in reply to Re^10: Perl vs C
in thread Perl vs C

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?

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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

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    However, @xyz does not. 

    @xyz = ( 'a', 'b', 'c' );
print(scalar( @xyz              ));   # 3

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      Well, 
      @xyz = ( 'a', 'b', 'c' );
print scalar (@xyz); #prints 3

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      is what I would have expected, the number of things in @xyz. 
      print(scalar( ( 'a', 'b', 'c' ) ));   # c

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      Is different that I would have expected.
      I will have to think more about how this occurs!
      But I will say that compared with scalar(@xyz) which happens very often, this is rare albeit interesting.
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