Return in generator together with yield

In Python 2 there was an error when return was together with yield in a function definition. But for this code in Python 3.3:

def f():
return 3
yield 2
  

x = f()
print(x.__next__())

there is no error that return is used in function with yield. However when the function __next__ is called then there is thrown exception StopIteration. Why there is not just returned value 3? Is this return somehow ignored?

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The return value is not ignored, but generators only yield values, a return just ends the generator, in this case early. Advancing the generator never reaches the yield statement in that case.

Whenever a iterator reaches the 'end' of the values to yield, a StopIteration must be raised. Generators are no exception. As of Python 3.3 however, any return expression becomes the value of the exception:

>>> def gen():
...     return 3
...     yield 2
...
>>> try:
...     next(gen())
... except StopIteration as ex:
...     e = ex
...
>>> e
StopIteration(3,)
>>> e.value
3

Use the next() function to advance iterators, instead of calling .__next__() directly:

print(next(x))
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最佳答案

This is a new feature in Python 3.3 (as a comment notes, it doesn't even work in 3.2). Much like return in a generator has long been equivalent to raise StopIteration(), return <something> in a generator is now equivalent to raise StopIteration(<something>). For that reason, the exception you're seeing should be printed as StopIteration: 3, and the value is accessible through the attribute value on the exception object. If the generator is delegated to using the (also new) yield from syntax, it is the result. See PEP 380 for details.

def f():
return 1
yield 2


def g():
x = yield from f()
print(x)


# g is still a generator so we need to iterate to run it:
for _ in g():
pass

This prints 1, but not 2.


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