When we instantiate a class, the methods get 'bound' to that particular instance for every instance created. What does the word 'bound' means here. I don't think a duplicate copy of the method object is created for every instance
I have read there is overhead associated with 'binding' the methods for every instance of the class. What kind of overhead is this
It means it is bound to an instance. ie. That instance gets passed as the self parameter
class Foo(object):
def bar(self):
print "The instance is:", self
foo = Foo()
The following have the same effect
foo.bar() # call bound method. instance gets passed in automatically
Foo.bar(foo) # unbound method. first parameter should be an instance
Simply, an unbound method requires you to pass an object in for self.
A bound method automatically passes the class instance as self.
Also see this answer and take a look at descriptors for a better understanding of methods.
In code:
# class definition
>>> class Foo:
>>> def bar(self):
>>> print 'self:', self
# class instance -> bound method
>>> a = Foo()
>>> print a, a.bar
<__main__.Foo instance at 0x107bcda70> <bound method Foo.bar of <__main__.Foo instance at 0x107bcda70>>
>>> a.bar()
self: <__main__.Foo instance at 0x107bcda70>
# class -> unbound method
>>> print Foo, Foo.bar
__main__.Foo <unbound method Foo.bar>
>>> Foo.bar()
TypeError: unbound method bar() must be called with Foo instance as first argument (got nothing instead)
# manually binding a method
>>> b = Foo.bar.__get__(a, Foo)
>>> print b
<bound method Foo.bar of <__main__.Foo instance at 0x107bcda70>>
>>> b()
self: <__main__.Foo instance at 0x107bcda70>
