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
