Re^2: Using & in function calls (&foo != foo(@_))

[lodin]

It passes the current @_ to the function called.

This is probably just a matter of wording, but I'd say it's sharing @_. &foo and foo(@_) are not equivalent. An example:

sub foo1 { &bar;    print "@_\n"; }
sub foo2 { bar(@_); print "@_\n"; }

sub bar { shift }

foo1(1, 2, 3);
foo2(1, 2, 3);

__END__
2 3
1 2 3

lodin