++ : undef; }; Invalid append 3 $VAR1 = 'foo'; $VAR1 = 'bar'; $VAR1 = 'baz'; $VAR1 = sub { use warnings; use strict 'refs'; shift @list; }; Invalid append 4 a

What gives?

Thanks for the assistance.

I've created a gist so you can download and run the code easier: https://gist.github.com/three18ti/7285070

Comment on [Solved][HOP] Working with Iterators, why do I suddenly get a coderef?
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Re: [HOP] Working with Iterators, why do I suddenly get a coderef?
by LanX (Canon) on Nov 03, 2013 at 00:42 UTC
    I'd suggest to use parens when calling list_iterator()

    Sorry can't test while mobile...

    Cheers Rolf

    ( addicted to the Perl Programming Language)

      Hey Rolf,

      Thanks, you nailed it! I was reading the parenthesis around qw() and thinking that was the parenthesis for the sub call, but it's not, it's the parenthesis around the list.

Re: [HOP] Working with Iterators, why do I suddenly get a coderef?
by choroba (Abbot) on Nov 03, 2013 at 00:42 UTC
    I am not sure I understand the question, but adding more parentheses might change the results you get. You can use B::Deparse (used by Data::Dumper in your code, anyway) to see how Perl adds the parentheses to your code. For example, in "Invalid append 1", the following line
    my $it3 = append(upto(1,3), list_iterator( qw(foo bar baz)), upto (7,9 +));

    changes the output to

    Invalid append 1 $VAR1 = 1; $VAR1 = 2; $VAR1 = 3; $VAR1 = 'foo'; $VAR1 = 'bar'; $VAR1 = 'baz'; $VAR1 = 7; $VAR1 = 8; $VAR1 = 9;
    لսႽ ᥲᥒ⚪⟊Ⴙᘓᖇ Ꮅᘓᖇ⎱ Ⴙᥲ𝇋ƙᘓᖇ

      Hello,

      Thanks for your reply. Interestingly, I'm not getting a different result when adding parenthisis for the append() call, but when I use parenthesis for the call to list_iterator, it DOES fix my problem.

      my $it3 = append (upto(1,3), list_iterator qw(foo bar baz), upto (7,9) +); print_it $it3;

      prints:

      Invalid append 1 $VAR1 = 1; $VAR1 = 2; $VAR1 = 3; $VAR1 = 'foo'; $VAR1 = 'bar'; $VAR1 = 'baz'; $VAR1 = sub { use warnings; use strict 'refs'; return $start <= $end ? $start++ : undef; };

      But when I use parenthesis around list_iterator:

      print "\n\nInvalid append 1\n"; my $it3 = append upto(1,3), list_iterator (qw(foo bar baz)), upto (7,9 +); print_it $it3;

      I get what I would expect

      Invalid append 1 $VAR1 = 1; $VAR1 = 2; $VAR1 = 3; $VAR1 = 'foo'; $VAR1 = 'bar'; $VAR1 = 'baz'; $VAR1 = 7; $VAR1 = 8; $VAR1 = 9;

      Thanks for the assist! I think the problem is trying to make too many calls to subs without parenthesis.

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