[PYTHON] a () and a.__ call__ () are not equivalent

As the title says, I didn't know it, so I made a note for myself.

When you override ʻa .__ call __ () and ʻa (), you can see that they actually have different __ behavior . It seems that ʻa ()callstype (a) . call __ (a)` to be exact.

python

>>> class A(object):
...     def __init__(self):
...         print 'init of A'
...         
...     def __call__(self):
...         print 'call of A'
...         
...     def method(self):
...         print 'method of A'
...         

Define __init__, __call__, method in the class A. Since method is an ordinary member function, you can override the method by setting ʻa.method = hoge`.

python

#Check the processing when calling normally
>>> a = A()
init of A
>>> a.method()
method of A
>>> a()
call of A

# a.Try overwriting method
>>> a.method = lambda: 'overriding method'
>>> a.method()
'overriding method'

# a.__call__Overwrite in the same way
>>> a.__call__ = lambda: 'overriding call'

# a.__call__()Is certainly overridden
>>> a.__call__()
'overriding call'

# a()It doesn't work
>>> a()
call of A
>>> type(a).__call__(a)
call of A

# type(a).__call__Forcibly overwrite
>>> type(a).__call__ = lambda a: 'new __call__'
>>> type(a).__call__(a)
'new __call__'

#Worked well
>>> a()
'new __call__'

reference

http://www.rakunet.org/tsnet/TSpython/43/1292.html