Why does a Python dataclass "persist" outside the scope of its defining function?

Question

Consider the following minimal example:

>>> from dataclasses import dataclass
>>> @dataclass
... class my_stuff:
...     my_variable: int = 0
...
>>> def my_func():
...     stuff = my_stuff
...     stuff.my_variable += 1
...     print(stuff.my_variable)
...
>>> my_func()
1
>>> my_func()
2
>>> my_func()
3

Why does the printed value increase with each call to my_func()? Should stuff not go out of scope once the execution of my_func() is complete? And should each call to my_func() not create a new instance with my_variable initialized to 0 and incremented to 1 each time?

How would I change this code to meet my (apparently irrational) expectation that 1 would be output each time my_func() is called?

Try replacing stuff = my_stuff by stuff = my_stuff() to instantiate a new object every time instead of modifying the class itself. — Seon, Jun 06 '21 at 19:24
`stuff.my_variable` where `stuff=my_stuff` modifies the class attribute value. I.e. your code is working as expected, and your expectation/assumption of something different is irrational. — DisappointedByUnaccountableMod, Jun 06 '21 at 19:36
To debug this sort of problem, in this case print `stuff` In the function; you’ll see it is the class defintion. — DisappointedByUnaccountableMod, Jun 06 '21 at 19:48
`stuff` is just another name for `my_stuff`. One name going out of scope does not affect the object itself. — jasonharper, Jun 06 '21 at 21:07

Hamed Rostami · Accepted Answer · 2021-06-07T11:06:19.910

You are using Class itself Rather than instance!

class human:
    pass


human_kind = human #Pointer to class

print(human_kind)

alex = human() #Instance initializing

print(alex)

Results:

<class 'main.human'>

<main.human object at 0x7fd11ce47100>

true code:

from dataclasses import dataclass


@dataclass
class my_stuff:
    my_variable: int = 0


def my_func():
    stuff = my_stuff()
    stuff.my_variable += 1
    print(stuff.my_variable)

Why does a Python dataclass "persist" outside the scope of its defining function?

1 Answers1