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It is easy when << returns self:
A << (B << C) \
<< (D << E)
This makes A with two children B and C each with a
If you want a path (tree without branches),
you can still do
A << (B << (C << D)))
Doh! Of course.
It has to return self.
I thin
Bugs item #21658, was opened at 2008-08-24 13:01
You can respond by visiting:
http://rubyforge.org/tracker/?func=detail&atid=1971&aid=21658&group_id=494
Category: None
Group: None
Status: Open
Resolution: None
Priority: 3
Submitted By: Nobody (None)
Assigned to: Nobody (None)
Summary: failure to
Bugs item #21658, was opened at 2008-08-24 13:01
You can respond by visiting:
http://rubyforge.org/tracker/?func=detail&atid=1971&aid=21658&group_id=494
Category: None
Group: None
Status: Open
Resolution: None
Priority: 3
Submitted By: Nobody (None)
Assigned to: Nobody (None)
Summary: failure to
Bugs item #21658, was opened at 2008-08-24 10:01
You can respond by visiting:
http://rubyforge.org/tracker/?func=detail&atid=1971&aid=21658&group_id=494
Category: None
Group: None
Status: Open
Resolution: None
Priority: 3
Submitted By: Nobody (None)
Assigned to: Nobody (None)
Summary: failure to
