Scalars in Perl - Can they be exported throughout the script

Question

Not sure how clear my title is, but im new to perl and had no other way to really describe it. What I'm trying to do is something like this:

if(condition){
     my $VAR = " ";
}

Then later use $VAR somewhere else...

if(! $USR){
     my $USR = "$VAR";
}

Is there a way to call a scalar that is referenced elsewhere that has been bracketed off? {}

Thanks in advance...

score 3 · Accepted Answer · answered Aug 29 '14 at 17:59

3

A variable that is declared inside of scope {} is destroyed outside of scope. You need to declare variable before. Always use strict and warnings it will save you a lot of time.

use strict;
use warnings; 

my $VAR;
my $USR;
if(condition){
     $VAR = " ";
}

if(! $USR){
    $USR = "$VAR";
}

answered Aug 29 '14 at 17:59

Andrey

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28

bummer, ok... yeah i use warning and strict. I was just hoping I could pull this little trick off. Anyway, thanks for the input – Nick Hatfield Aug 29 '14 at 18:37
2

@Nick Hatfield, The whole point of using `my` is to specifically disallow that. – ikegami Aug 29 '14 at 18:38

score 0 · Answer 2 · answered Aug 29 '14 at 18:05

In perl, the my variables are lexically scoped, so they can be accessed in that scope:

{ 
# start of lexical scope

my $VAR;

# end of lexical scope
}

In the above code $VAR can only be seen between the opening and closing braces, and after the ending brace $VAR no longer exists.

Scalars in Perl - Can they be exported throughout the script

2 Answers2