How do I pass a variable through multiple subroutines in perl? - perl

Question

I have a variable currentUser in a sub routine. It carries through to one subroutine, but does not carry through to another subroutine. How can I pass a variable through multiple subroutines while keeping the value?
sub login {
&app_header;
print <<EOF;
<form name="macform" method="POST" enctype="application/x-www-form-urlencoded" action="$fullurl">
...stuff
EOF
}
sub html_menu {
$me = $currentUser;
print $me;
print <<EOF;
<form name="menuform" method="POST" enctype="application/x-www-form-urlencoded" action="$fullurl">
..stuff
EOF
&app_list_button;
print "<br>";
&app_search_button;
print "<br>";
&app_edit_button;
print "</div>";
}
When I attempt to do the same thing the html_form sub does with currentUser in a new sub called after html_form), the variable does not display as what the user entered during login.

Answer 1

sub first {
my $currentUser = shift;
second($currentUser);
}
sub second {
my $currentUser = shift;
third($currentUser);
}
...and so on.
Passing a variable as argument is done through the #_ variable. You should not attempt to use global variables, it is not a good way to do it. You can do it like so:
my ($arg1, $arg2, #rest) = #_;
Or by using shift, pop and the other methods of manipulating arrays, like I above used shift.
If you are not already doing so, I recommend very strongly that you use
use strict;
use warnings;
It will help you solve many simple problems.

Answer 2

The safest way is to use the variable as an argument:
sub r1 {
my $arg = shift;
r2($arg);
}
sub r2 {
my $arg = shift;
print "$arg in r2\n";
}

Answer 3

When I attempt to do the same thing the html_form sub does with currentUser in a new sub called after html_form), the variable does not display as what the user entered during login.
Firstly, this is a rather complex example. You should have tried to break it down. To add to what others have said thus far, I'm going to tell you the & prefix on a subroutine's name without adding () changes the meaning. You use this in your examples, and I'm not sure if you know what it does. All of these have slightly different meanings.
foo() This simply calls sub foo with no arguments.
&foo This calls foo passing in #_ implicitly. Important note, if foo modifies #_ by reference then the #_ will change for the callee too.
&foo() This is just a relic of perl4. This calls foo with the explicit subref sigil. This is deprecated in all contexts.
there are also minor differences with prototypes and the & sigil that are outside of the bounds of the question
For more information see perldoc -q "calling a function" and perldoc perlsub