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Posted by PerlFAQ Server on March 3, 2008, 9:03 am
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This is an excerpt from the latest version perlfaq4.pod, which
comes with the standard Perl distribution. These postings aim to
reduce the number of repeated questions as well as allow the community
to review and update the answers. The latest version of the complete
perlfaq is at http://faq.perl.org .
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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
behavior.
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