Calling subs and typing in Perl 6

Calling subs and typing in Perl 6

In the ninth article in this series comparing Perl 5 to Perl 6, learn about the subtle differences in visibility of subroutines and the (gradual) typing core feature.

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This is the ninth article in a series about migrating code from Perl 5 to Perl 6. This article examines the subtle differences in visibility of subroutines between Perl 5 and Perl 6 and the (gradual) typing core feature of Perl 6.

This article assumes you're familiar with signatures—if you're not, read How subroutine signatures work in Perl 6, the fourth article in this series, before you continue.

Visibility of subroutines

In Perl 5, a named subroutine will, by default, be visible within the scope of the package in which it is defined, regardless of the scope where the definition takes place:

# Perl 5
{
    sub foo { "bar" }       # visible outside of this scope
}
say foo();
# bar

In Perl 6, a named subroutine is visible only within the lexical scope in which it is defined:

# Perl 6
{
    sub foo() { "bar" }     # only visible in this scope
}
say foo();
# ===SORRY!=== Error while compiling ...
# Undeclared routine:
#     foo used at line ...

Note that SORRY! in the Perl 6 error message means that the subroutine foo can't be found at compile time. This is a very useful feature that helps prevent typos in subroutine names when writing invocations of the subroutine.

You could consider subroutine definitions in Perl 6 to always have a my in front, similar to defining lexical variables. Perl 5 also has a (previously experimental) lexical subroutine feature, which has to be specifically activated in versions lower than Perl 5.26:

# Perl 5.18 or higher
no warnings 'experimental::lexical_subs';
use feature 'lexical_subs';
{
    my sub foo { "bar" }   # limit visibility to this scope
}
say foo();
# Undefined subroutine &main::foo called at ...

It is possible in both Perl 5 and Perl 6 to prefix the subroutine definition with an our scope indicator, but the result is subtly different: In Perl 5, this makes the subroutine visible outside the scope, but this isn't the case in Perl 6. In Perl 6, lookups of subroutines are always lexical: the use of our on subroutine declarations (regardless of scope) allows the subroutine to be called from outside the namespace in which it is defined:

# Perl 6
module Foo {
    our sub bar() { "baz" }  # make sub visible from the outside
}
say Foo::bar();
# baz

This would fail without the our. In Perl 5, any subroutine can be called from outside the namespace where it is defined:

# Perl 5
package Foo {
    sub bar { "baz" }     # always visible from the outside
}
say Foo::bar();
# baz

In Perl 5, the names of subroutines that are intended to be "private" (i.e., called from within that scope only and not from outside) usually start with an underscore. But that won't stop them from being called from the outside. In Perl 6, subroutines that are not intended to be called from the outside are simply invisible.

The our on a subroutine definition in Perl 6 not only indicates that the subroutine can be called from the outside; it also indicates that it will be exported if it is part of a module being loaded. In a future article about the creation of modules and module loading, I will go into exporting subroutines.

Calling a subroutine

When you call a subroutine in a Perl 5 without subroutine signatures enabled, it will call the subroutine if it exists (determined at runtime) and pass the parameters into @_ inside the subroutine. Whatever happens to the parameters inside the subroutine is entirely up to the subroutine (see How subroutine signatures work in Perl 6).

When a subroutine is called in Perl 6, it performs additional checks to see whether the arguments passed to the subroutine match the parameters the subroutine expects before it calls the subroutine's code. Perl 6 tries to do this as early as possible—if it determines that a call to a subroutine will never succeed, it will tell you at compile time:

# Perl 6
sub foo() { "bar" }    # subroutine not taking any parameters
say foo(42);           # try calling it with one argument
# ===SORRY!=== Error while compiling ...
# Calling foo(Int) will never work with declared signature ()

Note that the error message mentions the type of value (Int) being passed as an argument. In this case, calling the subroutine will fail because the subroutine doesn't accept any argument being passed to it (declared signature ()).

Other signature features

Apart from specifying positional and named parameters in a signature, you can also specify what type these parameters should be. If the parameter type doesn't smartmatch with the argument type, it will be rejected. In this example, the subroutine expects a single Str argument:

# Perl 6
sub foo(Str $who) { "Hello $who" }  # subroutine taking a Str parameter
say foo(42);                        # try calling it with an integer
# ===SORRY!=== Error while compiling ...
# Calling foo(Int) will never work with declared signature (Str $who)

It checks both the number of required parameters and the type. Unfortunately, it is not always possible to reliably see this at compilation time. But there's still the check done at runtime when binding the argument to the parameter:

# Perl 6
sub foo(Str $who) { "Hello $who" }  # subroutine taking a Str parameter
my $answer = 42;
say foo($answer);                   # try calling it with a variable
# Type check failed in binding to parameter '$who'; expected Str but got Int (42)
#   in sub foo at ...

However, if Perl 6 knows the type of variable being passed to the subroutine, it can determine at compile time that the call will never work:

# Perl 6
sub foo(Str $who) { "Hello $who" }  # subroutine taking a Str parameter
my Int $answer = 42;
say foo($answer);                   # try calling it with an Int variable
# ===SORRY!=== Error while compiling ...
# Calling foo(Int) will never work with declared signature (Str $who)

It should be clear that using typing in your variables and parameters enables Perl 6 to help you find problems quicker!

Gradual typing

The above is usually called gradual typing. Perl 6 always performs type checks at runtime (dynamic typing). But if it can determine at compile time that something will not work, it will tell you so. This is usually called static typing.

If you're coming from Perl 5 and have experience with Moose (and specifically MooseX::Types), you may worry about the performance implications of adding type information to your code. This is not a concern in Perl 6, as type checks always occur in Perl 6 with every assignment to a variable or binding of a parameter. That is because if you do not specify a type, Perl 6 will implicitly assume the Any type, which smartmatches with (almost) everything in Perl 6. So, if you're writing:

# Perl 6
my $foo = 42

You have in fact written:

# Perl 6
my Any $foo = 42;

And the same goes for a parameter to a subroutine:

# Perl 6
sub foo($bar) { ... }

Which is in fact:

# Perl 6
sub foo(Any $bar) { ... }

Adding type information not only helps find errors in your program, it also allows the optimizer to make better-informed decisions about what it can optimize and how to best optimize it.

Defined or not

If you specify a variable in Perl 5 but don't assign it, it contains the undefined value (undef):

# Perl 5
my $foo;
say defined($foo) ? "defined" : "NOT defined";
# NOT defined

This is not much different in Perl 6:

# Perl 6
my $foo;
say defined($foo) ?? "defined" !! "NOT defined";
# NOT defined

So, you can specify which types of values are acceptable in a variable and as a parameter. But what happens if you don't assign such a variable?

# Perl 6
my Int $foo;
say defined($foo) ?? "defined" !! "NOT defined";
# NOT defined

The value inside such a variable is still not defined, as in Perl 5. However, if you just want to show the contents of such a variable, it is not undef, as it would be in Perl 5:

# Perl 6
my Int $foo;
say $foo;
# (Int)

What you see is the representation of a type object in Perl 6. Unlike Perl 5, Perl 6 has a multitude of typed undefs. Each class that is defined, or which you define yourself, is a type object.

# Perl 6
class Foo { }
say defined(Foo) ?? "defined" !! "NOT defined";
# NOT defined

If, however, you instantiate a type object, usually with .new, it becomes a defined object, as expected:

# Perl 6
class Foo { }
say defined(Foo.new) ?? "defined" !! "NOT defined";
# defined

Type smileys

If you specify a constraint on a parameter in a subroutine, you can also indicate whether you want a defined value of that type or not:

# Perl 6
sub foo(Int:D $bar) { ... }

The :D combined with the Int indicates that you want a Defined value of the Int type. Because :D is also the emoji for a big smile, this decoration on the type is called a "type smiley." So what happens if you pass an undefined value to such a subroutine?

# Perl 6
sub foo(Int:D $bar) { ... }   # only accept instances of Int
foo(Int);                     # call with a type object
# Parameter '$bar' of routine 'foo' must be an object instance of
# type 'Int', not a type object of type 'Int'.  Did you forget a '.new'?

Careful readers may realize that this should create a compile-time error. But alas, it hasn't (yet). Although error messages are known to be pretty awesome in Perl 6, there is still a lot of work to make them even better (and more timely, in this case).

You can also use the :D type smiley on variable definitions to ensure that you provide an initialization for that variable:

# Perl 6
my Int:D $foo;                # missing initialization
# ===SORRY!=== Error while compiling ...
# Variable definition of type Int:D requires an initializer

Other type smileys are :U (for undefined) and :_ (for don't care, which is the default):

# Perl 6
sub foo(Int:U $bar) { ... }   # only accept Int type object
foo(42);                      # call with an instance of Int
# Parameter '$bar' of routine 'foo' must be a type object of type 'Int',
# not an object instance of type 'Int'.  Did you forget a 'multi'?

Hmmm… what's this multi that seems to be forgotten? Well, that's for the next article in this series!

Summary

Subroutines in Perl 6 are, by default, visible only in the lexical scope where they are defined. Even prefixing our will not make them visible outside of that lexical scope, but it does allow a subroutine to be called from outside the scope with its full package name (Foo::bar() for a subroutine bar in a package bar).

Perl 6 allows you to use gradual typing to ensure the validity of arguments to subroutines or assignments to variables. This does not incur any extra runtime costs. Adding typing to your code even allows the compiler to catch errors at compile time, and it allows the optimizer to make better decisions about optimizing during runtime.

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About the author

Elizabeth Mattijsen
Elizabeth Mattijsen - Elizabeth Mattijsen has been programming for a living since 1978 in various (mostly now defunct) programming languages before she started programming in Perl 4. In 1994 she started the first commercial web site development company in the Netherlands, using Perl 5 as the main programming language. From 2003 onwards, she was involved in the rapid growth of an online Hotel Reservation service. In 2012 she got more directly involved in the development of Rakudo Perl 6. From 2015 she has also...