comment on

Whilst the evaluation order is undefined, explaining the behaviour is fairly easy

P:\test>perl -MO=Deparse,-p
$a=$b=$c=1;
$a=$a++ + $a++;
$b=$b++ + $b;
$c=$c   + $c++;
print "$a $b $c\n";
^Z
($a = ($b = ($c = 1)));
($a = (($a++) + ($a++)));
($b = (($b++) + $b));
($c = ($c + ($c++)));
print("$a $b $c\n");
- syntax OK
[download]

If you work through unwinding the parens, then you will end up with the correct result. The behaviour is even definable and consistant.

Explaining why is this way is somewhat harder:)

Examine what is said, not who speaks.

"Efficiency is intelligent laziness." -David Dunham
"When I'm working on a problem, I never think about beauty. I think only how to solve the problem. But when I have finished, if the solution is not beautiful, I know it is wrong." -Richard Buckminster Fuller
If I understand your problem, I can solve it! Of course, the same can be said for you.

In reply to Re: Auto-increment frenzy by BrowserUk
in thread Auto-increment frenzy by Dist

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