python - TypeError: method() takes 1 positional argument but 2 were given -

if have class ...

class myclass:      def method(arg):         print(arg) 

... use create object ...

my_object = myclass() 

... on call method("foo") ...

>>> my_object.method("foo") traceback (most recent call last): file "<stdin>", line 1, in <module> typeerror: method() takes 1 positional argument (2 given) 

... why python tell me gave 2 arguments, when gave one?

in python, this:

my_object.method("foo") 

... syntactic sugar, interpreter translates behind scenes into:

myclass.method(my_object, "foo") 

... which, can see, indeed have 2 arguments - it's first 1 implicit, point of view of caller.

this because methods work object they're called on, there needs way object referred inside method. convention, first argument called self inside method definition:

class mynewclass:      def method(self, arg):         print(self)         print(arg) 

if call method("foo") on instance of mynewclass, works expected:

>>> my_new_object = mynewclass() >>> my_new_object.method("foo") <__main__.mynewclass object @ 0x29045d0> foo 

occasionally (but not often), don't care object method bound to, , in circumstance, can decorate method builtin staticmethod() function so:

class myotherclass:      @staticmethod     def method(arg):         print(arg) 

... in case don't need add self argument method definition, , still works:

>>> my_other_object = myotherclass() >>> my_other_object.method("foo") foo