I'm new at python I have some Question about Variable thing.
I've heard that s refers to a variable, and s[:] refers to copying a value, not a reference.
Why do both results in the same way?
I thought if [:] did this, b would be hello and a would still be hellQ, but the result is not
When you assigned a to b first, you only have one actual list data on the heap, you should try to assign another list data on heap to b first.
In your example, when you first assign ['h', 'e', 'l', 'l', 'o'] to a, you are creating a variable a pointing to the actual list data on heap. After that you assign a to b, which means creating a variable b pointing to the data that the a is pointing to, now the structure looks like this:
a ---> ['h', 'e', 'l', 'l', 'o']
^
|
b ------
Then, b[-1] = 'q' means change the last element of the list data which b is pointing to, which results in:
a ---> ['h', 'e', 'l', 'l', 'q']
^
|
b ------
After that, b[:] = a means change the data of the list to which b is pointing to the data of the list to which a is pointing, which in this situation does nothing (a and b are still pointing to the same list data). Therefore, the following a[-1] = 'Q' and b[-1] = 'o' are just changing the (only) list data, which looks like:
a ---> ['h', 'e', 'l', 'l', 'Q']
^
|
b ------
a ---> ['h', 'e', 'l', 'l', 'o']
^
|
b ------
Back to the difference of b = a and b[:] = a, if you want to see how b[:] = a is different from b = a, you could first assign ['h', 'e', 'l', 'l', 'o'] to a and assign another list [] to b, now the structure looks like:
a ---> ['h', 'e', 'l', 'l', 'o']
b ---> []
Then if you use b[:] = a, which change the data of the list to which b is pointing to the data of the list to which a is pointing, you will get a structure like:
a ---> ['h', 'e', 'l', 'l', 'o']
b ---> ['h', 'e', 'l', 'l', 'o']
In this state, any change you made to the list to which a is pointing won't affect the list to which b is pointing. For instance, use a[-1] = 'Q' now will only change the list to which a is pointing, which results in a structure like:
a ---> ['h', 'e', 'l', 'l', 'Q']
b ---> ['h', 'e', 'l', 'l', 'o']
@SliAce is right.
b[:] overwrites the contents of the list without creating a new reference.
a and b are refering to same list, So b[:] = a does noting. It copy values from it own list(using reference a) to it own list.
And it's no need to use copy.deepcopy, it's used to copy the element in the list which is mutable, like list or dict, in your situation, str is immutable, so it has created a new object.