in reply to Re^4: Pre vs Post Incrementing variables
in thread Pre vs Post Incrementing variables

Mind you, it might make explaining this one a tad awkward :) 

$i = 0; print ++${\++$i}, 0+ ++$i, 0+ ++$i, ++${\++$i};;
6 3 4 6

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Re^6: Pre vs Post Incrementing variables
by repellent (Priest) on Sep 13, 2010 at 04:20 UTC
    The argument list to print is evaluated from left-to-right. ++${\++$i} is merely incrementing $i twice and returning an alias to $i.
    $i = 0; print ++${\++$i}, 0+ ++$i, 0+ ++$i, ++${\++$i};;
              ^              ^        ^     ^
             (1)            (2)      (3)   (4)

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    (1) & (4) return aliases to $i. Since $i has been incremented six times, we get 6 for both of them.

    (2) & (3) return copies to $i because of the zero additions 0+. We get 3 for (2) after three increments, and 4 for (3) with one more increment.

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      You appear to be saying that the ability to describe what the implementation does, justifies why it does it that way.

      The fact that this:

      $i=0; print $n = ++$i, $m = ++$i;;
1 2

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      produces a different result to this:
      $i=0; print ++$i, ++$i;;
2 2

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      Is just plain weird. Some, including me, would say 'broken'.

      Justifying the weirdness by saying "that's what it does", kinda reminds of Apple's "You're holding it wrong." explanation of Antennagate.

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        No, I made no implied justification thus far.

        The root of my reply was merely to reveal my experimentation with assigning the return of (++$i) to some value (since I also ran into the same error message you got).

        Then, I attempted to explain the snippet you posted so that others may find it helpful to see what's going on.

        My opinion? Yeah, the lvalue stuff is confusing and I don't quite see its usefulness. I would expect the behavior the OP expected.
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