in reply to Re: why can't compare a variable with undef directly?
in thread why can't compare a variable with undef directly?

Actually that is invalid because exists doesn't operate on scalars.

exists argument is not a HASH or ARRAY element or a subroutine at ...

The valid equivalent of that is defined $a && defined $b but don't use $a and $b outside of a sort.

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Re^3: why can't compare a variable with undef directly?
by RonW (Parson) on Dec 10, 2014 at 21:40 UTC

    Actually, exists tests if the element is present. Even if the element exists, the value of the element could still be undef.

    Note that defined($array[$i]) will cause the element at index $i to be created. Likewise, defined($array{$k}) will create an element referred by key $k. In both cases, the new element with have the value undef.

    This "auto-vivification" may have undesirable side effects. This is most likely to become a problem where a hash is being used to hold options or named parameters. Often times, the absence of an option or parameter has a different meaning than when present with no value specified (ie, value is undef).

    When in doubt, it is generally better to test for existence then for a defined value:

    if (exists $params{'foo'}) { if (defined $params{'foo'}) { ...; } else { ...; } } else { ...; }

      Your response makes no sense to me since I was responding to AM's usage of exists with a scalar not a hash or array element.

        Meant to respond to the AM, not yours. Sorry for the confusion.
Re^3: why can't compare a variable with undef directly?
by Anonymous Monk on Dec 10, 2014 at 17:55 UTC
    Oops ... defined(), obviously. duh.