Uninitialized value

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;
[download]

Returns an unitialized value warning, as expected.

print 1+($junk ? $junk : 0);
[download]

Returns 1, no warning.

$junk = 11;
print 1+($junk ? $junk : 0);
[download]

Returns 12, seems to do what I want, but...

$junk=0;
print 1+($junk ? $junk : 15);
[download]

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

Comment on Uninitialized value Select or Download Code

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.	[reply] [d/l] [select]
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.` [download] The secret sauce is in the definedness of your variable.	[reply] [d/l]
Re: Uninitialized value by chromatic (Archbishop) on May 10, 2002 at 16:32 UTC
defined.	[reply]
Re: Uninitialized value by zzspectrez (Hermit) on May 10, 2002 at 16:41 UTC
`> my $junk; > print 1+$junk;` [download] > Returns an unitialized value warning, as expected. `> print 1+($junk ? $junk : 0);` [download] > 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);` [download] Then it will return a warning if $junk is undefined. zzspectrez	[reply] [d/l] [select]
Re: Uninitialized value by mephit (Scribe) on May 10, 2002 at 17:09 UTC
`print 1 + ($junk && ($junk != 0) ? $junk : 0);` Does that do what you want?	[reply] [d/l]
A reply falls below the community's threshold of quality. You may see it by logging in.


"be consistent"
	PerlMonks