The use of &foo in \&foo has nothing to do with dereferencing. You're not calling foo.

Also, in goto &$bar; you aren't derefencing - it's the magical goto that jumps to the subroutine pointed to by $bar. If you were derefencing, you'd jump to the label that was returned by &$bar;. And that's what's happening if you do goto $bar->();. Or goto &$bar();.

Another reason to use the ampersand version is to disable prototype checking, for those subroutines that are evil enough to use it without good reason.

But that's a difference between foo() and &foo(). Again, that has nothing to do with dereferencing. It can't even do prototype checking when doing dereferencing, as prototype checking is done at compile time. The following code runs without errors - despite violent prototype violations: 

!/usr/bin/perl

use strict;
use warnings;

my $ref = sub ($) {print "Hello, world\n"};

&$ref();
$ref->();

__END__

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Neither the &$ref();, nor the $ref->(); is checked for its prototype.

In reply to Re^3: Dereferencing code refs with & vs. ->() by Anonymous Monk
in thread Dereferencing code refs with & vs. ->() by doran

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