Re: [RFC] Generators

I think this doesn't bring anything further to talk about how to implement this in the parser. After reading the thread again, I would like to say that - either this should be implemented as currently defined - or it is implemented with the generator-keyword. But then it isn't a function anymore, because that's is ridiculous. In that case it should look like this: generator g() { yield; return; } 2012/8/20 Ángel González <keisial@gmail.com>: > 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. > > > > -- > PHP Internals - PHP Runtime Development Mailing List > To unsubscribe, visit: http://www.php.net/unsub.php > -- Freundliche Grüße Alex Aulbach