Re^8: Perl vs C

[JavaFan]

Now you're modifying an array, swapping one value (a list) for another (another list). Calling the array @list doesn't change the fact it's an array. The code below doesn't modify '3' either - it just replaces the value of a variable.

   my $var = 3;
   $var++;
   say $var;  # Prints 4.

[Marshall]

I don't understand what you mean. Sure it is possible to modify a list! Just like in C, I can pass a reference to a list to be modified and it will be modified! In the code below, I am modifying the actual list since the sub uses the reference to that list.

#!usr/bin/perl -w
use warnings;

sub modify_list
{
   my $listRef = shift;
   @$listRef = grep{!/^b/}@$listRef;
   #note there is no "return" value here
   #the list has been modified!!!!
}

my @list = ("a1", "b23", "c45", "d1", "b43", "b65");
print join (" ",@list),"\n"; 
#prints: a1 b23 c45 d1 b43 b65

modify_list(\@list);
print join (" ",@list),"\n"; 
#prints a1 c45 d1

[ikegami]

You've only modified array @list again. No lists were modified.

You're also wrong about there being no return value. It's just being ignored.

[Marshall]

1. Yes all Perl subs have a return value. Correct.

2. Again we are into semantics. I modified the data structure that was passed into the sub by reference. So write some code that modifies a "list". What would that be? Some authors say that @xyz is a list and that xyz is an array variable that defines a list. What's wrong with saying that @xyz is a list? Or that xyz in the context of @ is a list? Or that (@xyz >1) is a list in a scalar context?

[chromatic]

Just like in C, I can pass a reference to a list to be modified and it will be modified!

Please demonstrate, with working and tested Perl, how to take a reference to a list without using an array. At that point, I will believe that Perl supports mutable first-class lists as data structures (instead of multiple items on the Perl stack or a series of comma-separated expressions in source code).