Hi Midori,
I was going to, but then found that comprehensive guide for writing
tests, which has it all covered.
Current tests covering Zend are in Zend/tests, you can look in there
for inspiration, perhaps looking at the tests for a similar feature.
As for what to test, I can't say; I haven't seen a working
implementation yet.
Cheers
Joe
On Sun, Apr 3, 2016 at 5:22 PM, Midori Kocak <mtko...@gmail.com> wrote:
> Joe,
>
> Also that would be more helpful if you wrote some examples or guides, with
> your advises instead of writing one sentence emails. I would be more happy
> as a rookie that way.
>
> Yours,
> Midori
>
> On 03 Apr 2016, at 18:17, Midori Kocak <mtko...@gmail.com> wrote:
>
> Hello Joe,
>
> Those were examples for your feedback.
>
> Thanks,
> Midori
>
> On 03 Apr 2016, at 18:16, Joe Watkins <pthre...@pthreads.org> wrote:
>
> 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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>>
>>
>
>
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