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I tried to see if I could find a generic way of transforming a recursive function into an iterator. I found a method that's can be applied rather mechanically.

Let's say your recursive function visits a tree in preorder. The following would be the a non-iterative solution:

sub visit_preorder(&$) { my ($visitor, $node) = @_; my ($val, $l, $r) = @$node; $visitor->() for $val; &visit_preorder($visitor, $l) if defined $l; &visit_preorder($visitor, $r) if defined $r; }

Change the visitor code to return \$val; where $val is the data to return from the iterator. return \@val; is also acceptable if you wish to return more than one value.

Break up the function where a recursive call is made. Return each block as a function as follows:

sub visit_preorder { my ($node) = @_; my ($val, $l, $r) = @$node; return ( sub { return \$val; }, sub { return visit_preorder($l) if defined $l; return; }, sub { return visit_preorder($r) if defined $r; return; }, ); }

In this case, it can be simplified a little.

sub visit_preorder { my ($node) = @_; my ($val, $l, $r) = @$node; return ( \$val, defined($l) ? sub { return visit_preorder($l); } : (), defined($r) ? sub { return visit_preorder($r); } : (), ); }

Now we just need an engine to drive this.

sub make_iter { my $f = shift @_; my @todo = $f->(@_); return sub { while (@todo) { my $todo = shift @todo; if (ref($todo) eq 'CODE') { unshift @todo, $todo->(); } elsif (ref($todo) eq 'ARRAY') { return @$todo; } else { return $$todo; } } return; }; }

Finally, an example caller.

{ # a # / \ # b e # / \ \ # c d f my $tree = [ 'a', [ 'b', [ 'c', undef, undef, ], [ 'd', undef, undef, ], ], [ 'e', undef, [ 'f', undef, undef, ], ], ]; my $iter = make_iter(\&visit_preorder, $tree); while (my ($name) = $iter->()) { print($name); } print("\n"); }

In reply to Re: Question about recursively generated iterators by ikegami
in thread Question about recursively generated iterators by perlfan

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