Re^2: Ternary Quizical behaviour?

> But $m is a reference

yes and no ...

In proper CS speak it's a "call by reference", in Perl we tend to use "alias" for such passing.

But in Perl speak it's not a "reference" like created by the reference operator \

Cheers Rolf
_{(addicted to the Perl Programming Language :)

Wikisyntax for the Monastery}