如何可以访问的符号表当前包中的对象被实例化?例如,我有这样的事情:

my $object = MyModule->new;
# this looks in the current package, to see if there's a function named run_me
# I'd like to know how to do this without passing a sub reference
$object->do_your_job;

如果在do_your_job我使用__PACKAGE__的实施,它会在MyModule包搜索。我怎样才能使它看起来在正确的包?

编辑:我会尽力使这个更清楚。假设我有以下代码:

package MyMod;

sub new {
    return bless {},$_[0]
}

sub do_your_job {
    my $self = shift;
    # of course find_package_of is fictional here
    # just for this example's sake, $pkg should be main
    my $pkg = find_package_of($self);
    if(defined &{ $pkg . '::run_me' }) {
        # the function exists, call it.
    }
}

package main;

sub run_me {
   print "x should run me.\n";
}

my $x = MyMod->new;

# this should find the run_me sub in the current package and invoke it.
$x->do_your_job;

现在,$x应该以某种方式发现main是当前包,并搜索它的符号表。我试着用Scalar::Util的祝福,但它仍然给了<=>的<=>而不是我。我们希望,这是一有点清晰了。

有帮助吗?

解决方案

您只是想 caller

'owning_package'告诉您从它被称为包。 (在这里,我增加了一些标准的Perl。)

use Symbol qw<qualify_to_ref>;
#...
my $pkg = caller;

my $symb   = qualify_to_ref( 'run_me', $pkg );
my $run_me = *{$symb}{CODE};
$run_me->() if defined $run_me;

要看看它,看看它的定义,然后看看它调用它会复制它作为标准的Perl不做公共子表达式消除,所以你还不如1)检索它,和2)检查definedness槽,以及3)运行它,如果它被定义。

现在,如果你在一个包中创建一个对象,并在另一个使用它,这不会是太大的帮助。你可能会需要在构造函数中添加其他字段一样$x->{owning_package}

package MyMod;

#...
sub new { 
    #...
    $self->{owning_package} = caller || 'main';
    #...
}

现在'main'将包含<=>。

其他提示

请参阅的perldoc -f呼叫者

#!/usr/bin/perl

package A;
use strict; use warnings;

sub do_your_job {
    my ($self) = @_;
    my ($pkg) = caller;
    if ( my $sub = $pkg->can('run_me') ) {
        $sub->();
    }
}

package B;
use strict; use warnings;

sub test {
    A->do_your_job;
}

sub run_me {
    print "No, you can't!\n";
}

package main;

use strict; use warnings;

B->test;

输出:

C:\Temp> h
No, you can't!
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