Catching exception in context manager __enter__()

Question:

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

Asked By: tMC

||
Source

Answers:

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.

Answered By: Ignacio Vazquez-Abrams

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.

Answered By: Lauritz V. Thaulow

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.

Answered By: jfs

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
Answered By: newtover
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.

Answered By: Erich

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

Edit:
Unfortunately this approach still allows the user to create "dangling" resources that wont be cleaned up if he does something like:

foo = Foo() # this allocates resource without a with context.
raise ValueError("bla") # foo.__exit__() will never be called.

I am quite curious if this could be worked around by modifying the new implementation of the class or some other python magic that forbids object instantiation without a with context.

Answered By: Marti Nito

The docs contain an example that uses contextlib.ExitStack for ensuring the cleanup:

As noted in the documentation of ExitStack.push(), this method can be useful in cleaning up an already allocated resource if later steps in the __enter__() implementation fail.

So you would use ExitStack() as a wrapping context manager around the TstContx() context manager:

from contextlib import ExitStack

with ExitStack() as stack:
    ctx = TstContx()
    stack.push(ctx)  # Leaving `stack` now ensures that `ctx.__exit__` gets called.
    with ctx:
        stack.pop_all()  # Since `ctx.__enter__` didn't raise it can handle the cleanup itself.
        ...  # Here goes the body of the actual context manager.
Answered By: a_guest