Reputation: 15641
In Python 2 it used to cause an error when return
occurred together with yield
inside 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?
Upvotes: 102
Views: 64721
Reputation:
This is a new feature in Python 3.3. 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
.
Upvotes: 122
Reputation: 1124748
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))
Upvotes: 51