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
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