Re: Brain Teaser

[Abigail]

in reply to Brain Teaser

It evaluates to 0 (or rather, false) only if the print fails - otherwise, it will evaluate to true (1 in practise). ;-)

perl -e '* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
+ * *
         / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / /
+ / /
         % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % %
+ %;
         BEGIN {% % = ($ _ = " " => print "Just Another Perl Hacker\n"
+)}'

-- Abigail

[srawls]

Tricky Abigail, but I meant what does \\///\\\// evaluate to (not print)--mabey I should have explained myself better. And again, that was a slip, I meant to say that it evaluates to 0 in integer context, in numeric context it evaluates to just under 1.

The 15 year old, freshman programmer,
Stephen Rawls

[Abigail]

Eh, perhaps on *your* system, but look at this:

    $ perl -le 'print \\///\\\//'
    1.00000026558336981

What the result of \\///\\\// is depends on how memory is alloced. It'll depend on your malloc (which could be the Perl supplied, or the system one, depending on how Perl was build); it'll hand out addresses using some algorithm, based on various factors. Don't assume that when you run it once on your system, it will be the same universally.

-- Abigail

[srawls]

Well, I ran it on DOS and UNIX, and it worked. I was hoping that since the memory was allocated closely it would be below 1; oh well--nothing can be 100% portable : )

The 15 year old, freshman programmer,
Stephen Rawls