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

    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 is a considered a bug that we hope to fix.

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 behavior.

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