Workaround for python 2.4's yield not allowed in try block with finally clause

问题

I'm stuck on python2.4, so I can't use a finally clause with generators or yield. Is there any way to work around this?

I can't find any mentions of how to work around this limitation in python 2.4, and I'm not a big fan of the workarounds I've thought of (mainly involving __del__ and trying to make sure it runs within a reasonable time) aren't very appealing.

回答1:

You can duplicate code to avoid the finally block:

try:
  yield 42
finally:
  do_something()

Becomes:

try:
  yield 42
except:  # bare except, catches *anything*
  do_something()
  raise  # re-raise same exception
do_something()

(I've not tried this on Python 2.4, you may have to look at sys.exc_info instead of the re-raise statement above, as in raise sys.exc_info[0], sys.exc_info[1], sys.exc_info[2].)

回答2:

The only code that's guaranteed to be called when a generator instance is simply abandoned (garbage collected) are the __del__ methods for its local variables (if no references to those objects exist outside) and the callbacks for weak references to its local variables (ditto). I recommend the weak reference route because it's non-invasive (you don't need a special class with a __del__ -- just anything that's weakly referenceable). E.g.:

import weakref

def gen():
  x = set()
  def finis(*_):
    print 'finis!'
  y = weakref.ref(x, finis)
  for i in range(99):
    yield i

for i in gen():
  if i>5: break

this does print finis!, as desired.

来源：https://stackoverflow.com/questions/2339358/workaround-for-python-2-4s-yield-not-allowed-in-try-block-with-finally-clause