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












0














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.






python python-3.x exception with-statement contextmanager







share|improve this question






















  • assuming the resource is allocated either in the __init__ or __enter__ AND an allocation failure would raise an exception (which is what should happen for any not-totally-broken lib allocating resources of any kind) AND you don't do anything stupid (like silently swallowing this exception) in your Resource class, then you will NEVER enter the with block if the resource is not "correctly allocated". IOW, you have nothing special to do.
    – bruno desthuilliers
    Dec 27 at 13:53
















0














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.






python python-3.x exception with-statement contextmanager







share|improve this question






















  • assuming the resource is allocated either in the __init__ or __enter__ AND an allocation failure would raise an exception (which is what should happen for any not-totally-broken lib allocating resources of any kind) AND you don't do anything stupid (like silently swallowing this exception) in your Resource class, then you will NEVER enter the with block if the resource is not "correctly allocated". IOW, you have nothing special to do.
    – bruno desthuilliers
    Dec 27 at 13:53














0












0








0







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.






python python-3.x exception with-statement contextmanager







share|improve this question













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.






python python-3.x exception with-statement contextmanager




python python-3.x exception with-statement contextmanager






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Dec 27 at 13:32









Gulzar

7511520




7511520












  • assuming the resource is allocated either in the __init__ or __enter__ AND an allocation failure would raise an exception (which is what should happen for any not-totally-broken lib allocating resources of any kind) AND you don't do anything stupid (like silently swallowing this exception) in your Resource class, then you will NEVER enter the with block if the resource is not "correctly allocated". IOW, you have nothing special to do.
    – bruno desthuilliers
    Dec 27 at 13:53


















  • assuming the resource is allocated either in the __init__ or __enter__ AND an allocation failure would raise an exception (which is what should happen for any not-totally-broken lib allocating resources of any kind) AND you don't do anything stupid (like silently swallowing this exception) in your Resource class, then you will NEVER enter the with block if the resource is not "correctly allocated". IOW, you have nothing special to do.
    – bruno desthuilliers
    Dec 27 at 13:53
















assuming the resource is allocated either in the __init__ or __enter__ AND an allocation failure would raise an exception (which is what should happen for any not-totally-broken lib allocating resources of any kind) AND you don't do anything stupid (like silently swallowing this exception) in your Resource class, then you will NEVER enter the with block if the resource is not "correctly allocated". IOW, you have nothing special to do.
– bruno desthuilliers
Dec 27 at 13:53




assuming the resource is allocated either in the __init__ or __enter__ AND an allocation failure would raise an exception (which is what should happen for any not-totally-broken lib allocating resources of any kind) AND you don't do anything stupid (like silently swallowing this exception) in your Resource class, then you will NEVER enter the with block if the resource is not "correctly allocated". IOW, you have nothing special to do.
– bruno desthuilliers
Dec 27 at 13:53












1 Answer
1






active

oldest

votes


















1














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()





share|improve this answer





















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    1 Answer
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    active

    oldest

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    1














    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()





    share|improve this answer


























      1














      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()





      share|improve this answer
























        1












        1








        1






        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()





        share|improve this answer












        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()






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Dec 27 at 13:41









        Peter Wood

        16.2k33268




        16.2k33268






























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