in reply to Re^4: Perl vs C
in thread Perl vs C

Yes, a char is a character. But char*, is a pointer to an character. And char** is a pointer to a pointer to a character (array).

Ok some examples:
in Perl: 

#!usr/bin/perl -w
use warnings;
my @list = ("a1", "b23", "c45", "d1");
print join (" ",@list),"\n"; 
#prints: a1 b23 c45 d1

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In Perl, the list "knows" how big it is.
This is not true in C! In C we have to do stuff to help things out! Like this: 
include <stdio.h>
int main ()
{
   char *list[] = {"a1", "b23", "c45", "d1",'\0'};
   // list is an array of pointers to chars 
   char **mover;
  
   for (mover=list; *mover ;mover++)
       printf ("%s ",*mover);
   printf("\n"); 
}
// prints: a1 b23 c45 d1

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For me, the difference is obvious! The Perl version is cool!
The alternative way in C would be to keep track of the number of items in the list array instead of a NULL value sentinel (and yes, there are more ways than than one I showed above). All ways are a pain in the rear compared with Perl!

Perl lists may be changed at will just like in C. I don't understand this comment: "lists are unmutable". Please explain, I just don't understand what you mean.

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Re^6: Perl vs C
by JavaFan (Canon) on Mar 16, 2009 at 14:07 UTC
    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.
[reply]
      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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