1. yes
2. no and yes (perl treats while1 as one identifier)
3a. yes
8a. yes, cool, unexpected solution :-)
our official version uses a much much more obscure trick.. funny we never saw this one
Try this instead: eval(q[$$foo]) ne eval(q["$$foo"])
•Update: actually, no, the original was fine too.. even though $$foo and $$foo + 0 print differently, they don't compare numerically unequal, which is what the exercise was
11. doesn't appear to work for me..
% perl -e 'sub STDOUT{\*STDOUT} <STDOUT> eq <+STDOUT> or die'
Died at -e line 1.
12. yes, but why? :-)
•Update: fixed the "try this".. initially said "!=" instead of "ne"
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