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Bug #78387 references won't work for coalesce result
Submitted: 2019-08-08 21:17 UTC Modified: 2019-08-09 07:57 UTC
From: mpatnode at popsugar dot com Assigned:
Status: Not a bug Package: Scripting Engine problem
PHP Version: 7.4.0beta2 OS: any
Private report: No CVE-ID: None
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 [2019-08-08 21:17 UTC] mpatnode at popsugar dot com 
Description:
------------
Not sure if this is a bug, but the behavior is unexpected.  If a coalesce result is used in a foreach with a reference, the reference is ignored.   

If it's not legal syntax, then the reference should generate a warning or notice.

Behaved the same on every PHP version I tested.


Test script:
---------------
$foo = ['bar1' => 1, 'bar2' => 2];
foreach ($foo ?? [] as &$bar) { $bar = 3; }
var_dump($foo)

Expected result:
----------------
array(2) {
  'bar1' =>
  int(3)
  'bar2' =>
  int(3)
}

Actual result:
--------------
array(2) {
  'bar1' =>
  int(1)
  'bar2' =>
  int(2)
}

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 [2019-08-08 21:48 UTC] requinix@php.net
-Status: Open +Status: Not a bug -Package: Arrays related +Package: Scripting Engine problem
 [2019-08-08 21:48 UTC] requinix@php.net 
This is documented, more or less, with the ?? operator itself.

https://www.php.net/manual/en/language.operators.comparison.php#language.operators.comparison.coalesce
> Note: Please note that the null coalescing operator is an expression, and that it doesn't evaluate to a variable,
> but to the result of an expression. This is important to know if you want to return a variable by reference. The
> statement return $foo ?? $bar; in a return-by-reference function will therefore not work and a warning is issued.

Since $foo ?? [] is an expression, $bar will not be a reference to a value inside $foo but to a value inside the *copy of* $foo that was "returned" by the expression. It's legal syntax, just not what you wanted it to do.
 [2019-08-08 22:22 UTC] mpatnode at popsugar dot com 
You suggest a warning should of been issued for asking for a reference from an expression? We look for messages at the Notice level and didn't see anything in this case.
 [2019-08-08 22:30 UTC] requinix@php.net 
As the note says,
> The statement return $foo ?? $bar; in a return-by-reference function will therefore not work and a warning is
> issued.
the warning is when returning from a return-by-reference function.
 [2019-08-09 07:57 UTC] nikic@php.net 
The reason why this code doesn't generate a warning is that iterating a value by-reference can still be meaningful: Either because it has interior references, or because it's an iterator that produces references.

I believe this kind of code used to generate a compile-error prior to PHP 5.5, but the restriction was removed as part of generator support (which can yield by reference).
 
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