Greg Ewing writes: > Consider this: > > def g(): > return ((yield i) for i in range(10)) > > Presumably the yield should turn g into a generator, but... > then what? My brain is hurting trying to figure out what > it should do.
I don't understand why you presume that. The generator expression doesn't do that anywhere else. My model is that implicitly the generator expression is creating a function that becomes a generator factory, which is implicitly called to return the iterable generator object, which contains the yield. Because the call takes place implicitly = at compile time, all the containing function "sees" is an iterable (which happens to be a generator object). "Look Ma, no yields left!" And then g returns the generator object. What am I missing? In other words, g above is equivalent to def g(): def _g(): for i in range(10): # the outer yield is the usual implicit yield from the # expansion of the generator expression, and the inner # yield is explicit in your code. yield (yield i) return _g() (modulo some issues of leaking identifiers). I have not figured out why either your g or my g does what it does, but they do the same thing. _______________________________________________ Python-Dev mailing list [email protected] https://mail.python.org/mailman/listinfo/python-dev Unsubscribe: https://mail.python.org/mailman/options/python-dev/archive%40mail-archive.com