Re: [SPOILER] Re: New Quiz: "What does this code do?" (1-December-2006)

>>>>> "Jay" == Jay Savage <[email protected]> writes:
Jay> $1 is implicitly localized to the enclosing block, not to the operator
Jay> itself. That means that something like the following would be
Jay> dangerous if $1 weren't read-only:

Jay> {
Jay>     m/(.)/;
Jay>     &sub;
Jay>     print $1;
Jay> }

It's dangerous anyway.  Never use $1 unless you've tested the match!
$1 is *set* on a match, but not *reset* on a fail.

Consider:

        "xyz" =~ /(x..)/; # $1 is now xyz
        "xyz" =~ /(y..)/; # $1 is *still* xyz

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