in reply to Re^3: Perl::Improved Volume 0, Number 0
in thread Perl::Improved Volume 0, Number 0

Well, that would be: 
    $a [$J * ($i - 1) + $j]

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as opposed to 
    $a [($J + 1) * $i + $j]

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where $J is the highest second dimensional index. (Which if indices start with 1 happens to be identical to the number of indices).

If indices start at 1, taking a slice till the end of an array can be written as: 

    $a [5 .. @a]

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instead of having to write 
    $a [5 .. @a - 1]

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or to have to use the $#array form: 
    $a [5 .. $#a]

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Re^5: Perl::Improved Volume 0, Number 0
by ambrus (Abbot) on Aug 30, 2004 at 17:14 UTC
    If indices start at 1, taking a slice till the end of an array can be written as:

    The good solution for that imo would be to have the .. and ... operators as in ruby, that is, 3..8 means an inclusive range, but 3...8 means a range including 3 but not including 8.

    Update: I say this because I've never seen any good example of using $#{} other than a range ending in it.

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      Better still, have .. do some magic when unary to go right to the end of the array being indexed:

      $a[ 5 .. ]; # Equivilent to $a[ 5 .. $#a ]

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      Though I have no idea how hard that would be to add, or what compatibility issues come up.

      "There is no shame in being self-taught, only in not trying to learn in the first place." -- Atrus, Myst: The Book of D'ni.

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        I may be mistaken (I have a hard time following all the developments and keeping the status quo straight), but to my knowledge, that's supposed to work Perl6.

        Makeshifts last the longest.

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        That would be difficult, because 5..$#a is not always used in such a context (subscripting). It is used in other code like @data[sort { $data[$a]<=>$data[$b] } 0..$#data] where it would be difficult to derive the endpoint automatically.

        Btw I think perl6 will have infinite ranges, which is a good idea.

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