vyeddula has asked for the wisdom of the Perl Monks concerning the following question:

Hi perl monks.I am new to perl programming .I am a beginner.Here is my first question

#!/usr/bin/perl
use strict;
use warnings;
my $variable=total(1,2,3);
print"the variable is :$variable\n";
sub total
{
  my $element=0;
    foreach (@_)
    {
       $element=$element+$_;
    }
element;
}
Output is:
The variable is : 6

Now i changed the code to
#!/usr/bin/perl
use strict;
use warnings;
my $variable=total(1,2,3);
print"the variable is :$variable\n";
sub total
{
  my $element=0;
    foreach (@_)
    {
       $element=$element+$_;
    }
element;
print"the value of element is:$element\n";
}

The output is
the value of element is:6
the variable is :1
why the variable in this case is not showing as 6 is my question.

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Replies are listed 'Best First'.
Re: Doubt in functions
by BrowserUk (Patriarch) on Jul 06, 2012 at 00:28 UTC

    The value 1 in the second example is the successful return code from print;

    With the rise and rise of 'Social' network sites: 'Computers are making people easier to use everyday'
    Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.
    "Science is about questioning the status quo. Questioning authority".
    In the absence of evidence, opinion is indistinguishable from prejudice.

    The start of some sanity?
[reply]
Re: Doubt in functions
by toolic (Bishop) on Jul 06, 2012 at 00:50 UTC
    It is good practice to use an explicit return from a sub. See also perldoc perlsub.

    By the way, you really want to use a sigil in front of "element": $element

    Also, please use separate "code" tags so it is easier for us to distinguish your code from your output and your question.

[reply]
[d/l]

      Thank you all for the quick response The explicit usage with return command solved my problem Thanks again

[reply]
Re: Doubt in functions
by monsoon (Pilgrim) on Jul 06, 2012 at 00:18 UTC
    Quote from http://perldoc.perl.org/perlsub.html
    "If no return is found and if the last statement is an expression, its value is returned."
[reply]

      update: inadvertently answered monsoon instead of OP. My mistake. I read the doc quotation as the element about which OP had "doubt."</update>

      Perhaps the example will help:

      #!/usr/bin/perl
use 5.014;

# 980189

sub add_with_return {
    my $in_in_sub; my $in2_in_sub;
    $in_in_sub = $_[0];              # For utter clarity; no doubts; 
                                     # @_ is the array passed to the s
+ub
    $in2_in_sub = $_[1];             # $_[0] & $_[1] are the first two
+ elements of @_
    my $out_in_sub = $in_in_sub + $in2_in_sub; 
    return $out_in_sub;
}

sub concat_w_NO_return {
    my ($in_in_sub, $in2_in_sub);
    $in_in_sub = $_[0];              # For utter clarity; no doubts; 
                                     # but MANY simpler & BETTER ways 
+to do this
    my $in2_in_sub = $_[1];
    my $concat = "$in_in_sub" . "$in2_in_sub";  # Last statement, no r
+eturn; so the
}                                               # computed value of $c
+oncat gets returned

my $in = 1;
my $in2 = 2;
my $str1 = "foo";
my $str2 = "bar";

my $sum = add_with_return($in, $in2);
say "\t \$sum: $sum";

my $output = concat_w_NO_return($str1, $str2);  
say "\t \$output: $output";

      [download]

      Output:

               $sum: 3
         $output: foobar

      [download]

      P.S.: perldoc -f return puts the explanation this way:

      (In the absence of an explicit "return", a subroutine, eval, or do FILE automatically returns the value of the last expression evaluated.)
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