Python required positional argument when use "self"

Question

I read that the first parameter of methods is the instance the method is called on. I dont understand why when i write this

class A:
    def printName(self, name):
        print(name)

A.printName("asd")

I get that error TypeError: printName() missing 1 required positional argument: 'name'. What i'm missing?

score 3 · Accepted Answer · answered Jul 11 '17 at 01:09

3

You need to create a class instance and then call method on that instance:

class A:
    def printName(self, name):
        print(name)

class_instance = A()
class_instance.printName("asd")

answered Jul 11 '17 at 01:09

Andrea

thanks for the answer. I dont really get that A.printName("asd") is not an instance – akerbeltz Jul 11 '17 at 01:25
A is a class object while A() is a new class instance: – Andrea Jul 11 '17 at 09:46
Take a look to documentation: https://docs.python.org/2/tutorial/classes.html It's pretty important to know basis about classes. Have fun! – Andrea Jul 11 '17 at 09:49

score 0 · Answer 2 · answered Jul 11 '17 at 01:12

0

This would work, using staticmethod if you want to call printName() without creating an instance of that class

class A:
    @staticmethod
    def printName(name):
        print(name)

A.printName('hello')

answered Jul 11 '17 at 01:12

ksai

Thanks for the answer. Can you explain what is a static method in python and why it dont need the 'self'?, I tried your code and it works even without the @staticmethod, why is that? Sorry for the amount of questions but i'm having a lot of trouble to understant how works the 'self' argument – akerbeltz Jul 11 '17 at 01:22
1

@akerbeltz:) **staticmethods** are methods that are bound to a class rather than its object. They don't need class instance`@staticmethod` tells we are using `staticmethod`. In earlier versions, the syntax to write **staticmethod** was different. You can just google for more information – ksai Jul 11 '17 at 01:27
self is an equivalent of, for 'this' object. staticmethod is not for different instances it is bound to a class. – gout Jul 11 '17 at 01:29
Thanks for the answers! you help me a lot. There's something i don't understand at all. When i write `A.printName('hello')` or `a=A.printName('hello')` i'm not creating an instance of the class A? – akerbeltz Jul 11 '17 at 01:42
@akerbeltz :) No, you are not creating instance in either way. You can create an instance like `instance_A = A()`. if the answer was helpful, please mark the answer as _accept_ answer. – ksai Jul 11 '17 at 02:43

2 Answers2