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:

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
[download]

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

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?

<Reveal this spoiler or all in this thread>

Cheers Rolf

(addicted to the Perl Programming Language)