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.

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 {
    ...;
}
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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.

Oops ... defined(), obviously. duh.