in reply to Re^5: Perl vs C
in thread Perl vs C

I don't understand this comment: "lists are unmutable". Please explain, I just don't understand what you mean.
Try to make pure Perl program that modifies a list. Come back when you either have such a program, or when you are convince this isn't possible.

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Re^7: Perl vs C
by Marshall (Canon) on Mar 16, 2009 at 14:15 UTC
    Here is a simple thing that removes things from a Perl list that do not match a criteria. 
    #!usr/bin/perl -w
use warnings;

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

@list = grep{!/^b/}@list;  #remove things that start with b
print join (" ",@list),"\n"; 
#prints: a1 c45 d1

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      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.

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        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

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