ravendarke has asked for the wisdom of the Perl Monks concerning the following question:

Dearest Monks, I ran into a problem where I needed a function similar to Oracle's NVL function. For those of you who aren't Oracle people, the function substitutes a value for a returned NULL. A co-worker pointed me to the ( ? : ) operators. However, I've found a small problem with it. Well, let me show you. 
my $junk;
print 1+$junk;

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Returns an unitialized value warning, as expected. 
print 1+($junk ? $junk : 0);

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Returns 1, no warning. 
$junk = 11;
print 1+($junk ? $junk : 0);

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Returns 12, seems to do what I want, but... 
$junk=0;
print 1+($junk ? $junk : 15);

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returns 16, even though the value was defined as 0.
So, clearly this only worked because I didn't need a different number to be returned. I can see this coming to bite me later. So, I guess my question is, does anyone know of a function that will do what I'm looking for? Is this an obvious misuse of ? : operators?

I'm sorry this was so wordy for so small a question..
Marty

Replies are listed 'Best First'.
Re: Uninitialized value
by ariels (Curate) on May 10, 2002 at 16:32 UTC

    You want to know if $junk is defined, why not check if defined($junk)? Handles <samp>0</samp> and <samp>''</samp> correctly, is easily readable, and explains itself.

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Re: Uninitialized value
by boo_radley (Parson) on May 10, 2002 at 16:39 UTC
    ravendarke says :
    returns 16, even though the value was defined as 0.


    You've almost stumbled onto an answer by yourself, ravendarke.
    use warnings;
my $junk;
print 1+$junk;
print "\n";
#Returns an unitialized value warning, as expected.


print 1+(defined($junk) ? $junk : 0);
print "\n";
#Returns 1, no warning.

$junk = 11;
print 1+(defined($junk) ? $junk : 0);
print "\n";
#Returns 12, seems to do what I want, but...


$junk=0;
print 1+(defined($junk) ? $junk : 15);
print "\n";
# now prints out 1.

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    The secret sauce is in the definedness of your variable.
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Re: Uninitialized value
by chromatic (Archbishop) on May 10, 2002 at 16:32 UTC
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Re: Uninitialized value
by zzspectrez (Hermit) on May 10, 2002 at 16:41 UTC 

    > my $junk;
> print 1+$junk;

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    > Returns an unitialized value warning, as expected. 
    > print 1+($junk ? $junk : 0);

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    > Returns 1, no warning.

    This should not return a warning. This statement says, if $junk is true then return $junk add 1 to it and print the result. If $junk is false, which is the case since it is undefined, then return 0 and add 1 to it and print the result.
    I think what you meant is if you wrote it like: 

    print 1+ ( $junk ? 0 : $junk);

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    Then it will return a warning if $junk is undefined.

    zzspectrez

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Re: Uninitialized value
by mephit (Scribe) on May 10, 2002 at 17:09 UTC
     print 1 + ($junk && ($junk != 0) ? $junk : 0); Does that do what you want?
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