If Python slice copy the reference, why can't I use it to modify the original list?

Question

I know that Slicing lists does not generate copies of the objects in the list; it just copies the references to them.

But if that\'s the case, then why doesn\'t this wor

Answer 1

You are right that slicing doesn't copy the items in the list. However, it does create a new list object.

Your comment suggests a misunderstanding:

# Attempting to modify the element at index 1
l[0:2][-1] = 10

This is not a modification of the element, it's a modification of the list. In other words it is really "change the list so that index 1 now points to the number 10". Since your slice created a new list, you are just changing that new list to point at some other object.

In your comment to oldrinb's answer, you said:

Why are l[0:1] and l[0:1][0] different? Shouldn't they both refer to the same object, i.e. the first item of l?

Aside from the fact that l[0:1] is a list while l[0:1][0] is a single element, there is again the same misunderstanding here. Suppose that some_list is a list and the object at index ix is obj. This:

some_list[ix] = blah

. . . is an operation on some_list. The object obj is not involved. This can be confusing because it means some_list[ix] has slightly different semantics depending on which side of the assignment it is on. If you do

blah = some_list[ix] + 2

. . .then you are indeed operating on the object inside the list (i.e., it is the same as obj + 2). But when the indexing operation is on the left of the assignment, it no longer involves the contained object at all, only the list itself.

When you assign to a list index you are modifying the list, not the object inside it. So in your example l[0] is the same as l[0:2][0], but that doesn't matter; because your indexing is an assignment target, it's modifying the list and doesn't care what object was in there already.