FAQ 4.67 Why does passing a subroutine an undefined element in a hash create it?

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4.67: Why does passing a subroutine an undefined element in a hash create it?

    If you say something like:

        somefunc($hash{"nonesuch key here"});

    Then that element "autovivifies"; that is, it springs into existence
    whether you store something there or not. That's because functions get
    scalars passed in by reference. If somefunc() modifies $_[0], it has to
    be ready to write it back into the caller's version.

    This has been fixed as of Perl5.004.

    Normally, merely accessing a key's value for a nonexistent key does
    *not* cause that key to be forever there. This is different than awk's


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