I wasn't even trying to guess what it would really parse as so I just Deparse and run it

$ perl -MO=Deparse
print (two + two == five ? "true" : "false")
^Z
print two 'two' == 'five' ? 'true' : 'false';
- syntax OK

$ perl -MO=Deparse,-p
print (two + two == five ? "true" : "false")
^Z
print(two (('two' == 'five') ? 'true' : 'false'));
- syntax OK

$ perl -w
print (two + two == five ? "true" : "false")
print (...) interpreted as function at - line 1.
Unquoted string "two" may clash with future reserved word at - line 1.
Unquoted string "two" may clash with future reserved word at - line 1.
Unquoted string "five" may clash with future reserved word at - line 1
+.
^Z
Name "main::two" used only once: possible typo at - line 1.
Argument "five" isn't numeric in numeric eq (==) at - line 1.
Argument "two" isn't numeric in numeric eq (==) at - line 1.
print() on unopened filehandle two at - line 1.
[download]

I think B::Deparse gets it wrong -- as it sometimes does -- when you add -p:

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This is confirmed by the more low-level output of B::Concise:

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Dave

Look at the way Deparse uses whitespace, its subtle

My theory: When Deparse tries to add parens wherever it can, it treats the first two as a function call (which was my first guess too). But as tobyink correctly surmised, that first two is treated by print as a filehandle.

That is not the way I read it :) Usually, Deparse doesn't add extraneous space to function calls, its always "foo( ... )" its never "foo ( ... )"

so the way I read that deparse output is as filehandle not function call

For example

$ perl -MO=Deparse 2
sub two {
    rand;
}
print two(two() == 'five' ? 'true' : 'false');
2 syntax OK

$ perl -MO=Deparse,-p 2
sub two {
    (rand);
}
print(two(((two() == 'five') ? 'true' : 'false')));
2 syntax OK
[download]

When I look at Re^2: Tiny Perl puzzle I see  print( filehandle ... );

I don't think so.

did you recompile the output?

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Cheers Rolf

(addicted to the Perl Programming Language)