Basic List operation
Question:
I tried a very basic code in the python interactive shell
>>> a=[1,2,3]
>>> id(a)
36194248L
>>> a.append(4)
>>> a
[1, 2, 3, 4]
>>> id(a)
36194248L
>>>
>>> id([1,2,3])
36193288L
>>> id([1,2,3].append(4))
506033800L
>>> id([1,2,3].append(5))
506033800L
>>> id([1,2,3].append(6))
506033800L
Q: When i assign a list to a variable named ‘a’, and try to append more value, the id() does not change but if i try the same thing without assigning to a variable,the id() changes.Since lists are mutable(i.e. allow change at same memory address),why this behaviour is seen?
Answers:
list.append() returns None, not the list.
You are getting the id() result for that object, and None is a singleton. It’ll always be the same for a given Python interpreter run:
>>> id(None)
4514447768
>>> [].append(1) is None
True
>>> id([].append(1))
4514447768
Because list.append returns None, you are calling id(None), which there will only be one instance of.
I tried a very basic code in the python interactive shell
>>> a=[1,2,3]
>>> id(a)
36194248L
>>> a.append(4)
>>> a
[1, 2, 3, 4]
>>> id(a)
36194248L
>>>
>>> id([1,2,3])
36193288L
>>> id([1,2,3].append(4))
506033800L
>>> id([1,2,3].append(5))
506033800L
>>> id([1,2,3].append(6))
506033800L
Q: When i assign a list to a variable named ‘a’, and try to append more value, the id() does not change but if i try the same thing without assigning to a variable,the id() changes.Since lists are mutable(i.e. allow change at same memory address),why this behaviour is seen?
list.append() returns None, not the list.
You are getting the id() result for that object, and None is a singleton. It’ll always be the same for a given Python interpreter run:
>>> id(None)
4514447768
>>> [].append(1) is None
True
>>> id([].append(1))
4514447768
Because list.append returns None, you are calling id(None), which there will only be one instance of.