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Home/ Questions/Q 6848109
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Editorial Team
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Editorial Team
Asked: May 27, 20262026-05-27T00:49:39+00:00 2026-05-27T00:49:39+00:00

In Python, the object class serves as the root superclass for all the (new-style)

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In Python, the object class serves as the root superclass for all the (new-style) classes. By default at least, applying str and repr to the “class instance” of any subclass of object produces the same result:

>>> class spam(object): pass
... 
>>> str(spam)
"<class '__main__.spam'>"
>>> str(spam) == repr(spam)

I would like to define a subclass of object, say fancyobject, that is identical to object in every way, except that applying str and repr to fancyobject itself produces different outputs:

>>> class ham(fancyobject): pass
...
>>> str(ham)
'ham'
>>> repr(ham)
"<class '__main__.ham'>"

Is there a way to do this in Python?

PS: I’m aware of the __str__ special method, but it is my understanding that if class A overrides __str__, then the overriding method is called only when str is called on instances of A, not when it is called on A itself. I.e.:

>>> class A(object):
...     def __str__(self):
...         return 'from new __str__: ' + object.__str__(self)
... 
>>> str(A())
'from new __str__: <__main__.A object at 0x7f79c62a5310>'
>>> str(A)
"<class '__main__.A'>"
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  1. Editorial Team
    Editorial Team
    2026-05-27T00:49:40+00:00Added an answer on May 27, 2026 at 12:49 am

    Actually the same mechanism as for object instances applies for types. Types are just objects themselves, so they are converted to strings by calling the __str__() method on their type, which is called the “metaclass”. So you have to overwrite the __str__() method on the metaclass:

    class fancytype(type):
        def __str__(self):
            return self.__name__
    class ham(object):
        __metaclass__ = fancytype
    print ham
    

    prints

    ham
    
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