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()?

(contributed by brian d foy)

Calling a subroutine as &foo with no trailing parentheses ignores the
prototype of "foo" and passes it the current value of the argumet list,
@_. Here's an example; the "bar" subroutine calls &foo, which prints
what its arguments list:

sub bar { &foo }

sub foo { print "Args in foo are: @_\n" }

bar( qw( a b c ) );

When you call "bar" with arguments, you see that "foo" got the same @_:

Args in foo are: a b c

Calling the subroutine with trailing parentheses, with or without
arguments, does not use the current @_ and respects the subroutine
prototype. Changing the example to put parentheses after the call to
"foo" changes the program:

sub bar { &foo() }

sub foo { print "Args in foo are: @_\n" }

bar( qw( a b c ) );

Now the output shows that "foo" doesn't get the @_ from its caller.

Args in foo are:

The main use of the @_ pass-through feature is to write subroutines
whose main job it is to call other subroutines for you. For further
details, see perlsub.

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