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