Is it possible to ensure the __exit__() method is called even if there is an exception in __enter__()?
>>> class TstContx(object):
... def __enter__(self):
... raise Exception('Oops in __enter__')
...
... def __exit__(self, e_typ, e_val, trcbak):
... print "This isn't running"
...
>>> with TstContx():
... pass
...
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "<stdin>", line 3, in __enter__
Exception: Oops in __enter__
>>>
Edit
This is as close as I could get...
class TstContx(object):
def __enter__(self):
try:
# __enter__ code
except Exception as e
self.init_exc = e
return self
def __exit__(self, e_typ, e_val, trcbak):
if all((e_typ, e_val, trcbak)):
raise e_typ, e_val, trcbak
# __exit__ code
with TstContx() as tc:
if hasattr(tc, 'init_exc'): raise tc.init_exc
# code in context
In hind sight, a context manager might have not been the best design decision
Like this:
import sys
class Context(object):
def __enter__(self):
try:
raise Exception("Oops in __enter__")
except:
# Swallow exception if __exit__ returns a True value
if self.__exit__(*sys.exc_info()):
pass
else:
raise
def __exit__(self, e_typ, e_val, trcbak):
print "Now it's running"
with Context():
pass
To let the program continue on its merry way without executing the context block you need to inspect the context object inside the context block and only do the important stuff if __enter__ succeeded.
class Context(object):
def __init__(self):
self.enter_ok = True
def __enter__(self):
try:
raise Exception("Oops in __enter__")
except:
if self.__exit__(*sys.exc_info()):
self.enter_ok = False
else:
raise
return self
def __exit__(self, e_typ, e_val, trcbak):
print "Now this runs twice"
return True
with Context() as c:
if c.enter_ok:
print "Only runs if enter succeeded"
print "Execution continues"
As far as I can determine, you can't skip the with-block entirely. And note that this context now swallows all exceptions in it. If you wish not to swallow exceptions if __enter__ succeeds, check self.enter_ok in __exit__ and return False if it's True.
No. If there is the chance that an exception could occur in __enter__() then you will need to catch it yourself and call a helper function that contains the cleanup code.
You could use contextlib.ExitStack (not tested):
with ExitStack() as stack:
cm = TstContx()
stack.push(cm) # ensure __exit__ is called
with ctx:
stack.pop_all() # __enter__ succeeded, don't call __exit__ callback
Or an example from the docs:
stack = ExitStack()
try:
x = stack.enter_context(cm)
except Exception:
# handle __enter__ exception
else:
with stack:
# Handle normal case
See contextlib2 on Python <3.3.
if inheritance or complex subroutines are not required, you can use a shorter way:
from contextlib import contextmanager
@contextmanager
def test_cm():
try:
# dangerous code
yield
except Exception, err
pass # do something
I suggest you follow RAII (resource acquisition is initialization) and use the constructor of your context to do the potentially failing allocation. Then your __enter__ can simply return self which should never ever raise an exception. If your constructor fails, the exception may be thrown before even entering the with context.
class Foo:
def __init__(self):
print("init")
raise Exception("booh")
def __enter__(self):
print("enter")
return self
def __exit__(self, exc_type, exc_val, exc_tb):
print("exit")
return False
with Foo() as f:
print("within with")
Output:
init
Traceback (most recent call last):
File "<input>", line 1, in <module>
...
raise Exception("booh")
Exception: booh
class MyContext:
def __enter__(self):
try:
pass
# exception-raising code
except Exception as e:
self.__exit__(e)
def __exit__(self, *args):
# clean up code ...
if args[0]:
raise
I've done it like this. It calls __exit__() with the error as the argument. If args[0] contains an error it reraises the exception after executing the clean up code.
