On 3 March 2016 at 11:52, 刘琦帆 <lqf....@gmail.com> wrote:
>
> "A yield statement is not allowed in the try clause of a try/finally
> construct. The difficulty is that there's no guarantee the generator will
> ever be resumed, hence no guarantee that the finally block will ever get
> executed; 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 case.
> 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However that information is out of date.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 = sys.stdin
try:
sys.stdin = newstdin
yield
finally:
sys.stdin = 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
--
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