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From: Oscar Benjamin <oscar.j.benjamin@gmail.com>
Newsgroups: comp.lang.python
Subject: Re: yield in try/finally case
Date: Thu, 3 Mar 2016 12:13:36 +0000
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On 3 March 2016 at 11:52, =E5=88=98=E7=90=A6=E5=B8=86 <lqf.txx@gmail.com> w=
rote:
>
> "A yield statement is not allowed in the try clause of a try/finally cons=
truct.  The difficulty is that there's no guarantee the generator will ever=
 be resumed, hence no guarantee that the finally block will ever get execut=
ed; that's too much a violation of finally's purpose to bear." from https:/=
/www.python.org/dev/peps/pep-0255/
>
> But, meanwhile, the code showed on that page use yield in a try/finally c=
ase.
> It really puzzles me. Is there anything wrong?

I think what it means is that you can put a yield in the finally block
but not the try block so:

# Not allowed
def  f():
    try:
        yield 1
    finally:
        pass

# Allowed
def f():
    try:
        pass
    finally:
        yield 1

However that information is out of date. The restriction was removed
in some later Python version. Actually the construct is quite common
when using generator functions to implement context managers:

@contextlib.contextmanager
def replace_stdin(newstdin):
    oldstdin =3D sys.stdin
    try:
        sys.stdin =3D newstdin
        yield
    finally:
        sys.stdin =3D oldstdin

Although the restriction was removed the problem itself still remains.
There's no guarantee that a finally block will execute if there is a
yield in the try block. The same happens if you use a context manager
around a yield statement: the __exit__ method is not guaranteed to be
called. One implication of this is that in the following code it is
not guaranteed that the file will be closed:

def upperfile(filename):
    with open(filename) as fin:
        for line in fin:
            yield line.upper()

--
Oscar