Returning generator from a function

Question

I'm slowly getting to wrap my head around Python generators.

While it's not a real life problem for now, I'm still wondering why I can't return a generator from a function.

When I define a function with yield, it acts as a generator. But if I define it inside another function and try to return that instead, I get an ordinary function, i.e. not a generator with next method.

In other words, why the give_gen() approach in code below does not work?

#!/usr/bin/python import time def gen(d): n = 0 while True: n = n + d time.sleep(0.5) yield n def give_gen(d): def fn(): n = 0 while True: n = n + d time.sleep(0.5) yield n return fn if __name__ == '__main__': g = give_gen(3) # does not work g = gen(3) # works well while True: print g.next() # AttributeError: 'function' object has no attribute 'next' # in case of give_gen

Why can't I return a generator from a function?

Martijn Pieters · Accepted Answer · 2013-08-15 12:24:46Z

A generator function returns a generator only when called. Call fn to create the generator object:

return fn()

or call the returned object:

g = give_gen(3)()

You did call gen(); had you referred to just gen without calling it you'd have a reference to that function.

Collectives™ on Stack Overflow

Returning generator from a function

1 Answer 1

Comments

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Related