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

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;

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print on an EBCDIC platform?

Abigail

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Re: Re: $a++ allowed by $a-- is not ! why?
by Anonymous Monk on Sep 02, 2003 at 16:07 UTC
    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.

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[d/l]
      Sorry, totally wrong, it should be: 
      'ab' = (1*(26**1)) + (2*(26**0))

'ab' -- = (1*(26**1)) + ((2*(26**0)) - (1*(26*0)))

        = (1*(26**1)) + (1*(26**0))

        = 'aa'

      [download]
      But ideally a string containing only alphanumerics with an alphabet in upper and lower case could be handled as a Base 62 number which would also allow decrement, because the question with the above is how does it handle numerics and upper case. Base 62 would solve that.
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[d/l]

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