This is probably completely wrong, but...

What i think is happening is this: the reference to the variable $i is being modified by the auto-(in|de)crement operators before the addition operation because of the precedence table. The reference then gets updated for the left $i for a right-side auto-increment because we're dealing with a reference to the actual variable, not a hard-coded number -- and in the case of the post-increment, because of a bug.

Examples:
Let's say we have the expression

$a + ++$a
[download]

This should leave $a at the value 1, but it should evaluate to 1. It evaluates to 2! Because the pre-increment updates the reference to $a on the left side of the addition operator as well. This is fine if it gets documented, but the following is a straight bug:

0 + $a++
[download]

We get a return value of 0, and $a is set to 1 as it should be. However if $a is on the left side:

$a + $a++
[download]

We get a return value of 1! $a is still set to 1 as it should be. If instead of $a we use some other (zeroed) variable it evaluates to 0 correctly. Somewhere Perl is messing up what happens on a right-side post-increment if the same variable is being added to itself. Maybe it's adding 1 to the left side even if it's a post-increment?

So, to explain the strange ++$a + $a++ i think it works like this:

1. the code is parsed, the ++'s get the precedence to go first
2. addition being a left-wise operator, ++$a gets evaluated first; $a gets ++'d to 1
3. $a++ generates a post-decrement after the statement is over and increments the left-side (erroneously)
4. the values (2 + 1?) are added returning 3
5. the post-decrement code is run, and $a is now 2

i could be wrong :/

Here's what i ran with the results i obtained:

#!/usr/bin/perl -w
use strict;

for $b (
    '$a++       ',
    '0    + $a++',
    '0    + ++$a',
    '++$a + $a',
    '$a   + ++$a',
    '$a   + $a++',
    '++$a + $a++',
    '$a++ + ++$a',
    '$a+=$a++'
) {
  $a = 0;
  print "$b\t=", eval($b), "\t\$a= $a\n";
}
__END__
# results:
$a++           =0    $a= 1
0    + $a++    =0    $a= 1
0    + ++$a    =1    $a= 1
++$a + $a    =2    $a= 1
$a   + ++$a    =2    $a= 1
$a   + $a++    =1    $a= 1
++$a + $a++    =3    $a= 2
$a++ + ++$a    =2    $a= 2
$a+=$a++    =1    $a= 1
[download]

That's a bug imho (though i could be wrong; better explenations anyone?)

hope this helps,
jynx

update: Hmmm, even stranger.

If we take

$a + $a + $a++ # = 0 (correct)
$a + $a + ++$a # = 1 (correct)
$a + $a++ + $a # = 2 (the trailing $a has no effect maybe?)
$a + ++$a + $a # = 3 (the ++ effects previous $a as before?)
[download]

Interesting...

So the first addition is probably becoming zero before the second addition is messing it up? And to further confuse things:

$a + $a + ++$a + $a++ # = 2 (correct!)
0 + ++$a + $a++ # = 3 (d'oh!)
[download]

i think at this point i've gone too far...

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