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What's the difference between calling a function as &foo and foo()?

by faq_monk (Initiate)
on Oct 08, 1999 at 00:27 UTC ( #699=perlfaq nodetype: print w/ replies, xml ) Need Help??

Current Perl documentation can be found at

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

When you call a function as &foo, you allow that function access to your current @_ values, and you by-pass prototypes. That means that the function doesn't get an empty @_, it gets yours! While not strictly speaking a bug (it's documented that way in the perlsub manpage), 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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