I know that "variable assignment" in python is, in fact, a binding / re-binding of a name (the variable) to an object.
b = [1,2,3]
a = b[2] # binding a to b[2] ?
a = 1000
b is [1, 2, 3]
after this change, why b is not changed?
here is another example:
b = [1,2,3]
a = b
a[0] = 1000
this case b is [1000, 2, 3]
isn't assignment in Python reference binding?
Thank you
a = b[2] # binding a to b[2] ?
Specifically, this binds the name a to the same value referenced by b[2]. It does not bind a to the element in the list b at index 2. The name a is entirely independent of the list where it got its value from.
a = 1000
Now you bind the name a to a new value, the integer 1000. Since a has no association with b, b does not change.
In your second example:
a = b
Now a is bound to the same list value that b is bound to. So when you do
a[0] = 1000
You modify an element in the underlying list that has two different names. When you access the list by either name, you will see the same value.
a[0] = ... is a special form of assignment that desugars to a method call,
a.__setattr__(0, ...)
You aren't reassigning to the name a; you are assigning to a "slot" in the object referenced by a (and b).
Lists are mutable objects, ints aren't. 4 can't become 5 in python, but a list can change its contents.
1,2,3are constance not can change the value, but list its an object then when you assignbtoab you are creating a reference to b calledaanother words it same that a pointer to, then a is same b if you mutate one mute the reference "the memory"