Closure question

Pug has asked for the wisdom of the Perl Monks concerning the following question:

I have a question concerning Perl's closures.

The code below should, to me, completely work. But when I call doesnt_work @r is not initilized. 8( Which is surpising to me because perl knows that sub doesnt_work exists so why doesn't it init @r when it is called?

#!/usr/bin/perl -w
{       my @s=("This works\n");
    sub works { shift @s}
}
#This works..
print &works;

# Error!!
# This gives an error of Use of uninitialized value in print at ....
print &doesnt_work;
{       my @r=("Should this work..\n");
    sub doesnt_work{ pop @r;    }
}
[download]

--
Pug

Comment on Closure question Download Code

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Re: Closure question by broquaint (Abbot) on Mar 02, 2004 at 17:44 UTC
This is the nature of the execution flow of perl - `my @r` hasn't been 'executed' yet, so you'll want something along the lines of a `BEGIN` block if you want it to be initialized and have it after the subroutine call e.g `foo(); BEGIN { my $str = 'a string'; sub foo { print "\$str is $str\n" } } __output__ $str is a string` [download] HTH `_________ broquaint`	[reply] [d/l]
Re: Closure question by Limbic~Region (Chancellor) on Mar 02, 2004 at 18:05 UTC
Pug, What you are using is a one-use closure. Because it is in a naked block it can not be used multiple times. And as broquaint points out, it must be "seen" prior to being used. Here is an example of a re-usable closure that can be used before or after the sub has been "seen". `#!/usr/bin/perl use strict; use warnings; my $count_1 = iterator( 1 ); print $count_1->(), "\n" for 1 .. 10; my $count_2 = iterator( 'a' ); print $count_2->(), "\n" for 1 .. 10; sub iterator { my $iterator = shift; return sub { $iterator++ }; }` [download] Cheers - L~R	[reply] [d/l]
Re: Closure question by fletcher_the_dog (Friar) on Mar 02, 2004 at 18:05 UTC
I have run into this problem before and it has to do with the fact that the statement "my @r=..." hasn't been executed yet. Perl lets you use the @r under strict because your definition for "doesnt_work" is within the lexical scope of @r. But it does not automatically execute code just because it uses a variable that might possible be in a closure or just because it is in a block that contains a definition of a sub.	[reply]


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