His code assumes @a is empty, which means $#a returns -1, so he's doing (...)[-1 - 2 .. -1] which is just (...)[-3 .. -1]. Your case cannot be handled in a similar manner -- you'd need to know the size of the list being returned by split(). Your first approach is fine.
Jeff japhy Pinyan, P.L., P.M., P.O.D, X.S.: Perl, regex, and perl hacker
How can we ever be the sold short or the cheated, we who for every service have long ago been overpaid? ~~ Meister Eckhart

In reply to Re^3: How to cut strings from the end by japhy
in thread How to cut strings from the end by Bharath

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