Perl：数组中的 undef 元素

Question

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?

Answer 1

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.