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Re: How can I tell if I am inside a context manager?

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  Re: How can I tell if I am inside a context manager? Gerald Britton <gerald.britton@gmail.com> - 2011-02-01 13:38 -0500
    Re: How can I tell if I am inside a context manager? alex23 <wuwei23@gmail.com> - 2011-02-02 20:38 -0800

#55719 — Re: How can I tell if I am inside a context manager?

"Perhaps something like this:"

>>x = open('somefile')
>>if hasattr(x, '__enter__'):
>>    return false
>>with open('somefile') as x:
>>    do_something()

>>> class f():
    def __init__(s): pass
    def __enter__(s): return s
    def __exit__(s,a,b,c): return None

>>> x = f()
>>> hasattr(x, '__enter__')
True
>>> with f() as x:
	hasattr(x,'__enter__')

	
True
>>>

As you can see, the object has a '__enter__' method regardless of how
it was created.  Whatever the test, it needs to return False in the
first case and True in the second case, without modifying the class
definition.


Gerald Britton

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#55924

Gerald Britton <gerald.brit...@gmail.com> wrote:
> >>> x = f()
> >>> hasattr(x, '__enter__')
> True
> >>> with f() as x:
>         hasattr(x,'__enter__')
> True
>
> As you can see, the object has a '__enter__' method regardless of how
> it was created.  Whatever the test, it needs to return False in the
> first case and True in the second case, without modifying the class
> definition.

If you control all instantiation and this is to simplify code further
on, could you have a ContextF that's a child of F? The only difference
would then be the protocol methods.

A use case could also help here.

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