如何为类定义 __str__ 方法？

Question

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. Ie:

>>> 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'>"

Answer 1

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

Answer 2

You can also set the default metaclass for a whole module like this

class fancytype(type):
    def __str__(self):
        return self.__name__

__metaclass__ = fancytype

class ham:
    pass
print ham

Answer 3

Here's the new answer for Python 3. Basically, you pass in a metaclass as a keyword parameter to the class definition.

class fancytype(type):
    def __str__(self):
        return self.__name__
class ham(metaclass=fancytype):
    pass
print(ham)

prints

ham