I agree that it was not what was expected but it does fulfill the criteria as stated.

No, as juerd pointed out, you're not changing $x[0] nor $x[1]: they both remain equal to \$_. All you show is that if you change ${$x[0]} then ${$x[1]} changes, which wasn't the exercise.

Note that I have no problem with unexpected solutions — I love the $[ one.. didn't think of that — but they do need to be valid.   :-)

In reply to Re[7]: How's your Perl? by xmath
in thread How's your Perl? by xmath

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