如何将Perl数组引用中的元素作为子例程的单独参数传递？

Question

I have a list that contains arguments I want to pass to a function. How do I call that function?

For example, imagine I had this function:

sub foo {
  my ($arg0, $arg1, $arg2) = @_;
  print "$arg0 $arg1 $arg2\n";
}

And let's say I have:

my $args = [ "la", "di", "da" ];

How do I call foo without writing foo($$args[0], $$args[1], $$args[2]) ?

Answer 1

You dereference an array reference by sticking @ in front of it.

foo( @$args );

Or if you want to be more explicit:

foo( @{ $args } );

Answer 2

This should do it:

foo(@$args)

That is not actually an apply function. That syntax just dereferences an array reference back to plain array. man perlref tells you more about referecences.

Answer 3

尝试这个：

foo(@$args);

Answer 4

foo(@$args);

Or, if you have a reference to foo :

my $func = \&foo;
...
$func->(@$args);

Answer 5

It's simple. foo(@{$args})