Re: Combine ??= and ?:= assignment operator rfc's.

Well, now the RFC has passed so congratulation on that one! I can agree a bit that the RFC itself is a bit under specified, but what to do about it? Some options: - Expand it in the PHP documentation, but then the documentation becomes the master not the RFC itself. - Update the RFC itself with some more examples, maybe not feasible given that the voting is over. - Update RFC with link to Ruby. What do you think? Anyway, I'm not so privy to the RFC process so maybe this is a non issue... Regards //Björn Larsson Den 2016-03-25 kl. 13:46, skrev Midori Kocak:

http://www.rubyinside.com/what-rubys-double-pipe-or-equals-really-does-5488.html <http://www.rubyinside.com/what-rubys-double-pipe-or-equals-really-does-5488.html>

On 25 Mar 2016, at 13:42, Nikita Popov <nikita.ppv@gmail.com> wrote: On Fri, Mar 25, 2016 at 11:59 AM, Midori Kocak <mtkocak@gmail.com <mailto:mtkocak@gmail.com>> wrote: Hi Everyone, I think it's better idea to combine those two assignment operator RFC’s. So I am going to close the current one and open ??= with ?:= What do you think? And we have to find better names. Wishes, Midori Kocak I'd prefer to keep them separate, or at least keep their votes separate. The ??= operator vote is currently unanimous at 24:0, while the ?:= vote was closed at something like 9:2, so there clearly are differences of opinion regarding these two operators. I'll use this chance for some comments on the proposal. I can see the general usefulness of ??=, but right now the RFC is severely underspecified and I'm uncomfortable voting on it in it's current form as so much will depend on the final implementation. So, what do I mean by underspecified? The only statement the RFC essentially makes is that $a ??= $b will be the same as $a = $a ?? $b, for variable-expression $a and expression $b. This statement, while a good high-level illustration, does not explain the exact behavior of this operator. For example, consider the expression $a[print 'X'] ??= $b. A simple desugaring into $a[print 'X'] = $a[print 'X'] ?? $b will result in 'X' being printed twice. However, this is not how all other existing compound assignment operators behave: They will print X only once, as the LHS is only evaluated once. I assume that ??= would behave the same way. However, with ??= the problem becomes more complicated. Let us assume that $a is an ArrayAccess object and consider the expression $a[0] ??= $b. Let us further assume that $x = $a->offsetGet(0) is non-null. Will $a[0] ??= $b result in a call to $a->offsetSet(0, $x)? This is what would normally happen with a compound assignment operator and what would be implied by the desugaring $a[0] = $a[0] ?? $b. However this assignment is not really necessary, as we're just reassigning the same value. So, does the call happen or not? Is the proper desugaring maybe if (!isset($a[0])) $a[0] = $b? Let us now assume that $a is a recursive ArrayAccess object with by-reference offsetGet() and consider the expression $a[0][1] ??= expr. For a normal compound assignment operator, this would issue the call sequence

     $b = expr;
     $x =& $a->offsetGet(0);
     $y = $x->offsetGet(1);
     $y OP= $b;
     $x->offsetSet(1, $y);

Note that we only issue one offsetSet() at the end. We do not refetch $x via $a->offsetGet(0). How would the same work with the ??= operator? As the RHS is evaluated lazily, it is my opinion that only performing the offsetSet() call without refetching $x beforehand would violate PHP's indirection memory model. Additionally as ??= has to fetch offsets in BP_VAR_IS mode, we likely wouldn't be able to write them without refetching anymore. So, what would be the desugared call sequence for $a[0][1] ??= expr? Something like this?

     if (!$a->offsetHas(0)) {
         goto assign;
     }
     $x = $a->offsetGet(0);
     if (x === null) {
         goto assign;
     }
     if (!$x->offsetHas(0)) {
         goto assign;
     }
     $y = $x->offsetGet(0);
     if ($y === null) {
         goto assign;
     }
     goto done;

assign:

     $b = expr;
     $x =& $a->offsetGet(0);
     $x->offsetSet(1, $b);

done: That would be some first thoughts on the issue, though I'm sure there are more subtleties involved. I'd like to see the exact behavior of ??= (and ?:=) specified. I'm also pretty sure that writing a patch for this will not be entirely easy. The combination of execute-once LHS side-effects and lazy RHS execution does not translate well to PHP's VM constraints. Regards, Nikita