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

by faq_monk (Initiate)
on Oct 08, 1999 at 00:20 UTC ( #634=perlfaq nodetype: print w/ replies, xml ) Need Help??

Current Perl documentation can be found at perldoc.perl.org.

Here is our local, out-dated (pre-5.6) version:

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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