On Aug 17, David T-G said:
>% > 45 my $body =
>% > 46 &parseit
>% > 47 (
>% > 48 {ASCII=>$ascii,HTML=>$html},
>% > 49 {flag=>$flag,EMAIL=>$email,NAME_FIRST=>$fn,NAME_LAST=>$ln}
>% > 50 ) ;
>%
>% First, you can drop the & on the function call. It's not necessary.
>
>Interesting. I thought it was a good thing for clarification. Not
>needed because I'm passing params, maybe?
Ther 'perlsub' documentation sets the record straight. Basically, a '&'
is only needed...
1. when calling a user-defined function with the same name as a built-in
2. when taking a reference to a function
3. to avoid the function's prototype
4. when used with goto &function to avoid another stack level
5. in conjunction with no argument list, to pass the current @_
>% ($a, $b, @c) = (1, 2, 3, 4, 5); # $a=1, $b=2, @c=(3,4,5)
>% ($a, @b, $c) = (1, 2, 3, 4, 5); # $a=1, @b=(2,3,4,5) $c=undef
>% ($a, @b, @c) = (1, 2, 3, 4, 5); # $a=1, @b=(2,3,4,5) @c=()
>
>All of that makes sense. I guess it can only follow for passing to a
>function, eh?
Exactly, since the arguments to a function are placed into an ordinary
array (with a special name).
>% sub parse_it {
>% my ($tmpl, $user) = @_; # hash references are scalars
>% my %template = %$tmpl; # remember, hashes slurp, so we couldn't
>% my %userdata = %$user; # have done (%a, %b) = (%$c, %$d)
>
>Right; I can see that. So I still get a local %template that I could
>change as I like without trashing the original. Yay.
Very important thing you've noted. Yes, %template is just a COPY of the
hash stored in $tmpl. However, if any of the values in %$tmpl are
references themselves, changing them in %template would change them in
%$tmpl, because a "copy" of a reference is just the reference.
my $data = {
name => [ "Jeffrey", "Pinyan" ],
age => 21,
};
my %copy = %$data;
++$copy{age}; # in november
$copy{name}[0] = "Jeff"; # please, call me Jeff
print "@{ $data->{name} } is $data->{age} years old\n";
# "Jeff Pinyan is 21 years old"
print "@{ $copy{name} } is $copy{age} years old\n";
# "Jeff Pinyan is 22 years old"
You can see from that example that, since 'age' is not a reference,
changing it in %copy doesn't affect it in %$data. However, 'name' points
to an array reference, so $data->{name} and $copy{name} are really the
same array reference. (Read more about this in perlref and the like.)
--
Jeff "japhy" Pinyan [EMAIL PROTECTED] http://www.pobox.com/~japhy/
RPI Acacia brother #734 http://www.perlmonks.org/ http://www.cpan.org/
<stu> what does y/// stand for? <tenderpuss> why, yansliterate of course.
[ I'm looking for programming work. If you like my work, let me know. ]
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