FAQ 7.22 What's the difference between calling a function as &foo and foo()?

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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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