CODEWITHSUNDEEP
pythonpython-3.xexceptionwith-statementcontextmanager
Gulzar
Gulzar

Reputation: 28014

How to check if resouce was allocated correctly in a python ContextDecorator?

I have a class 

class Resource:
    __init__(self):
        self.resource = ...
    __enter__(self):
        #can fail, such as file open returning error, memory allocation fail, or any more complicated failure
    __exit__(self,*args):
        ...

Now I want

with Resource() as r:
    r.do_stuff()

But if r has failed to __enter__() successfully, this fails.

What is the correct, pythonic way of handling this?

I didn't want to use some is_allocated_correctly

like so

with Resource() as r:
    if r.is_allocated_correctly():
        r.do_stuff()

as it breaks the point of the with statement.

Please give me some idea what to do here.

Upvotes: 0

Views: 43

Answers (1)

Open AI - Opting Out
Open AI - Opting Out

Reputation: 24153

The purpose of the with statement is correctly de-allocate resources or reset state after the block has finished.

If you can't enter the context block without an error, that needs handling outside of the with statement.

Surround the whole thing with a try/except:

try:
    with Resource() as r:
        r.do_stuff()
except ResourceException as error:
    handle_error(error)

Or if there's nothing you can do about the error, just let it pass:

with Resource() as r:
    r.do_stuff()

Upvotes: 2

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