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Posted by PerlFAQ Server on March 25, 2008, 3:03 am
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This is an excerpt from the latest version perlfaq7.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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7.22: What's the difference between calling a function as &foo and foo()?
When you call a function as &foo, you allow that function access to your
current @_ values, and you bypass prototypes. The function doesn't get
an empty @_--it gets yours! While not strictly speaking a bug (it's
documented that way in perlsub), it would be hard to consider this a
feature in most cases.
When you call your function as "&foo()", then you *do* get a new @_, but
prototyping is still circumvented.
Normally, you want to call a function using "foo()". You may only omit
the parentheses if the function is already known to the compiler because
it already saw the definition ("use" but not "require"), or via a
forward reference or "use subs" declaration. Even in this case, you get
a clean @_ without any of the old values leaking through where they
don't belong.
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