Re^9: Perl vs C

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]

Comment on Re^9: Perl vs C Download Code

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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
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?	[reply]
Re^12: Perl vs C by ikegami (Patriarch) on Mar 16, 2009 at 16:17 UTC
So write some code that modifies a "list". What would that be? We're still waiting for you to show it's possible. Some authors say that @xyz is a list and that xyz is an array variable that defines a list. So far, I've only seen you, and that's the problem. What's wrong with saying that @xyz is a list? It's very confusing. It leads to contradictory statements. If we went by your definition, all of the following statements are true depending on whether you are talking about `(foo,bar)` or `@foo`. A list can't be modified. A list can be modified. A list in scalar context evaluates to its last item. A list in scalar context evaluates to the number of elements it contains. Nested lists are flattened into a single list. Nested lists aren't flattened into a single list. [What's wrong with saying] that xyz in the context of @ is a list? Now you want to redefine "context" too! What do you have against being understood. [What's wrong with saying] that (@xyz >1) is a list in a scalar context? You wouldn't be talking about Perl. In Perl, lists evaluate to their last element in scalar context. `print(scalar( ( 'a', 'b', 'c' ) )); # c` [download] However, `@xyz` does not. `@xyz = ( 'a', 'b', 'c' ); print(scalar( @xyz )); # 3` [download]	[reply] [d/l] [select]
Re^13: Perl vs C by Marshall (Canon) on Mar 16, 2009 at 17:02 UTC
Re^14: Perl vs C by ikegami (Patriarch) on Mar 16, 2009 at 17:21 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]