How to write Python generator function that never yields anything

问题

I want to write a Python generator function that never actually yields anything. Basically it's a "do-nothing" drop-in that can be used by other code which expects to call a generator (but doesn't always need results from it). So far I have this:

def empty_generator():
    # ... do some stuff, but don't yield anything
    if False:
        yield

Now, this works OK, but I'm wondering if there's a more expressive way to say the same thing, that is, declare a function to be a generator even if it never yields any value. The trick I've employed above is to show Python a yield statement inside my function, even though it is unreachable.

回答1:

Another way is

def empty_generator():
    return
    yield

Not really "more expressive", but shorter. :)

Note that iter([]) or simply [] will do as well.

回答2:

An even shorter solution:

def empty_generator():
  yield from []

回答3:

Edit:

This doesn't work- read comments

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can't you just yield None? Seems like it would be easier to read to me.

来源：https://stackoverflow.com/questions/6266561/how-to-write-python-generator-function-that-never-yields-anything