No it isn't. )/g; is not valid Perl so you'd get an error message about that in that case. perl is very smart about inferring the correct delimiter in the overwhelming majority of cases anyway; please check your assumptions next time.

Makeshifts last the longest.

In reply to Re^2: Why doesn't non-greediness work? (not two commands) by Aristotle
in thread Why doesn't non-greediness work? by kiat

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