How python's standard library functions are called without creating an object?

Question

In below example, the object of random is not created

from random import randint

print(randint(0, 10))

the definition of the randint function is as follow

import _random

class Random(_random.Random):
     def randint(self, a, b):
        return self.randrange(a, b+1)

Now in my case:

class XYZ:
     def func(self):
         do something

to call func, I have to create object of XYZ and then call function XYZ().func()else it throws error...

But I want to do it like

XYZ.func()

simply like how standard library function works. is there any way to achieve this?

chepner · Accepted Answer · 2020-05-16 23:22:29Z

2

The random module defines an instance of Random at the module level:

# Line 786
_inst = Random()

and random.randint is just a reference to a bound method for that instance:

# Line 791
randint = _inst.randint

answered May 16, 2020 at 23:22

chepner

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Comments

J_H · Accepted Answer · 2020-05-16 23:25:09Z

0

You wrote this:

class XYZ:
     def func(self):
         do something

You want this:

class XYZ:
     @classmethod
     def func(cls):
         do something

(or use @staticmethod if you prefer).

answered May 16, 2020 at 23:25

J_H

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