[Omaha.pm] not defined $bucket || $bucket eq '???'

Jay Hannah jay at jays.net
Wed Dec 3 16:50:32 PST 2008

From: brandonglesmann at gmail.com
Date: December 3, 2008 4:42:30 PM CST
> Disclaimer: I have been out of perl programming for over two years.  
> So, I am really rusty. Also I never used the perl debugger so I  
> could be misinterpreting the output. Lastly, I am a coward which is  
> why I did not reply all.

Well, thanks for saying it was OK for me to drag you into the  
sunshine so all can benefit from my stupidity.  :)

> WOW! Ok, I read that 10 times and have asked other perl programmers  
> in the area to read it as well.

Other Perl programmers? Sweet! Are they on the Omaha Perl Mongers  
mailing list? Recruit them!  :)

> couple questions:
> 1. How are the cases different? If ' ' is true and then both cases  
> of your example should be the same.......right?!?! Obviously not  
> since you changed the code but I don't know why.

' ' is very different from ''.   :)

I'm often a hunt and peck coder, so whatever perl does is the right  
answer. I don't know that there necessarily is a "why". (Other than  
"because that's what perl's C source code says" -grin-). So let's try  
some things...

First, note that '' is false and ' ' is true.

$ cat j.pl
print ''  ? 'yes ' : 'no ';
print ' ' ? 'yes ' : 'no ';
$ perl j.pl
no yes

But my thing was more like this:

$ cat j.pl
print ((not defined $j || 1) ? 'yes ' : 'no ');
print ((not defined $j or 1) ? 'yes ' : 'no ');
$ perl j.pl
no yes


I'm betting this happens due to operator precedence. Reading "perldoc  
perlop" doesn't help me much at a glance since I'm not sure where  
'defined' falls in the documented precedence order stack. But let's  
assume for a second that the precedence of 'defined' is higher than  
that of 'or' but lower than that of '||'. If that's true, then:

    not defined $j || 1

would be processed as:

    $j || 1          true
    defined true     true
    not true         false

So if 'defined' is a "nonassoc list operators (rightward)" then this  
is probably the correct answer and I feel all smart and stuff.   :)

> 2. How do you get a eq to return ' ' ?

You don't? Apparently it returns '' if false, 1 if true:

$ perl -d -e 1
   DB<1> x $j eq 'blah'
0  ''
   DB<2> x $j eq ''
0  1

perldoc perlop says:

   Binary "eq" returns true if the left argument is stringwise equal to
   the right argument.

And '' is false, and '1' is true, so there you have it...  :)

Did that help at all?



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