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Calls the caller's method

by anazawa (Scribe)
on Jul 10, 2012 at 15:33 UTC ( #980893=perlquestion: print w/replies, xml ) Need Help??
anazawa has asked for the wisdom of the Perl Monks concerning the following question:

Hi Monks, I'm working on the following situation:
# package Foo; use strict; use warnings; use Bar; sub foo { warn 'Foo::foo was called' } Bar->bar; 1;
A package Foo calls Bar's bar() method. Foo defines foo() method which prints a message to STDERR. On the other hand, looks as follows:
# package Bar; use strict; use warnings; sub bar { Foo->foo } 1;
Bar defines bar() method which calls Foo's foo() method. I executed and confirmed the script output 'Foo::foo was called':
$ ls $ perl Foo::foo was called

This result shows us that Bar can call Foo's foo() method without loading (I mean, without 'use Foo;') In other words, Bar can call the caller's method ( Foo::foo() ). If so, why is this possible? Is this one of well-known specifications? Which perldoc helps me understand this result?


Replies are listed 'Best First'.
Re: Calls the caller's method
by tobyink (Abbot) on Jul 10, 2012 at 16:16 UTC

    If you're calling a function as a method (i.e. with ->) or calling it using a fully-qualified name (e.g. Scalar::Util::blessed()), then you do not need to use the module in question - all that is important is that it's been loaded (i.e. use, require or do) somewhere in this process. (And your caller has always been at least partially loaded before you were.)

    perl -E'sub Monkey::do{say$_,for@_,do{($monkey=[caller(0)]->[3])=~s{::}{ }and$monkey}}"Monkey say"->Monkey::do'
      Thank for your suggestion. I totally agree with you. I was blind to this issue :-)
Re: Calls the caller's method
by daxim (Chaplain) on Jul 10, 2012 at 15:56 UTC
    When sub Bar::bar executes, the package Foo is already loaded and the symbol Foo::foo is already installed (as subroutine).
      That makes sense. Foo::foo() is globalized, right? I remember 'our' declaration globalizes variables.

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