in reply to Re: Re: $a++ allowed by $a-- is not ! why?
in thread $a++ allowed by $a-- is not ! why?

However, why is the auto-decrement operator not magical in nature?

Suppose it was, what would the following print: 

    my $a = "a";
    $a --;
    print $a, "\n";

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Abigail

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Re: Re: $a++ allowed by $a-- is not ! why?
by Anonymous Monk on Sep 02, 2003 at 13:52 UTC
    It would output "\`\n" if it acted upon the ordinal value.
[reply]
      It would output "\`\n" if it acted upon the ordinal value.

      But since ++ doesn't act on ordinal values, why should --? Furthermore, what should: 

          my $a = "a";
    $a --;
    print $a;

      [download]

      print on an EBCDIC platform?

      Abigail

[reply]
[d/l]
        It depends upon which EBCDIC variant the platform was running but IIRC with Western European platforms it would be a speech mark. But I take your point.

        But since the sequence is a b c d e f g h i etc. why not apply magic to the -- operation and use the numbers as Base 26.

        So a decrement would of ab would become:

        
'ab' = (1*(25*1))+(2*(25*0)) 

'ab'-- = (1*(25*1))+((2*(25*0))-(1*(25*0))) 
       = (1*(25*1))+((1*(25*0))
       = 'aa'

        [download]

        Then 'a'-- would always = -1 and decrements become OK with a predictable behaviour.

[reply]
[d/l]

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