trim spaces

ok, thanks, although i am abit unclear or what the shift does...(the purpose of using shift in the function)
thanks anyway

$ perldoc -f shift
       shift ARRAY
       shift   Shifts the first value of the array off and returns it, short-
               ening the array by 1 and moving everything down.  If there are
               no elements in the array, returns the undefined value.  If
               ARRAY is omitted, shifts the @_ array within the lexical scope
               of subroutines and formats, and the @ARGV array at file scopes
               or within the lexical scopes established by the "eval ''",
               "BEGIN {}", "INIT {}", "CHECK {}", and "END {}" constructs.

               See also "unshift", "push", and "pop".  "shift" and "unshift"
               do the same thing to the left end of an array that "pop" and
               "push" do to the right end.

If you have only one arg, then whether you do

my $arg = shift;

or

my $arg = @_;

is a matter of preference.

By default the arguments to a function are passed in the array named @_. TinTin's shift is shifting the first variable off the array.  The following is functionally equivalent:

sub trim {
 my $string = shift(@_);
 $string =~ s/\s//g;
 return $string;
}

The difference is that I explicitly specified the array, but Tintin uses one of Perl's niceties; if the array to shift is not specified, @_ will be used. This may be a little cryptic, but you better get used to it; it is a very common practice. Actually, I see arguments processed Tintin's way more often than I see yours. You have used the same concept in the line
    s/^\s+//;
Only that operates on the implicit scalar variable named $_

Lets analyze your regexes to see why they didn't work:

   1.  s/^\s+//;
        Will only remove the white space at the begining of a line
        '^' says to match the begining of a line
        "\s+" says to match one or more whitespaces
    2. s/\s+$//;
        Will only remove the white space at the end of a line
        "\s+" says to match one or more whitespaces(same as the first regex)
        '$' says to match the end of a line

You must have been mistaken when you said that this will not delete whitespace from the end. If you are correct, then there is something you haven't shown us.

Here is how Tintin's regex, "s/\s//g" works: The "\s" says match whitespace, no surprise there, you already used it. The g at the end stands for global. Thus, the regex is applied to the entire line until no more matches occur.

Very well explained djplaistow.

I signed up 'cause I just had to say "excellent explanation!"

Additional Resources:
http://en.wikipedia.org/wiki/Trim_(programming)