Why does popping from the original list make reversed(original_list) empty?

Taking a look at the cpython code on GitHub, we can get some intuition as to why it no longer works.

The iterator that is returned essentially requires knowing the position of the last index and the length of the array. If the size of the array is changed, the iterator will no longer work.

Test 1: Increasing the array length

This will not produce the correct results either, but the iterator does run:

s = [1,2,3]
t = reversed(s)
s.append(4)

for i in t:
    print(i)
# output: [3, 2, 1]

Test 2: Decreasing, then increasing the length

s = [1,2,3]
t = reversed(s)
s.pop()
s.append(4)

for i in t:
    print(i)
# output: [4, 2, 1]

It still works!

So there's an internal check to see whether or not the last index is still valid, and if it is, it's a simple for loop down to index 0.

If it doesn't work, the iterator returns empty.

calling reversed return a iterator over that list, which a special object that allow you iterate in reverse order over the original list, is not a new list and is a one time use only

>>> s= [1,2,3]
>>> t = reversed(s)
>>> t
<list_reverseiterator object at 0x00000261BE8F0C40>
>>> list(t)
[3, 2, 1]
>>> list(t)
[]
>>> 

and because this iterator reference the original list, any change on it is reflected when you iterate over the iterator later.

Update

In particular and as MZ explain, if that change is such that the state of the list is different from when the iterator was created you get nothing if the size decreases or an incomplete version of the list if increased

>>> s= [1,2,3]
>>> t = reversed(s)
>>> s.insert(0,23)
>>> s
[23, 1, 2, 3]
>>> list(t)
[2, 1, 23]
>>> t = reversed(s)
>>> s.append(32)
>>> list(t)
[3, 2, 1, 23]
>>> s
[23, 1, 2, 3, 32]
>>> t = reversed(s)
>>> s.pop()
32
>>> list(t)
[]
>>>