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Date: Thu, 31 Mar 2016 18:23:32 +0100
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Subject: Re: [PHP-DEV] Copy On Write and Assign By Reference perform different
 on PHP5 and PHP7
From: rowan.collins@gmail.com (Rowan Collins)

Huqiu Liao wrote on 31/03/2016 17:47:
> Hi, internals,
>
> I have a question about Assign By Reference and I posted on StackOverflow,
> I'd like to know the reason behind it, and I did not get any this kind of
> answer, can anyone give me some clues.
>
> ---
>
> We have a piece of simple code:
>
> 1    <?php
> 2    $i = 2;
> 3    $j = &$i;
> 4    echo (++$i) + (++$i);
> On PHP5, it outputs 8, because:
>
> $i is a reference, when we increase $i by ++i, it will change the zval
> rather than make a copy, so line 4 will be 4 + 4 = 8. This is Assign By
> Reference.

To me, that sounds like a fixed bug. It's noteworthy in the 3v4l results 
that HHVM also gives 7 as the answer, and I can't see any logical reason 
for the answer to be 8 - regardless of execution order, the result of 
the two calls to ++$i should never be the same.

It would be interesting to see the VLD dump in PHP 5, with and without 
reference, because looking at the PHP 7 one, it's clear that the two 
increments result in two temp vars so it shouldn't make any difference 
what type of zval was given as input:

    4     2        PRE_INC $2      !0     # !0 is compiled $i, $2 is a 
new value returned
          3        PRE_INC $3      !0     # !0 is referenced again, but 
$3 is unrelated to $2
          4        ADD ~4      $2, $3    # the two values are added, 
with no reference to !0

Maybe in PHP 5 the opcodes are the same, but $2 and $3 somehow end up as 
references to !0, rather than new zvals, as though the code were:

$i = 2;
$i++; $temp1 =& $i;
$i++; $temp2 =& $i;
echo $temp1 + $temp2;

rather than, as seems logical to me, HHVM, and PHP7:

$i = 2;
$i++; $temp1 = $i;
$i++; $temp2 = $i;
echo $temp1 + $temp2;

Regards,
-- 
Rowan Collins
[IMSoP]