Global dictionaries don't need keyword global to modify them?

Question

I wonder why I can change global dictionary without global keyword? Why it’s mandatory for other types? Is there any logic behind this?

E.g. code:

#!/usr/bin/env python3

stringvar = "mod"
dictvar = {'key1': 1,
           'key2': 2}

def foo():
    dictvar['key1'] += 1

def bar():
    stringvar = "bar"
    print(stringvar)

print(dictvar)
foo()
print(dictvar)

print(stringvar)
bar()
print(stringvar)

Gives following results:

me@pc:~/$ ./globalDict.py 
{'key2': 2, 'key1': 1}
{'key2': 2, 'key1': 2}  # Dictionary value has been changed
mod
bar
mod

where I would expect:

me@pc:~/$ ./globalDict.py 
{'key2': 2, 'key1': 1}
{'key2': 2, 'key1': 1}  # I didn't use global, so dictionary remains the same
mod
bar
mod

Asked By: Jovik

||

Source

Answer 1

You can modify any mutable object without using global keyword.

This is possible in Python because global is used when you want to reassign new objects to variable names already used in global scope or to define new global variables.

But in case of mutable objects you’re not re-assigning anything, you’re just modifying them in-place, therefore Python simply loads them from global scope and modifies them.

As docs say:

It would be impossible to assign to a global variable without global.

In [101]: dic = {}

In [102]: lis = []

In [103]: def func():
    dic['a'] = 'foo'
    lis.append('foo') # but  fails for lis += ['something']
   .....:     

In [104]: func()

In [105]: dic, lis
Out[105]: ({'a': 'foo'}, ['foo'])

dis.dis:

In [121]: dis.dis(func)
  2           0 LOAD_CONST               1 ('foo')
              3 LOAD_GLOBAL              0 (dic)     # the global object dic is loaded
              6 LOAD_CONST               2 ('a')
              9 STORE_SUBSCR                         # modify the same object

  3          10 LOAD_GLOBAL              1 (lis)    # the global object lis is loaded
             13 LOAD_ATTR                2 (append)
             16 LOAD_CONST               1 ('foo')
             19 CALL_FUNCTION            1
             22 POP_TOP             
             23 LOAD_CONST               0 (None)
             26 RETURN_VALUE

Answered By: Ashwini Chaudhary

Answer 2

The reason is that the line

stringvar = "bar"

is ambiguous, it could be referring to a global variable, or it could be creating a new local variable called stringvar. In this case, Python defaults to assuming it is a local variable unless the global keyword has already been used.

However, the line

dictvar['key1'] += 1

Is entirely unambiguous. It can be referring only to the global variable dictvar, since dictvar must already exist for the statement not to throw an error.

This is not specific to dictionaries- the same is true for lists:

listvar = ["hello", "world"]

def listfoo():
    listvar[0] = "goodbye"

or other kinds of objects:

class MyClass:
    foo = 1
myclassvar = MyClass()

def myclassfoo():
    myclassvar.foo = 2

It’s true whenever a mutating operation is used rather than a rebinding one.

Answered By: David Robinson

Global dictionaries don't need keyword global to modify them?

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