Setting a dynamic type's docstring in Python 3

Question

I am dynamically creating some classes and I want them to have different docstrings. I have:

def make_class(class_docstring):
    class X:
        pass
    X.__doc__ = class_docstring
    return X

That didn't work because docstrings are read-only. Then, I tried:

def make_class(class_name, class_docstring):
    class X:
        def __init__(self):
            super().__init__()

    d = {'__doc__': class_docstring}
    d.update(X.__dict__)
    return type(class_name, (), d)

ClassName = make_class(
    'ClassName',
    """
    Some docstring...
    """)

which worked until it had to call super.

What is the correct way to dynamically set the docstring attribute?

Your first approach works perfectly fine. Why do you think doc strings are read-only? They aren't. — Sven Marnach
– Sven Marnach, Commented Nov 16, 2011 at 1:20

hwiechers · Accepted Answer · 2011-11-16 01:43:40Z

5

You can set the docstring inside the class.

>>> def make_class(class_docstring):
...     class X:
...         __doc__ = class_docstring
...     return X
...
>>> x = make_class('test doc')
>>> x
<class '__main__.X'>
>>> xx = x()
>>> xx.__doc__
'test doc'

I'm not sure why your 2nd attempt is failing.

answered Nov 16, 2011 at 1:43

hwiechers

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3 Comments

Neil G Over a year ago

The second attempt fails because the call to super() is binding to the type X (it's being translated to super(X)) and then the method is being copied to the dynamically-created type, which doesn't have X in its __mro__.

Neil G Over a year ago

Would it be okay for me to use: globals()[class_name] = X instead of return X so that I can just write make_class('name', 'docstring') instead of name = make_class(…?

Thomas K Over a year ago

@NeilG: It might work, but injecting things into globals is almost always bad practice. It's much clearer to use name = ... .

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