Comparing two class types in python
Explanation
This is why your comparison doesn't work as expected
>>> class ClassA(object):
... pass
...
>>> class ClassB(object):
... pass
...
>>> type(ClassB)
<class 'type'>
>>> type(ClassA)
<class 'type'>
>>> type(ClassA) == type(ClassB)
True
But why do ClassA
and ClassB
have the same type type
? Quoting the docs:
By default, classes are constructed using type(). The class body is executed in a new namespace and the class name is bound locally to the result of type(name, bases, namespace).
Example:
>>> ClassB
<class '__main__.ClassB'>
>>> type('ClassB', (), {})
<class '__main__.ClassB'>
>>> type(ClassB)
<class 'type'>
>>> type(type('ClassB', (), {}))
<class 'type'>
Getting the type of ClassB
is exactly the same as getting the type of type('ClassB', (), {})
, which is type
.
Solutions
Compare them directly (w/out using the type()
function):
>>> ClassA
<class '__main__.ClassA'>
>>> ClassB
<class '__main__.ClassB'>
>>> ClassA == ClassB
False
or initialize them and compare the types of their objects:
>>> a = ClassA()
>>> b = ClassB()
>>> type(a)
<class '__main__.ClassA'>
>>> type(b)
<class '__main__.ClassB'>
>>> type(a) == type(b)
False
FWIW you can also use is
in place of ==
(for classes).
If you want to check if types are equal then you should use is
operator.
Example: we can create next stupid metaclass
class StupidMetaClass(type):
def __eq__(self, other):
return False
and then class based on it:
in Python 2
class StupidClass(object): __metaclass__ = StupidMetaClass
in Python 3
class StupidClass(metaclass=StupidMetaClass): pass
then a simple check
StupidClass == StupidClass
returns False
, while the next check returns an expected True
value
StupidClass is StupidClass
So as we can see ==
operator can be overridden while there is no simple way to change is
operator's behavior.
You're comparing the type of the class object, which are all of type 'type'
.
If you just want to compare the classes, compare them directly:
print Class3 == Class4