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"
+)}'

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-- Abigail

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Re: Re: Brain Teaser
by srawls (Friar) on Jun 11, 2001 at 00:26 UTC
    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

[reply]
[d/l]
      Eh, perhaps on *your* system, but look at this: 
          $ perl -le 'print \\///\\\//'
    1.00000026558336981

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      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

[reply]
[d/l]
[select]
        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

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