http://qs1969.pair.com?node_id=333312

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; } }

--
Pug

Replies are listed 'Best First'.
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
    HTH

    _________
    broquaint

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++ }; }
    Cheers - L~R
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.