in reply to passing hashes and hash reference
To answer the question of why your code snippet won't work, remember that Perl flattens subroutine arguments into a single list of scalars; your code shifts the first element off that list ( a hash key in this case ) and takes a reference to it, i.e. you end up with a reference to a scalar and the remainder of your hash is thrown away.
Others have posted the correct solution.
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