AW: passing an argument to a subroutine

[B. Fongo]

I assigned it to scalar to get the number of elements contained in the
array @x. 
Printing @x 0utside the subroutine gives me the right answer (5), but
within the sub, I get 1; which is wrong.

I need the correct number of elements within the sub to use it for a
database entry.
Why does Perl give me 5 elements outside the sub, but 1 within the sub?

This is exactly what my original sub does:


#!/usr/bin/perl -w

@group = param->('group');
# let's @group contains "John, mark, Peter"

# I 'm passing exactly "John, mark, Peter" to the sub here.
do_db(@group);


sub do_db {

      @x = @_; 
     # @x should be "John, mark, Peter" here too, but I get only John. 
     

      foreach (@x){
           insert into group_table (blah blah) values(blah blah);
     }

}









-----Ursprüngliche Nachricht-----
Von: Freddy Söderlund [mailto:[EMAIL PROTECTED] 
Gesendet: Donnerstag, 4. September 2003 11:46
An: [EMAIL PROTECTED]
Betreff: Re: passing an argument to a subroutine


----- Original Message ----- 
From: "B. Fongo" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>; <[EMAIL PROTECTED]>
Sent: Thursday, September 04, 2003 11:34 AM
Subject: passing an argument to a subroutine


> Hello
>
> An argument passed to a subroutine returns wrong value.

What value do you want it to print?

>
> Code example:
>
> @x = (1..5);
> $x = @x;

Are you trying to get $x to contain "12345"?

Let's assume you want to pass the @x array into your subroutine and have
it
print out it's contents from there.
Then I think this is what you want.

#!perl -w

@x = (1..5);

showValue (@x);

sub showValue {
  my @forwarded = @_;
  foreach(@forwarded){
   print "$_\n";
 }
}

What it really comes to is what value you want to pass to your
subroutine.
>From reading your code, I can see that you are converting an array into
a
scalar but is that really what you want?

Some more explanation of what you would like to acomplish would be nice.

>
> showValue ($x);   # or showValue (\$x);
>
>
> sub showValue {
>
>   my $forwarded = @_;
>   print $forwarded;  # print ${$forwarded};
>
> }
>
> In both cases, the script prints out 1.
> What is going on here?
>
> Thanks
>
> Babs
>



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