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.

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

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