+ 1
Why the assignment of b value doesn't change when a change.
Can someone explain me that when I changed the a[0] value, b[0] also changed but when I changed the entire a array, b doesn't change, why ? https://code.sololearn.com/cQ4zn2wdG15z/?ref=app
7 Réponses
+ 2
a = [1,2,3]
b =a
print(id(b),id(a))
a[0] = 5
print(b[0], end = "")
a = [6,7,8]
print(id(a))
print (b[0], end = "")
Run this and see for yourself.
+ 3
at the beginning a and b refer to the same object,then
you changed a,but b is still [5, 2, 3]
+ 3
B still refers to the first object [5, 2, 3], but a refers to [6, 7, 8]
+ 2
Hima , now I understand that the object id remain same as long long we modified the content of it.
The moment I redefined object, the object id changed and it is no longer associated with the previous id, so it is now a new object. Thanks for help.
+ 1
But, what about in next step. When I completely changed a, that should also changed b as well, but that doesn't happen, why ? If they belong to same object, it should be consistent.
+ 1
I just noticed that even you redefined a value as same as earlier, this also changed the id of a.
This means, we can edit the object, that doesn't change it store locations but redefining will immediately tag it to new location even though the redefined values are exactly same as earlier.
https://code.sololearn.com/cQuiNk2Rh2Ln/?ref=app
0
When you write a = something , you name something as 'a' and you can access 'something' by calling a. When you write a[0] = other , you are calling an item stored at a[0] and replacing it with other.
Kinda confusing to word it I must say.