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Perl, using variable from within While loop outside of the loop?

This seems really simple but it is giving me a hard time figuring it out as I'm new to perl.. I've been looking through a lot of documentation now about loops and I am still stumped by this... I have a sub that contains a while loop and I want to use a variable value from within the loop outside of the loop (after the loop has run), however when I try to print out the variable, or return it out of the sub, it doesn't work, only when I print the variable from within the loop does it work.. I would appreciate any advice as to what I'm doing wrong.

Doesn't work (doesn't print $test ):

sub testthis {    
    $i = 1;
    while ($i <= 2) {    
        my $test = 'its working' ;    
        $i++ ;
    }
    print $test ;
}

&testthis ;

Works, prints $test:

sub testthis {
    $i = 1;
    while ($i <= 2) {
        my $test = 'its working' ;
        $i++ ;
        print $test ;
    }
}

&testthis ;
like image 851
Rick Avatar asked Jul 14 '10 09:07

Rick


3 Answers

You declare variable test inside the loop, so it scope is the loop, as soon as you leave the loop the variable is not longer declared.
Add my $test; just between $i=1 and while(..) and it will work. The scope will now be the entire sub instead of only the loop

like image 195
radius Avatar answered Nov 04 '22 00:11

radius


Place my $test before the while loop. Note that it will only contain the last value that is assigned in the while loop. Is that what you are after?

// will print "it's working" when 'the loop is hit at least once,
// otherwise it'll print "it's not working"
sub testthis {
    $i = 1;
    my $test = "it's not working";

    while ($i <= 2) {
        $test = "it's working";
        $i++ ;
    }
    print $test ;
}
like image 26
Abel Avatar answered Nov 04 '22 00:11

Abel


you can try this one:

sub testthis {
my $test
$i = 1;
while ($i <= 2) {

$test = 'its working' ;

$i++ ;

print $test ;
}

}

&testthis ;

Note: whenever write perl code, its better to add use strict; and use warning in the beginning of the code.

like image 29
Nikhil Jain Avatar answered Nov 03 '22 23:11

Nikhil Jain