Difference in type() and hasattr() between Python 3 and Python 2

Question

When I am trying to check type of class declaration in Python 3 and Python 2, I get strange results, as followings show,

>>> #python 2.7.6
>>> class MyClass:
...     pass
... 
>>> type(MyClass)
<type 'classobj'>
>>> hasattr(MyClass, '__call__')
False

The type of MyClass in Python 2 is classobj and MyClass has no __call__ attribute. This is what I expect.

>>> #python 3.3.3
>>> class MyClass:
...     pass
... 
>>> type(MyClass)
<class 'type'>
>>> hasattr(MyClass, '__call__')
True

However, in Python 3, the type of MyClass is class 'type', and MyClass has __call__ attribute although I don't declare it within MyClass.

I guess the reason is that in Python 3, MyClass is a type, and type has __call__ attribute, am I right? and why Python 3 change the behavior of type function like this?

tyteen4a03 · Accepted Answer · 2014-01-16 23:01:57Z

11

What changed is that all classes in 3.x are new-style classes by default. Your Python 2 classes did not inherit from object, so Python thought it was an old-style class.

In 2.7.5:

>>> class MyClass(object):
...   pass
... 
>>> type(MyClass)
<type 'type'>
>>> hasattr(MyClass, '__call__')
True

edited Jan 16, 2014 at 23:01

tyteen4a03

1,9432 gold badges27 silver badges51 bronze badges

answered Jan 16, 2014 at 22:55

Ignacio Vazquez-Abrams

804k160 gold badges1.4k silver badges1.4k bronze badges

Sign up to request clarification or add additional context in comments.

5 Comments

mitchelllc Over a year ago

Thanks for your prompt reply! I have another question. If I set c = MyClass(), and callable(c) will return false, but hasattr(MyClass, '__call__') returns true. So why callable(c) returns false?

tyteen4a03 Over a year ago

@mitchelllc You did not override the default __call__, so callable(c) shows as False. However, if you inherit from a function, it will be callable() by default.

mitchelllc Over a year ago

@tyteen4a03 well, I am a little confused. callable(c) returns false because I don't override the default __call__, but why hasattr(MyClass, '__call__') return true? Big thanks!

user395760 Over a year ago

@mitchelllc You're mixing up the class and its instances. callable(MyClass) is true. hasattr(c, '__call__') is false.

mitchelllc Over a year ago

Hi @delnan, thank you for the point. I find these words help me understand it.

classes are callable (calling a class returns a new instance); class instances are callable if they have a __call__() method.

from docs.python.org/2/library/functions.html#callable

Collectives™ on Stack Overflow

Difference in type() and hasattr() between Python 3 and Python 2

1 Answer 1

5 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

5 Comments

Your Answer

Sign up or log in

Post as a guest

Related