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My python utility functions

Project description

noglobal

This python package provides the @noglobal decorator which prohibits the use of global variables from the function scope.

Install

pip install noglobal

Usage

import os.path as osp
from noglobal import NoGlobal
noglobal = NoGlobal(globals()).noglobal

foo = 3

def bar():
    return "bar"

@noglobal
def f():
    print(osp.join("foo", "bar"))
    print(bar())
    print(foo)

f()
foo/bar
bar
---------------------------------------------------------------------------
NameError                                 Traceback (most recent call last)
Input In [1], in <cell line: 16>()
     13     print(bar())
     14     print(foo)
---> 16 f()

Input In [1], in f()
     12 print(osp.join("foo", "bar"))
     13 print(bar())
---> 14 print(foo)

NameError: name 'foo' is not defined

FAQ

How all this works?

The function with @noglobal decorator are re-created as a type.FunctionType object, with no global variables.

All globals are affected?

No. Althogh this package name is "noglobal", the globals below are not affected:

  1. modules
  2. callable objects (which means the objects which have __call__ method)

Thus you can not bother importing already imported modules everytime, and so on for the funcsions or classes.

Can we use explicit global variables?

No. Functions with the @noglobal decorator also disallow the explicit use of global variables. That is, if you update a variable bound by a global declaration within a function, the original global variable will not be affected. The following code prints 10 on the standard output.

bar = 10

@noglobal
def foo():
    global bar
    bar = 11
foo()
print(bar)

What happens for the nested functions?

Nested functions in the function decorated with @noglobal are also affected. For example, the code below returns error:

buzz = 11

@noglobal
def foo():
    def bar():
        print(buzz)
    bar()

foo()
---------------------------------------------------------------------------
NameError                                 Traceback (most recent call last)
Cell In [8], line 1
----> 1 foo()

Cell In [7], line 5, in foo()
      3 def bar():
      4     print(buzz)
----> 5 bar()

Cell In [7], line 4, in foo.<locals>.bar()
      3 def bar():
----> 4     print(buzz)

NameError: name 'buzz' is not defined

Attribution

Based on the source codes:

Project details


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