Instead, use: my ($foo, $bar, $baz) = ParseForExpDB2Params(@params);

References are scalars. Assigning a (part of a) list to an array or hash puts all remaining scalars in that list into that array or hash, even if the left-hand side of the assignment has more than one variable.

Example: ($foo, @bar, @baz) = (1..10); results in $foo containing 1, @bar containing the list equivalent to (2..10), and @baz containing nothing (it was assigned an empty list, so any contents were overwritten).

Additionally, with warnings enabled, you would receive a warning about an odd (3) number of elements assigned to a hash, which requires an even number of elements (key-value pairs).

In reply to Re: Returning multiple Hash references from a sub() by Anonymous Monk
in thread Returning multiple Hash references from a sub() by Xhings

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