Why assigning list to another list produce 'None'?

Question

I am trying to understand why printing a list that was assigned using another list will give None. The code is as follows.

A=[]
for i in range(12,4,-2):
    A.append(i)
A_A=A.append(4)
print(A)
print(A_A)

In above code, why print(A_A) gives None?.

Thank you.

It's specifically with the `list.append` method. It returns None instead of another list. — TrebledJ, May 18 '19 at 05:32
because `A.append(4)` returns None. You want to do `A_A = A + [4]` perhaps — Devesh Kumar Singh, May 18 '19 at 05:32
It is a function that only has a `side-effect` (appending), nothing is returned. — Nishant, May 18 '19 at 05:41

score 1 · Answer 1 · answered May 18 '19 at 05:34

1

Append does not return anything, it just modifies the original list.

answered May 18 '19 at 05:34

Sergiu

130
6

score 1 · Accepted Answer · answered May 18 '19 at 05:35

1

The function

list.append()

modifies the list that it's called on. It doesn't return anything - thus, when you try to assign its return value to A_A, you get None.

You'll notice that, in your code, 4 was successfully appended to A, as you'd expect. The rationale here is, if you already know what you're trying to add, why bother returning it? Or, put another way - what could list.append() possibly return that would be useful? The conclusion was that it couldn't; thus, list.append() doesn't return anything.

answered May 18 '19 at 05:35

Green Cloak Guy

23,793
4
33
53

Although I marked the answer correct, the question is why it printed `A' then? – laplaceable May 18 '19 at 06:48
Well, `print(None)` prints `None`, it's as simple as that. Since `A_A` is `None`, then `print(None)` prints `None`. – Green Cloak Guy May 18 '19 at 17:01

Why assigning list to another list produce 'None'?

2 Answers2