global / local namespace

[nzcan]

hello,
recently i have 'discovered' a little bit 'strange' block of code.

a = 5
print(id(a))

if a == 5:
	a = a + 2
	print(a)
	print(id(a))

print(id(a)) 

the output:

Output:
10924160
7
10924224
10924224 

it turns out that the interpreter see the ( already ... 'localized'!? ) object 'a' in the global name space also.
should the 'a' not be encapsulated in the local scope of the 'if-statement' and not accessible from the global name space?

thanks in advance!

[Larz60+]

throw in a twist:

>>> a = 5
>>> b = a
>>> print(id(a))
9342464
>>> print(id(b))
9342464
>>> if a == 5:
...     a = a + 2
...     print(a)
...     print(id(a))
... 
7
9342528
>>> print(id(a))
9342528
>>> print(id(b))
9342464
>>>

[Gribouillis]

should the 'a' not be encapsulated in the local scope of the 'if-statement' and not accessible from the global name space?

There is no local namespace for block statements such as if, while, for. They use the surrounding namespace. Function bodies have a local namespace that is different for each execution of the function. Class definitions bodies also have their own namespace

>>> x = object()
>>> class A:
...     x = object()
...     print(id(x))
... 
139880381153440
>>> print(id(x))
139880381153648

Note that for small integers, things are different because python (at least CPython) keeps a single instance
for each of them. On my version of python, it goes up to 256 included

>>> a = 256
>>> b = 256
>>> a is b # the names a and b share the same integer instance
True
>>> a = 257
>>> b = 257
>>> a is b # two integer instances were created
False

[woooee]

The code just shows that strings are not mutable, and has nothing to do with local vs global namespaces. This gives the same results

a = 5
print(id(a))
 
##if a == 5:
a = a + 2
print(a)
print(id(a))
 
print(id(a))