Re^7: Perl vs C

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
[download]

Comment on Re^7: Perl vs C Download Code

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Re^8: Perl vs C by JavaFan (Canon) on Mar 16, 2009 at 14:23 UTC
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.` [download]	[reply] [d/l] [select]
Re^9: Perl vs C by Marshall (Canon) on Mar 16, 2009 at 15:00 UTC
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` [download]	[reply] [d/l]
Re^10: Perl vs C by ikegami (Patriarch) on Mar 16, 2009 at 15:34 UTC
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.	[reply] [d/l]
Re^11: Perl vs C by Marshall (Canon) on Mar 16, 2009 at 15:58 UTC
Re^12: Perl vs C by ikegami (Patriarch) on Mar 16, 2009 at 16:17 UTC
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Re^10: Perl vs C by chromatic (Archbishop) on Mar 16, 2009 at 18:13 UTC
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).	[reply]