Why does next raise a 'StopIteration', but 'for' do a normal return?

自闭症网瘾萝莉.ら 提交于 2019-12-17 06:34:27

问题


In this piece of code, why does using for result in no StopIteration or is the for loop trapping all exceptions and then silently exiting? In which case, why do we have the extraneous return?? Or is the raise StopIteration caused by: return None?

#!/usr/bin/python3.1
def countdown(n):
    print("counting down")
    while n >= 9:
        yield n
        n -= 1
    return

for x in countdown(10):
    print(x)

c = countdown(10)
next(c)
next(c)
next(c)

Assuming StopIteration is being triggered by: return None. When is GeneratorExit generated?

def countdown(n):
    print("Counting down from %d" % n)
    try:
        while n > 0:
            yield n
            n = n - 1
    except GeneratorExit:
        print("Only made it to %d" % n)

If I manually do a:

c = countdown(10)
c.close() #generates GeneratorExit??

In which case why don't I see a traceback?


回答1:


The for loop listens for StopIteration explicitly.

The purpose of the for statement is to loop over the sequence provided by an iterator and the exception is used to signal that the iterator is now done; for doesn't catch other exceptions raised by the object being iterated over, just that one.

That's because StopIteration is the normal, expected signal to tell whomever is iterating that there is nothing more to be produced.

A generator function is a special kind of iterator; it indeed raises StopIteration when the function is done (i.e. when it returns, so yes, return None raises StopIteration). It is a requirement of iterators; they must raise StopIteration when they are done; in fact, once a StopIteration has been raised, attempting to get another element from them (through next(), or calling the .next() (py 2) or .__next__() (py 3) method on the iterator) must always raise StopIteration again.

GeneratorExit is an exception to communicate in the other direction. You are explicitly closing a generator with a yield expression, and the way Python communicates that closure to the generator is by raising GeneratorExit inside of that function. You explicitly catch that exception inside of countdown, its purpose is to let a generator clean up resources as needed when closing.

A GeneratorExit is not propagated to the caller; see the generator.close() documentation.



来源:https://stackoverflow.com/questions/14413969/why-does-next-raise-a-stopiteration-but-for-do-a-normal-return

易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!