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

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

am 09.01.2008 15:03:03 von PerlFAQ Server

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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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