“<method> takes no arguments (1 given)” but I gave none

Question

I am new to Python and I have written this simple script:

#!/usr/bin/python3
import sys

class Hello:
    def printHello():
        print('Hello!')

def main():
    helloObject = Hello()
    helloObject.printHello()   # Here is the error

if __name__ == '__main__':
    main()

When I run it (./hello.py) I get the following error message:

Traceback (most recent call last):
  File "./hello.py", line 13, in <module>
    main()
  File "./hello.py", line 10, in main
    helloObject.printHello()
TypeError: printHello() takes no arguments (1 given)

Why does Python think I gave printHello() an argument while I clearly did not? What have I done wrong?

Answer 1

The error is referring to the implicit self argument that is passed implicitly when calling a method like helloObject.printHello(). This parameter needs to be included explicitly in the definition of an instance method. It should look like this:

class Hello:
  def printHello(self):
      print('Hello!')

Answer 2

If you want printHello as instance method, it should receive self as argument always(ant python will pass implicitly) Unless you want printHello as a static method, then you'll have to use @staticmethod

#!/usr/bin/python3
import sys

class Hello:
    def printHello(self):
        print('Hello!')

def main():
    helloObject = Hello()
    helloObject.printHello()   # Here is the error

if __name__ == '__main__':
    main()

As '@staticmethod'

#!/usr/bin/python3
import sys

class Hello:
    @staticmethod
    def printHello():
        print('Hello!')

def main():
    Hello.printHello()   # Here is the error

if __name__ == '__main__':
    main()

Answer 3

Calling a method on the instance of an object returns the object itself (usually self) to the object. For example, calling Hello().printHello() is the same as calling Hello.printHello(Hello()), which uses an instance of a Hello object as the first argument.

Instead, define your printHello statement as def printHello(self):