Is there any benefit (w.r.t performance/memory usage) in including use mylibrary conditionally (assuming mylibrary is used only if condition is true) compared to adding use mylibrary on top of the script unconditionally?
# Script 1 (Unconditional use)
use mylibrary;
if($condition)
{
# Do something with mylibrary
}
# Script 2 (Conditional use)
if($condition)
{
use mylibrary;
# Do something with mylibrary
}
use is a compile-time construct. In your two cases, mylibrary is actually being imported in both of your "Unconditional" and "Conditional" cases. If you want to import a library conditionally, use require, a run-time construct, instead.
if ($condition) {
require mylibrary;
# mylibrary->import;
# ...
}
In such a case you lose some of the compile-time benefits of use. For example, require does not call mylibrary->import at compile time, as use does. You can call import yourself if you want, as I show above, but anything import does that has an effect at compile time will not have that effect when called at run time.
Suppose your module mylibrary exports a function foo. Then this works:
use strict;
use mylibrary; # exports function foo()
foo;
But this is an error:
use strict;
require mylibrary;
mylibrary->import; # too late to notify Perl's parser about the foo() function
foo; # error; unknown function
As to whether there's any benefit to doing so, there can be if mylibrary is expensive to import. Most of the time, probably not.
You can achieve conditional loading, including the side benefits of the use (the calling of ->import, and execution at compilation time) with the if pragma: http://perldoc.perl.org/if.html
use if $Config{usethreads}, 'Library::Requiring::Threads';
One caveat, of course, is that the if expression is executed at compile time, so everything it depends on must also be known and available at compile time.
Related
I have a set of tests, always named Module.t, each one starts like this:
use 5.026;
use strict;
use warnings;
use Test::Perl::Critic (-severity => 3);
use Module::Path 'module_path';
use Test::More tests => 8;
use Test::Log4perl;
Test::Log4perl->suppress_logging;
BEGIN { use_ok("My::Module") }
critic_ok(module_path("My::Module"));
... actual tests for this module ...
It's done this way because a bunch of modules are not coded very nicely and in effort to refactor stuff as we go, I'm trying to write tests for individual modules over time. Eg. I can't just enable Perl::Critic for all sources cause it will blow up in my face.
I would like to ideally make a "parent" test for all of these so that when me or a different developer wants to write a new test they will always have all the required stuff. Something like:
use 5.026;
use strict;
use warnings;
# 6 tests because 2 (use_ok and critic_ok) are already in the parent
use parent ParentTest("My::Module", tests => 6);
... actual tests for this module ...
Does perl have a way of doing that?
Disclaimer: I'm a perl noob, so maybe this has a better solution :-)
Sounds like you just want a helper module that loads some other modules and runs some initial tests for you.
Something like:
# ParentTest.pm
package ParentTest;
use strict;
use warnings;
use Test::Perl::Critic (-severity => 3);
use Module::Path 'module_path';
use Test::More;
use Test::Log4perl;
sub import {
my (undef, $module, %args) = #_;
$args{tests} += 2;
plan %args;
Test::Log4perl->suppress_logging;
use_ok $module;
critic_ok module_path $module;
#_ = 'Test::More';
goto +Test::More->can('import');
}
1
Usage would be:
use ParentTest "My::Module", tests => 6;
This is all untested, but the idea is:
We want to run some code to set up the initial test plan and run some tests.
We also want to export everything that Test::More exports, so our caller doesn't have to use Test::More themselves.
use Some::Module #args is equivalent to BEGIN { require "Some/Module.pm"; Some::Module->import(#args); }, so we can just put our custom logic in the import method.
We start by ignoring the first argument (which is a class name because import is called as a class method) and assigning the remaining arguments to $module and %args.
We increment $args{tests} by 2 to account for the two extra tests we perform automatically (if tests wasn't passed in, it is implicitly created here).
We pass %args to plan from Test::More, which is nice for setting up a test plan outside of the initial use line.
We perform the initial tests.
We tail call Test::More::import, erasing our own stack frame. This makes it look like our caller did Test::More->import(), which exports all the Test::More utility functions to them.
The unary + in goto +Test::More->... has no real effect, but it helps distinguish between the goto LABEL and goto EXPRESSION syntactic forms. We want the latter interpretation.
I was appalled to discover today that one of my perl modules uses another module, but doesn't import it.
However, I was shocked when I realized that this has never caused any problems!
The code is something like:
package This;
# no import for OTHER !!
sub new {
... implementation ...
my $something = OTHER->new(#arguments); # no error!
... more implementation ...
}
So what gives? None of This's imports import OTHER -- that was the first thing I checked.
Could it be that if the code that imports This also imports OTHER, OTHER is available to This?
Clarification of question:
X, Y, and Z are modules.
in X: use Y;
in X: use Z;
Y does not use Z
Does Y have access to the functions and methods in Z (and vice versa)?
Your use of the term "import" is a bit misleading.
Does Y have access to Z without importing it (and vice versa)?
Yes, more or less. useing a module just loads it into the current script; all used modules are loaded into the same script, so can see each other (provided they're used in the right order).
But true imports — the things that a module actually exports, so that they're copied into the useing module's namespace — will only be copied into the useing module's namespace.
For example, consider this Perl script:
use Time::HiRes 'time';
package Foo;
sub time1() { return time(); } # calls the built-in time() function
sub time2() { return Time::HiRes::time(); }
package main;
print Foo::time1(), "\n"; # prints (e.g.) 1323048440
print Foo::time2(), "\n"; # prints (e.g.) 1323048440.80571
print time(), "\n"; # prints (e.g.) 1323048440.8061
It uses the Time::HiRes module, and tells it to export Time::HiRes::time into the current namespace (the main namespace), overwriting the built-in time. (Time::HiRes::time is like time, but it has sub-second resolution; it'll give something like 1323048440.80571 instead of just 1323048440.) So package Foo can see Time::HiRes, and anything within Time::HiRes, such as its time; but it has to explicitly specify that, by explicitly writing Time::HiRes::time.
(Note: I wrote the above as a single script, for simplicity of description, but the same thing happens when Foo is defined in Foo.pm and loaded using use Foo;.)
Yes, at least when you use indirect package and object syntax. From calls like
OtherPackage->method()
$anOtherPackageObject->method()
Perl can resolve a fully qualified subroutine name (&OtherPackage::method in both of the above cases) and invoke the subroutine from anywhere.
Lets say you have a parent Perl class in one file:
#!/usr/bin/perl
package Foo;
use strict;
use warnings;
use Data::Dumper;
sub new{
my $class = shift;
my %self = ();
return bless %self, $class;
}
1;
and a subclass in a different file:
#!/usr/bin/perl
package Bar;
use base "Foo";
1;
Will the subclass inherit the use statements from the parent? I know the method new will be inherited.
Basically I am trying to reduce the amount of boilerplate in my code and I can't find a clear answer to this question.
You asked in a comment about Test::Most and how it reduces boilerplate. Look at its import method. It's loading the modules into its namespace, adding those symbols to #EXPORT, then re-calling another import through a goto to finally get them into the calling namespace. It's some serious black magic that Curtis has going on there, although I wonder why he just didn't use something like import_to_level. Maybe there are some side effects I'm not thinking about.
I talk quite a bit about this sort of thing in Avoid accidently creating methods from module exports in The Effective Perler. It's in a different context but it's some of the same issues.
Here's a different example.
If some other module loads a module, you have access to it. It's not good to depend on that though. Here are three separate files:
Top.pm
use 5.010;
package Top;
use File::Spec;
sub announce { say "Hello from top!" }
1;
Bottom.pm
package Bottom;
use parent qw(Top);
sub catfiles { File::Spec->catfile( #_ ) }
1;
test.pl
use 5.010;
use Bottom;
say Bottom->catfiles( qw(foo bar baz) );
say File::Spec->catfile( qw( one two three ) );
I only load File::Spec in Top.pm. However, once loaded, I can use it anywhere in my Perl program. The output shows that I was able to "use" the module in other files even though I only loaded it in one:
Bottom/foo/bar/baz
one/two/three
For this to work, the part of the code that loads the module has to load before any other part of the code tries to use that module. As I said, it's a bad idea to depend on this: things break if the loading sequence changes or the loading module disappears.
If you want to import symbols, however, you have to explicitly load the module you want while you are in the package you want to import into. That's just so the exporting module defines the symbols in that package. It's not something that depends with scope.
Ah, good question!
Will the subclass inherit the use statements from the parent?
Well this depends on what you mean by inherit. I won't make any assumptions until the end, but the answer is maybe. You see, perl mixes the ideas of Classes, and Namespaces -- a package is a term that can describe either of them. Now the issue is the statement use all it does is force a package inclusion, and call the targets import() sub. This means it essentially has unlimited control over your package - and by way of that your class.
Now, compound this with all methods in perl being nothing more than subs that take $self as a first argument by convention and you're left with perl5. This has an enormous upside for those that know how to use it. While strict is a lexical pragma, what about Moose?
package BigMooseUser;
use Moose;
package BabyMooseUser;
our #ISA = 'BigMooseUser';
package Foo;
my $b = BabyMooseUser->new;
print $b->meta->name;
Now, where did BabyMooseUser get the constructor (new) from? Where did it get the meta class from? All of this is provided from a single use Moose; in the parent class (namespace). So
Will the subclass inherit the use statements from the parent?
Well, here, in our example, if the effects of the use statement are to add methods, than certainly.
This subject is kind of deep, and it depends if you're talking about pragmas, or more obscure object frameworks, or procedural modules. If you want to mitigate a parents namespace from affecting your own in the OO paradigm see namespace::autoclean.
For boilerplate reduction, I have a couple of strategies: Most of my classes are Moose classes, which takes care of OO setup and also gives me strict and warnings. If I want to have functions available in many packages, I'll create a project specific MyProject::Util module that uses Sub-Exporter to provide me with my own functions and my own interface. This makes it more consistent, and if I decide to change the Dumper (for example) later for whatever reason, I don't have to change lots of code. That'll also allow you to group exports. A class then usually looks something like this:
package Foo;
use Moose;
use MyProject::Util qw( :parsing :logging );
use namespace::autoclean;
# class implementation goes here
1;
If there's other things you regard as boilerplate and want to make simpler to include, it of course depends on what those things are.
A pragmatic answer to your problem: Either use, or look at how Modern::Perl does it to enforce strict and warnings.
You can get a definitive answer by examining the symbol tables for each package:
# examine-symbol-tables.pl
use Bar;
%parent_names = map{$_ => 1} keys %Foo::;
%child_names = map{$_ => 1} keys %Bar::;
delete $parent_names{$_} && ($common_names{$_} = delete $child_names{$_}) foreach keys %child_names;
print "Common names in symbol tables:\n";
print "#{[keys %common_names]}\n\n";
print "Unique names in Bar symbol table:\n";
print "#{[keys %child_names]}\n\n";
print "Unique names in Foo symbol table:\n";
print "#{[keys %parent_names]}\n\n";
$ perl inherit.pl
Common names in symbol tables:
BEGIN
Unique names in Bar symbol table:
ISA isa import
Unique names in Foo symbol table:
Dumper new VERSION
I am writing a Perl module Galaxy::SGE::MakeJobSH with OO.
I want to use MakeJobSH->new() instead of Galaxy::SGE::MakeJobSH->new(),
or some other shortnames. How can I do that?
You can suggest that your users use the aliased module to load yours:
use aliased 'Galaxy::SGE::MakeJobSH';
my $job = MakeJobSH->new();
Or you could export your class name in a variable named $MakeJobSH;
use Galaxy::SGE::MakeJobSH; # Assume this exports $MakeJobSH = 'Galaxy::SGE::MakeJobSH';
my $job = $MakeJobSH->new();
Or you could export a MakeJobSH function that returns your class name:
use Galaxy::SGE::MakeJobSH; # Assume this exports the MakeJobSH function
my $job = MakeJobSH->new();
I'm not sure this is all that great an idea, though. People don't usually have to type the class name all that often.
Here's what you'd do in your class for the last two options:
package Galaxy::SGE::MakeJobSH;
use Exporter 'import';
our #EXPORT = qw(MakeJobSH $MakeJobSH);
our $MakeJobSH = __PACKAGE__;
sub MakeJobSH () { __PACKAGE__ };
Of course, you'd probably want to pick just one of those methods. I've just combined them to avoid duplicating examples.
I don't bother with aliasing. I think it's the wrong way to go. If you're just looking for less to type, it might be the answer (but is a new dependency more benefit than risk?). I don't like the idea of tricking a maintenance programmer by hiding the real name from him since the aliasing happens a long way away from its use and there's no indication that what looks like a class name isn't a real class.
I'm mostly looking for easy subclassing, so I let the class decide for itself which module will implement a part.
For instance, I might start with a class that wants to use Foo to handle part of the job. I know that I might want to subclass Foo later, so I don't hard-code it:
package Foo::Bar;
sub foo_class { 'Foo' }
sub new {
....
eval "require $self->foo_class";
$self->foo_class->do_something;
}
In the application, I choose to use 'Foo::Bar':
#!perl
use Foo::Bar;
my $obj = Foo::Bar->new();
Later, I need to specialise Foo, so I create a subclass overrides the parts I need:
package Foo::Bar::Baz;
use parent 'Foo::Bar';
sub foo_class { 'Local::Foo::SomeFeature' }
1;
Another application uses almost all of the same stuff, but with the small tweak:
#!perl
use Foo::Bar::Baz;
my $obj = Foo::Bar::Baz->new();
You can also do a similar thing at the application level if you want to write one program and let users choose the class through configuration.
Thanks cjm.
I just choose to inline aliased.
require Exporter;
our #ISA = qw(Exporter);
our #EXPORT = qw(MakeJobSH);
sub MakeJobSH() {return 'Galaxy::SGE::MakeJobSH';}
aliased works well when you want to only affect calls from packages that explicitly request the aliasing. If you want global aliasing of one namespace to another, use Package::Alias instead.
It is almost exactly same approach as aliased but using standard Perl module:
use constant MakeJobSH => 'Galaxy::SGE::MakeJobSH';
my $job = MakeJobSH->new();
What is the difference between package, module and class in object oriented Perl?
Modules are a single file, a .pm file that provides code. That could be no packages, a single package, or more than one package. A module doesn't really care what is in it, so it can be code that inserts itself into the same namespace, a more-traditional set of subroutines in a library or define Perl's idea of a class.
A package, also known as a namespace, contains its own variables and subroutines. It's a way of segregating different parts of your program. You create the package and put your code into it:
package SomePackage;
sub some_subroutine { ... } # really SomePackage::some_subroutine
You load the module to get access to the package:
use SomePackage; # read and compile the module file
SomePackage::some_subroutine( ... );
A Perl class is a package and its associated behavior. The methods in a class are just normal subroutines, although when we treat the subroutines as methods, the first parameter is the thing (a package name or object, also known as the referent) that called method:
package SomeClass;
sub class_method { my( $class, #args ) = #_; ... }
sub instance_method { my( $self, #args ) = #_; ... }
Since the class is just a package like any other package and probably lives in a module, you access it the same way with use:
use SomeClass;
my $i = SomeClass->class_method( ... );
The OO arrow syntax does some special stuff to let the some_method subroutine know that it's being called as a method. Perl puts the referent (the SomeClass in this case) as the first argument. Additionally, when using the OO syntax, Perl knows to use its inheritance features.
Methods called with '->' get the referent as the first parameter to the method, so this call:
SomeClass->new('world');
is syntactically the same as if you had called it with the class name as the first parameter:
SomeClass::new( 'SomeClass' ,'world'); # no inheritance this way
That works the same for objects too. When an object is the referent:
my $i = SomeClass->new();
$i->bar( 'world');
the object is the first parameter as the method:
SomeClass::bar($i, 'world');
Perl doesn't have classes. It has namespaces that you change with package. For the complete details of Perl OOP, see Intermediate Perl or Object Oriented Perl. You can also see the perltoot and perlboot documentation. In short, Perl fakes what people expect "real" classes to be with packages, normal subroutines, and references.
A module is a distributable piece of code contained in a file. See perlmod.
I say more about this in my post for The Effective Perler, Find a module's release managers. I don't get into the OO stuff, but I talk about the other terms around "module".