> 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

In reply to Re: Uninitialized value by zzspectrez
in thread Uninitialized value by ravendarke

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