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Date: Mon, 20 Aug 2012 22:07:14 +0200
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To: Nikita Popov <nikita.ppv@gmail.com>
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Subject: Re: [PHP-DEV] [RFC] Generators
From: keisial@gmail.com (=?ISO-8859-1?Q?=C1ngel_Gonz=E1lez?=)

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 variables
to a class. But it's not trivial for any parser.