How many ways can you call a subroutine and ignore its prototype in Perl?

Question

We should all be familiar with the problems related to prototypes in Perl. Here are the two biggies:

they don't work like prototypes in other languages, so people misunderstand them.
they don't apply to all ways of calling a subroutine.

The second item is the one I am curious about at the moment.

I know of two ways to subvert/work around/ignore prototype enforcement when calling a subroutine:

Call the sub as a method. Foo->subroutine_name();
Call the sub with a leading & sigil. &subroutine_name();

Are there any other interesting cases I've missed?

Udpate:

@brian d foy, I don't particularly want to evade prototypes, but I wondered "how many ways are there to do it?" I ask this question out of curiosity.

@jrockway, I agree with you, and I believe that you have more explicitly and more concisely described my first point regarding the problems with prototypes, that people misunderstand them. Perhaps the problem lies in programmer expectations and not in the feature. But that is really a philosophical question of the sort I don't want to have.

This is not a "problem". Prototypes are hints for the lexer, not a data validation framework. Understand that and you won't have problems. — jrockway, Sep 28 '09 at 14:37
Cross-reference to the question where this is discussed in more detail: http://stackoverflow.com/questions/297034/why-are-perl-function-prototypes-bad — Ether, Sep 28 '09 at 17:11

score 12 · Accepted Answer · answered Sep 28 '09 at 07:51

12

Call it via a subroutine reference.

sub foo($) { print "arg is $_[0]\n" }
my $sub = \&foo;
$sub->();

Call it before Perl has seen the prototype (important because perl doesn't make you declare subs before use):

foo();
sub foo($) { print "arg is $_[0]\n" }

answered Sep 28 '09 at 07:51

derobert

49,731
15
94
124

3

Interestingly, `print prototype($sub), "\n";` still prints "$" (the correct prototype). – Chris Lutz Sep 28 '09 at 07:55
1

The second one does give a warning: main::foo() called too early to check prototype... – oylenshpeegul Sep 28 '09 at 11:04
It looks like any call that is determined at runtime steps around them. This makes sense since they are processed at compile time. – daotoad Sep 28 '09 at 21:07

score 3 · Answer 2 · answered Oct 22 '09 at 06:35

3

Using the goto &name syntax.

answered Oct 22 '09 at 06:35

Eric Strom

39,821
2
80
152

score 3 · Answer 3 · answered Oct 22 '09 at 07:08

3

Type-glob.

sub foo($) { print "arg is $_[0]\n" }
*foo{CODE}()

Also, consider Bar->foo() rewritten in indirect notation:

foo Bar

Any more? c'mon, bring it.

answered Oct 22 '09 at 07:08

How many ways can you call a subroutine and ignore its prototype in Perl?

3 Answers3

Linked