I defined some elements in an array:
my @arrTest;
$arrTest[0]=1;
$arrTest[2]=2;
#Result for @arrTest is: (1, undef, 2)
if($arrTest[1]==0)
{
print "one\n";
}
elsif($arrTest[1] == undef)
{
print "undef\n";
}
I think it should print "undef". But it prints "one" here...
Does it mean $arrTest[1]=undef=0?
How can I modify the "if condition" to distinguish the "undef" in array element?
The operator == in the code $arrTest[1] == 0 puts $arrTest[1] in numeric context and so its value gets converted to a number if needed, as best as the interpreter can do, and that is used in the comparison. And when a variable in a numeric test hasn't been defined a 0 is used so the test evaluates to true (the variable stays undef).
Most of the time when this need be done we get to hear about it (there are some exceptions) -- if we have use warnings; that is (best at the beginning of the program)† Please always have warnings on, and use strict. They directly help.
To test for defined-ness there is defined
if (not defined $arrTest[1]) { ... }
† A demo
perl -wE'say "zero" if $v == 0; say $v; ++$v; say $v'
The -w enables warnings. This command-line program prints
Use of uninitialized value $v in numeric eq (==) at -e line 1.
zero
Use of uninitialized value $v in say at -e line 1.
1
Note how ++ doesn't warn, one of the mentioned exceptions.
use strict; and use warnings; as the first two lines. The only acceptable situation to omit those is when working with extremely legacy code that doesn't follow strict guidelines.
strict as well (which has nothing to do with this question so it's not clear how best to weave it in)
if (@ary) (it has something), or if (not @ary) (empty). Note that if (@ary) is true even if the array has an element which is undef or an empty string. It tests whether there is anything in it
elsif( ! defined($arrTest[1]) ) {