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