Morning Midori,
PHP doesn't use PHPUnit tests.
Please see: https://qa.php.net/write-test.php
Cheers
Joe
On Sun, Apr 3, 2016 at 3:41 PM, Midori Kocak <mtko...@gmail.com> wrote:
> Yes, I think I should too. But still no feedbacks :(
>
> > On 03 Apr 2016, at 13:15, Björn Larsson <bjorn.x.lars...@telia.com>
> wrote:
> >
> > Hi Midori,
> >
> > Will you update the RFC also? Even if it's not the normal way of doing
> > things, one should keep in mind that RFC's are often listed as references
> > in books about PHP, being the first piece of documentation. Two such
> > examples are:
> > - https://daveyshafik.com/archives/book/upgrading-to-php-7
> > # In Appendix B
> > - http://www.php7book.com
> > # At the end of every chapter
> >
> > Regards //Björn Larsson
> >
> > PS Maybe best to finish implementation and tests first.
> >
> > Den 2016-04-03 kl. 03:17, skrev Midori Kocak:
> >> Dear All,
> >>
> >> Based on the concerns I wrote some tests. Can you check those and give
> feedback? Also, in ruby, $a ||= $b, the implementation is not equal to $a =
> $a || $b, but is equal to $a || $a = $b; I am a little bit confused, I am
> not entirely sure, but I guess this approach would solve our problems.
> >>
> >> https://gist.github.com/midorikocak/abc9fd9b6ca30359d201bc859edba9ee <
> https://gist.github.com/midorikocak/abc9fd9b6ca30359d201bc859edba9ee>
> >>
> >> We can use these examples as the part of the new documentation and as a
> guideline for implementation tests. Can you add also any extreme cases that
> should raise errors to my test?
> >>
> >> Yours,
> >> Midori
> >>
> >>> On 25 Mar 2016, at 13:42, Nikita Popov <nikita....@gmail.com> wrote:
> >>>
> >>> On Fri, Mar 25, 2016 at 11:59 AM, Midori Kocak <mtko...@gmail.com
> <mailto:mtko...@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
> >>
> >
>
>
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