If for some reason everything's fine, there's a couple other ways to try:

sub example {
    my ( $scalar1, $scalar2, @info1 ) = @_;
    ...
}
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&example( $scalar1, $scalar2, \@info1 ) = @_;
...
sub example {
    my $scalar1 = shift;
    my $scalar2 = shift;
    my $arrref = shift;
    my @info1 = @$arrref;
    ...
}
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-----------------------------------------------------
Dr. Michael K. Neylon - mneylon-pm@masemware.com || "You've left the lens cap of your mind on again, Pinky" - The Brain
It's not what you know, but knowing how to find it if you don't know that's important

$s1 = "first scalar";
$s2 = "second scalar";
@a1 = ("first item","second item","third item","fourth item");

&example($s1, $s2, @a1);

sub example {
    $es1 = shift;$es2 = shift;@ea1 = @_;
    print "S1: $es1\n";
    print "S2: $es2\n";
    print "ARRAY:\n";
    for $i (@ea1) {print "$i ; "}
}
#prints the following:
#S1: first scalar
#S2: second scalar
#ARRAY:
#first item ; second item ; third item ; fourth item ;
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shame on my empty head, the problem was as you both stated, in the code before passing the variables. the array was empty, so i couldnt get any value

Setting values to feed example() and printing inside the function works. You are courting trouble by reusing $scalar1 and friends as names in the sub and not making them lexical with my.

Please say what the symptoms of 'not working' are, and show how the variables are declared and assigned.

$ cat pit.pl
#!/usr/bin/perl --
use strict;
use warnings;

my $scalar1 = 'foo';
my $scalar2 = 'ferall';
my @info1 = ( 0..10 );

&example($scalar1, $scalar2, @info1);

sub example {
      $scalar1 = shift;$scalar2 = shift;@info1 = @_; # bad
# prefer lexical scope, ought not to reuse names
#     my ($scalar1, $scalar2, @info1) = @_;
print $scalar1,$/;
print $scalar2,$/;
print @info1,$/;
#      .....
}
$ ./pit.pl
foo
ferall
012345678910
$
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After Compline,
Zaxo