print two( 0==0 ? "true" : "false" );
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Good puzzle. This sort of thing is precisely the reason various best practices have arisen.

This surprises any even moderately experienced Perl programmer?

Update: It appears that my test didn't sufficiently accurately replicate the original program text. Though I admit this particular quirk didn't and doesn't bother me much as I avoid barewords exactly because they lead to lots of ambiguities.

Can't test ATM :)

update

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

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.

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

$ perl -MO=Concise,-exec, -e 'print ( two + two == five ? "true" : "fa
+lse" )'
1  <0> enter 
2  <;> nextstate(main 1 -e:1) v:{
3  <0> pushmark s
4  <$> gv(*two) s
5  <1> rv2gv sKR/1
6  <$> const(PV "two") s/BARE
7  <$> const(PV "five") s/BARE
8  <2> eq sK/2
9  <|> cond_expr(other->a) lK/1
a      <$> const(PV "true") s
           goto b
d  <$> const(PV "false") s
b  <@> print vKS
c  <@> leave[1 ref] vKP/REFC
-e syntax OK
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It's easy to see from this output that the first two is treated as a typeglob; a filehandle. The second two and the five are treated as barewords; constants in this case, which are then compared to one another. They evaluate to the same constant defined value, which is probably an empty string, equating to 0 for the purpose of numeric equality. So "true" will result, but it's printed to a filehandle that hasn't been attached to anything.




Dave

print (two + two == five)
  ? ( user == politician) 
     ? "Re-elect me and I will fix it."
     : "It's not a bug, it's a feature!"
  : (user == accountant) 
    ? "What do you -want- the answer to be?"
    : "Don't worry, sonny, no child will be left behind.™" 
;
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Another incorrect Perl puzzle?

(hint: print greedily loves brackets! :)

Cheers Rolf

(addicted to the Perl Programming Language)

(Please mark spoilers accordingly.)

sundialsvc4, why not honor the OPs request and use spoiler tags?