python 2.7 - how to call parent class constructor

Question

I have base class like below

class FileUtil:
    def __init__(self):
        self.outFileDir = os.path.join(settings.MEDIA_ROOT,'processed')
        if not os.path.exists(outFileDir):
            os.makedirs(outFileDir)
    ## other methods of the class

and I am extending this class as below:

class Myfile(FileUtil):
    def __init__(self, extension):
        super(Myfile, self).__init__()
        self.extension = 'text'
    ## other methods of class

But i am getting below error?

super(Myfile, self).__init__()
TypeError: super() takes at least 1 argument (0 given)

I gone through many documents and found that there are different sytex of calling super() in 2.x and 3.x. I tried both ways but getting error.

Possible duplicate of [Python, invoke super constructor](http://stackoverflow.com/questions/2399307/python-invoke-super-constructor) — Paul Rooney, Feb 05 '16 at 03:24
I am using this code in Django..Although I can understand there is nothing Django specific but still asking if there is something i am missing. — Gaurav Pant, Feb 05 '16 at 03:29

score 8 · Accepted Answer · answered Feb 05 '16 at 03:31

8

You have 2 options

old style class, you should call the super constructor directly.

class FileUtil():
    def __init__(self):
        pass

class Myfile(FileUtil):
    def __init__(self, extension):
        FileUtil.__init__(self)

new style class, inherit from object in your base class and your current call to super will be processed correctly.

class FileUtil(object):
    def __init__(self):
        pass

class Myfile(FileUtil):
    def __init__(self, extension):
        super(Myfile, self).__init__()

answered Feb 05 '16 at 03:31

Paul Rooney

20,879
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They differ as described. The first one calls the super constructor differently and the second one inherits from `object`. – Paul Rooney Feb 05 '16 at 03:40
Second example is what OP's using and having issues. So it's not an answer to OP's question – midori Feb 05 '16 at 03:42
2

@minitoto in his code `class FileUtil:` in my code `class FileUtil(object):` – Paul Rooney Feb 05 '16 at 03:43

score 0 · Answer 2 · answered Feb 05 '16 at 03:25

0

You may need to create the FileUtil class using the super() function as well:

class FileUtil(object):
    def __init__(self):
        super(FileUtil, self).__init__()
        ...

answered Feb 05 '16 at 03:25

tjohnson

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1
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python 2.7 - how to call parent class constructor

2 Answers2