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Date: Mon, 20 Aug 2012 23:55:27 +0200
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To: =?ISO-8859-1?Q?=C1ngel_Gonz=E1lez?= <keisial@gmail.com>
Cc: PHP internals <internals@lists.php.net>
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Subject: Re: [PHP-DEV] [RFC] Generators
From: nikita.ppv@gmail.com (Nikita Popov)

On Mon, Aug 20, 2012 at 10:07 PM, =C1ngel Gonz=E1lez <keisial@gmail.com> wr=
ote:
> On 20/08/12 15:12, Nikita Popov wrote:
>>> You could not decorate it and rely instead on the presence of the yield
>>> keyword, but parsers will thank knowing about it from the start rather
>>> than realising at mid-parsing that the function is a completely
>>> different beast.
>> No, parsers don't care about this. It's trivial to detect in both cases.
>>
>> Nikita
> Yes, it's trivial to detect, but how much work is for it to do so?
> Suppose the parser reads php code and outputs machine instructions. When
> it encouters a function(), it adds the function prologue, goes reserving
> stack
> space for the variables it encounters, and so on. Then you find a yield
> keyword.
> Options:
> - Discard the generated code and go back to the function begin, using
> this time
> an storage into an object instead of the stack.
> - Copy all the state to an object and return it.
> - Add an initial pass checking which functions are generators.
>
> PHP being a dynamic language, it may be able to move the function variabl=
es
> to a class. But it's not trivial for any parser.

Sorry, I don't understand what we are arguing about here. I
implemented this. I did the parser changes for this feature. And I
told you that they were simple either way (both with keyword, as in
the initial implementation, and without keyword, as in the current
implementation). For 3rd parties (like IDEs) it is even simpler
because their parsers are AST-based (unlike PHP's own parser), so they
shouldn't care at all.

Honestly, I stopped understanding what this whole discussion is about
around ~100 mails ago.

Nikita