Python [0]*n syntax works only on immutable object?

Question

a = [[0]*3] * 4
print(a[0] is a[1]) # True

when I initialize a two dimensional array that way, things went wrong. Took me a bit time to find this unexpected behavior. So is this syntax only work on immutable object?

Answer 1

It "worked" in your example too, in its way. This is just how the implementation of list.__mul__ interprets what you want. The list can't construct new objects, it doesn't know how to create new objects from whatever objects it happens to contain. It expands itself with new references to those objects.

You get the same behavior with immutable integers

>>> x = [0] * 3
>>> x[0] is x[1]
True

You can get the two dimensional array with

>>> a = [[0]*3 for _ in range(4)]
>>> a[0] is a[1]
False

The reason why [0] * 3 does what you want is that it is creating list that contains 3 references to the same immutable 0.